ch06

STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS

152

Chapter 6 6-1 Draw a free-body diagram of the angle bracket

shown in Fig. P6-1. SOLUTION

6-2 Draw a free-body diagram of the diving board shown in Fig. P6-2. The surface at B is smooth. Neglect the weight of the diving board.

SOLUTION

6-3 Draw a free-body diagram of the lawn mower shown in Fig. P6-3. The lawn mower has a weight W and is resting on a rough surface.

SOLUTION

6-4 Draw a free-body diagram of the sled shown in Fig. P6-4.

SOLUTION

6-5 Draw a free-body diagram of the bracket shown in Fig. P6-5. The contact surfaces between the cylinders and bracket are smooth.

SOLUTION

6-6 Forces P are applied to the handles of the pipe pliers shown in Fig. P6-6. Draw free-body diagrams of each handle.

SOLUTION


153

6-7 The man shown in Fig. P6-7 has a weight W1; the beam has a weight W2 and a center of gravity G. Draw a free-body diagram of the beam.

SOLUTION

6-8 Forces P are applied to the handles of the bolt cutter shown in Fig. P6-8. Draw a free-body diagram of

(a) The lower handle. (b) The lower cutter jaw. SOLUTION

6-9 Draw a free-body diagram of the door shown in Fig. P6-9. The homogeneous door has weight W.

SOLUTION

6-10 Draw a free-body diagram of the bent bar shown in Fig. P6-10. The support at A is fixed, and the bar has negligible mass.

SOLUTION

6-11 Draw a free-body diagram of the bar shown in Fig. P6-11. The support at A is a journal bearing and the supports at B and C are ball bearings.

SOLUTION

6-12 Draw a free-body diagram of the shaft shown in Fig. P6-12. The bearing at A is a thrust bearing, and the bearing at D is a ball bearing. Neglect the weights of the shaft and the levers.

SOLUTION


154

6-13 A beam is loaded and supported as shown in Fig. P6-13. Determine the reactions at supports A and B. SOLUTION From a free-body diagram of the beam, the equilibrium equations are solved to get the forces

0 :xF→ Σ = 0xA =

0 :AMΣ =�

( ) ( )( ) ( )

15 3 500 6 800

9 700 12 400 0

B − −

− − =

1160 lbB =

0 :yF↑ Σ = 500 800 700 400 0yA B+ − − − − =

1240 lbyA =

1240 lb = ↑A ............................................................................................................................ Ans.

1160 lb = ↑B ............................................................................................................................ Ans.

6-14 A beam is loaded and supported as shown in Fig. P6-14. Determine the reaction at support A. SOLUTION From a free-body diagram of the beam, the equilibrium equations are solved to get the forces and moment

0 :xF→ Σ = 0xA =

0 :yF↑ Σ = 2 0yA − =

2 kNyA =

0 :AMΣ =� ( )2 4 3 0AM − − =

11 kN mAM = ⋅

2 kN = ↑A ................................................................................................................................Ans.

11 kN m A = ⋅M � ..................................................................................................................... Ans.

6-15 A 30-lb force P is applied to the brake pedal of an automobile as shown in Fig. P6-15. Determine the force Q applied to the brake cylinder and the reaction at support A.

SOLUTION From a free-body diagram of the brake pedal, the equilibrium equations are solved to get the forces

0 :AMΣ =� ( )( ) ( ) ( )5.5 30cos30 11 30sin 30 4 0Q − ° − ° =

62.871 lbQ =

0 :xF→ Σ = 30cos30 0xA Q− + ° =

36.890 lbxA =

0 :yF↑ Σ = 30sin 30 0yA − ° =

15 lbyA =

39.8 lb=A 22.13° ............................................................Ans.

62.9 lb = ←Q ........................................................................Ans.


155

6-16 A beam is loaded and supported as shown in Fig. P6-16. Determine the reactions at supports A and B. SOLUTION From a free-body diagram of the beam, the equilibrium equations are solved to get the forces

0 :xF→ Σ = 0xA =

0 :AMΣ =�

( )( ) ( )

( )( ) ( )

4.5 300 1.5 0.75

1 400 1.5 3 0.5 02

B −

− + =

308.333 NB =

0 :yF↑ Σ = ( )( ) ( )( )1300 1.5 400 1.5 02yA B − − + =

441.667 NyA =

442 N = ↑A .............................................................................................................................. Ans.

308 N = ↑B .............................................................................................................................. Ans.

6-17 A rope and pulley system is used to support a body W as shown in Fig. P6-17. Each pulley is free to rotate and the rope is continuous over the pulleys. Determine the tension T in the rope required to hold body W in equilibrium if the weight of body W is 400 lb. Assume that all rope segments are vertical.

SOLUTION The pulleys are rigidly connected together and can be treated as a single body for the purpose of drawing a free-body diagram. From a free-body diagram of the pulleys, the equilibrium equations are solved to get the tension

0 :yF↑ Σ = 4 400 0T − =

100 lbT = ................................................................................................ Ans.

6-18 Pulleys 1 and 2 of the rope and pulley system shown in Fig. P6-18 are connected and rotate as a unit. The radii of pulleys 1 and 2 are 100 mm and 300 mm, respectively. Rope A is wrapped around pulley 1 and is fastened to pulley 1 at point A�. Rope B is wrapped around pulley 2 and is fastened to pulley 2 at point B�. Rope C is continuous over pulleys 3 and 4. Determine the tension T in rope C required to hold body W in equilibrium if the mass of body W is 225 kg.

SOLUTION

( )225 9.81 2207.25 NW = =

From a free-body diagram of the pulley 4, the equilibrium equations are solved to get

0 :yF↑ Σ = 2 0BT T W+ − =

2BT W T= −

Next, from a free-body diagram of the pulley 3, the equilibrium equations are solved to get

0 :yF↑ Σ = 2 0AT T− =

2AT T=


156

Finally, from a free-body diagram of the pulleys 1 and 2, the equilibrium equations are solved to get

0 :axleMΣ =� 100 300 0A BT T− =

100 300A BT T=

( ) ( )100 2 300 2T W T= −

8 3 6621.75 NT W= =

828 NT = ................................................................................................ Ans.

6-19 Three pipes are supported in a pipe rack as shown in Fig. P6-19. Each pipe weighs 100 lb. Determine the reactions at supports A and B.

SOLUTION From a free-body diagram of the pipes and the rack, the equilibrium equations are solved to get the forces

0 :AMΣ =�

( ) ( ) ( ) ( )( ) ( ) ( )( )

18 100cos30 9 100cos30 15

100cos30 21 3 100sin 30 4 0

B − ° − °

− ° − ° =

249.840 lbB =

0 :xF→ Σ = sin 30 0xA B+ ° =

124.920 lbxA = −

0 :yF↑ Σ = 300 cos30 0yA B− + ° =

83.632 lbyA =

150.3 lb=A 33.80° ............................................................................................................Ans.

250 lb=B 60° ..................................................................................................................... Ans.

6-20 The man shown in Fig. P6-20 has a mass of 75 kg; the beam has a mass of 40 kg. The beam is in equilibrium with the man standing at the end and pulling on the cable. Determine the force exerted on the cable by the man and the reaction at support C.

SOLUTION From a free-body diagram of the man and the beam, the equilibrium equations are solved to get the forces

0 :xF→ Σ = 0xC =

0 :CMΣ =�

( ) ( ) ( ) ( )75 9.81 3 40 9.81 1.5

3 1.5 0T T

+

− − =

621.300 NT =

0 :yF↑ Σ = ( ) ( )2 75 9.81 40 9.81 0yT C+ − − =

114.450 NyC = −

114.5 N = ↓C ........................................................................................................................... Ans.

621 NT = ....................................................................................................................................Ans.


157

6-21 A 75-lb load is supported by an angle bracket, pulley, and cable as shown in Fig. P6-21. Determine (a) The force exerted on the bracket by the pin at C. (b) The reactions at supports A and B of the bracket. (c) The shearing stress on a cross section of the ¼-in.-diameter pin at C, which is in double shear. SOLUTION (a) First, from a free-body diagram of the pulley, the equilibrium

equations are solved to get the axle forces

0 :xF→ Σ = ( )( )4 5 75 0xC− =

60 lbxC =

0 :yF↑ Σ = ( ) ( )75 3 5 75 0yC − − =

120 lbyC =

134.2 lb=C 63.43 on the pulley°

134.2 lb=C 63.43 on the bracket° ..........................Ans. (b) Next, from a free-body diagram of the bracket, the equilibrium

equations are solved to get the forces

0 :xF→ Σ = 60 0xA + =

60 lbxA = −

0 :AMΣ =� ( )10 18 60 0B − =

108 lbB =

0 :yF↑ Σ = 120 0yA B+ − =

12 lbyA =

61.2 lb=A 11.31° ...........................................................................................................Ans.

108.0 lb = ↑B ....................................................................................................................... Ans. (c) Finally, dividing the shear force on the pin by twice its cross-sectional area (since the pin is in double shear)

gives the shear stress

( )2

134.2 1367 psi2 2 0.25 4

Cs

CA

τπ

= = =

...........................................................................Ans.

6-22 A pipe strut BC is loaded and supported as shown in Fig. P6-22. Determine (a) The reactions at supports A and C. (b) The shearing stress on a cross section of the 10-mm-diameter pin at C, which is in double shear. (c) The elongation of cable AB if it is made of aluminum alloy (E = 73 GPa) and has a diameter of 6 mm. SOLUTION (a) From a free-body diagram of the pipe strut, the equilibrium equations are solved to get the forces

0 :CMΣ =� ( )1000 800 750 0ABT − =

600 NABT = ...................................................................Ans.

0 :xF→ Σ = 0x ABC T− =

600 NxC =


158

0 :yF↑ Σ = 750 0yC − =

750 NyC =

960.47 N 960 N= ≅C 51.34° .....................................................................................Ans. (b) Dividing the shear force on the pin by twice its cross-sectional area (since the pin is in double shear) gives the

shear stress

( )

6 22

960.47 6.11 10 N/m 6.11 MPa2 2 0.010 4

Cs

CA

τπ

= = = × =

...................................Ans.

(c) Finally, the stretch of the cable AB is given by

( )

( ) ( )29

600 16000.465 mm

73 10 0.006 4AB

PLEA

δπ

= = = ×

.....................................................Ans.

6-23 The lawn mower shown in Fig. P6-23 weighs 35 lb and has a center of gravity at G. Determine (a) The magnitude of the force P required to push the mower at a constant velocity. (b) The forces exerted on the front and rear wheels by the inclined surface. (c) The shearing stresses in the ½-in.-diameter shoulder bolts of the front and rear wheels. SOLUTION (a) From a free-body diagram of the lawn mower, the equilibrium equations are solved to get the forces

0 :OMΣ =� ( ) ( ) ( ) ( )227 35cos15 13 35sin15 4 0N − ° + ° =

2 14.936 lbN = .................................................Ans.

0 :xFΣ = cos30 35sin15 0P ° − ° =

10.460 lbP = ...................................................Ans.

0 :yFΣ = 1 2 35cos15 sin 30 0N N P+ − ° − ° =

1 24.101 lbN = .................................................Ans.

(c) Finally, dividing the shear forces on the pins by their cross-sectional areas gives the shear stresses

( )

11 2

1

24.101 122.7 psi0.5 4s

NA

τπ

= = = ......................................................................................Ans.

( )

22 2

2

14.936 76.1 psi0.5 4s

NA

τπ

= = = .......................................................................................Ans.

6-24 The coal wagon shown in Fig. P6-24 is used to haul coal from a mine. If the mass of the coal and wagon is 2000 kg (with its center of mass at G), determine

(a) The magnitude of the force P required to move the wagon at a constant velocity. (b) The force exerted on each of the front wheels by the inclined surface. (c) The shearing stress in each of the 25-mm-diameter front axles. SOLUTION

( )2000 9.81 19,620 NW = =

(a) From a free-body diagram of the coal wagon, the equilibrium equations are solved to get the forces

0 :xFΣ = sin 30 0P W− ° =


159

9810 NP = ....................................................................Ans.

1 0 :MΣ =�

( )( ) ( ) ( )( )( ) ( )

22 3 1 cos30 2

sin 30 2 0

N P W

W

− − °

+ ° =

2 4028.81 NN = ............................................................Ans.

0 :yFΣ = 1 22 2 cos30 0N N W+ − ° =

1 4466.90 NN =

(c) Finally, dividing the shear force on the front axle by its cross-sectional area gives the shear stress

( )

6 222 2

2

4028.81 8.21 10 N/m 8.21 MPa0.025 4s

NA

τπ

= = = × = ............................................Ans.

6-25 A lever is loaded and supported as shown in Fig. P6-25. Determine (a) The reactions at A and C. (b) The normal stress in the ½-in.-diameter rod CD. (c) The shearing stress in the ½-in.-diameter pin at A, which is in double shear. (d) The change in length of rod CD, which is made of a material with a modulus of elasticity of 30(106) psi. SOLUTION

( )1tan 6 12 26.565θ −= = ° 2 26 12 13.416 in.CDL = + =

(a) From a free-body diagram of the lever, the equilibrium equations are solved to get the forces

0 :AMΣ =� ( ) ( ) ( ) ( ) ( )cos 6 40 4 60 8 80 12 0CDT θ − − + =

59.628 lbCDT = −

0 :xF→ Σ = 125 cos 0x CDA T θ+ + =

71.667 lbxA = −

0 :yF↑ Σ = 40 60 80 sin 0y CDA T θ− − + − =

6.666 lbyA = −

72.0 lb=A 5.31° ..................................Ans.

59.6 lb (C)CDT = .........................................Ans.

(b) Dividing the normal force in the rod CD by its cross-sectional area gives the normal stress

( )2

59.628 304 psi 304 psi (C)0.5 4

CDCD

n

TA

σπ

−= = = − = .......................................................Ans.

(c) Dividing the shear force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )2

72.0 183.3 psi2 2 0.5 4A

As

VA

τπ

= = =

............................................................................Ans.

(d) Finally, the change in length of the rod CD is given by


160

( )( )

( ) ( )26

59.628 13.416

30 10 0.5 4CD

PLEA

δπ

−= =

×

4 41.358 10 in. 1.358 10 in. (shrink)− −= − × = × ................................................................Ans.

6-26 The wood plane shown in Fig. P6-26 moves with a constant velocity when subjected to the forces shown. Determine

(a) The shearing force of the wood on the plane. (b) The normal force, and its location, of the wood on the plane. SOLUTION From a free-body diagram of the plane, the equilibrium equations are solved to get the forces

0 :xF→ Σ = 40cos 70 70cos16 0F − ° − ° =

0 :yF↑ Σ = 40sin 70 70sin16 0N − ° − ° =

0 :AMΣ =� ( )( ) ( ) ( )70cos16 75 70sin16 220° − °

( ) ( )40cos 70 60 0Nd+ ° + =

81.0 NF = ............................................................................................................Ans.

56.882 NN = .......................................................................................................Ans.

28.5 mmd = −

31.5 mm (from front of plane) ........................................................................Ans.

6-27 A bracket of negligible weight is used to support the distributed load shown in Fig. P6-27. Determine (a) The reactions at the supports A and B. (b) The shearing stress in the 0.25-in.-diameter pin at A, which is in single shear. (c) The bearing stress between the bracket and the 1-in.×1-in. bearing plate at B. SOLUTION (a) From a free-body diagram of the bracket, the equilibrium equations are solved to get the forces

0 :AMΣ =� ( )( ) ( )18 12 10 2 12 3 0B − =

0 :xF→ Σ = 0xA B+ =

0 :yF↑ Σ = ( )( )12 10 2 0yA − =

13.333 lbB =

13.333 lbxA = − 60.0 lbyA =

61.464 lb 61.5 lb= ≅A 77.47° ............................................Ans.

13.33 lb = →B ...............................................................................Ans. (b) Dividing the shear force on the pin by its cross-sectional area gives the shear stress

( )2

61.464 1252 psi0.25 4

AA

s

VA

τπ

= = = .....................................................................................Ans.

(c) Finally, the bearing stress is given by

13.333 13.33 psi

1 1bb

BA

σ = = =×

.............................................................................................Ans.


161

6-28 Determine the force P required to push the 135-kg cylinder over the small block shown in Fig. P6-28. SOLUTION

1 220 75cos 48.769220

θ − −= = °

90 20 21.231φ θ= ° − − ° = °

From a free-body diagram of the cylinder, the equilibrium equations are solved to get the forces

0 :centerMΣ =� 220 0F− =

0 NF =

0 :xF→ Σ = cos 0P N φ− =

0 :yF↑ Σ = ( )sin 135 9.81 0N φ − =

3657.2 NN =

3410 N P = → ...................................................................................................................... Ans.

6-29 The electric motor shown in Fig. P6-29 weighs 25 lb. Due to friction between the belt and pulley, the belt forces have magnitudes of T1 = 21 lb and T2 = 1 lb. Determine

(a) The support reactions at A and B. (b) The shearing stress in the ¼-in.-diameter pin at A, which is in single shear. SOLUTION (a) From a free-body diagram of the motor and support,

the equilibrium equations are solved to get the forces

0 :xF→ Σ = 21 1 0xA − − =

22 lbxA =

0 :AMΣ =� ( ) ( ) ( )12 8 25 5.5 1 10.5 21 0B − + + =

2.167 lbB = −

0 :yF↑ Σ = 25 0yA B+ − =

27.167 lbyA =

34.958 lb 35.0 lb= ≅A 51.00° ...................................................................................Ans.

2.17 lb = ↓B ......................................................................................................................... Ans. (b) Dividing the shear force on the pin by its cross-sectional area gives the shear stress

( )2

34.958 712 psi0.25 4

AA

s

VA

τπ

= = = ......................................................................................Ans.

6-30 Bar AB of Fig. P6-30 has a uniform cross section, a mass of 25 kg, and a length of 1 m. Determine the angle θ for equilibrium.

SOLUTION From a free-body diagram of the bar, the equilibrium equations are

0 :xF→ Σ = sin 30 sin 45 0A B° − ° =

0 :yF↑ Σ = ( )cos30 cos 45 25 9.81 0A B° + ° − =

0 :AMΣ =� ( )( ) ( )( ) ( ) ( )cos 45 1cos sin 45 1sin 25 9.81 0.5cos 0B Bθ θ θ° + ° − =


162

Rearranging the first equation gives

1.41421A B= Substituting this into the second equation gives

126.951 NB =

179.535 NA = Finally, the third equation gives

tan 0.366θ =

20.10θ = ° ................................................................................................................................Ans.

6-31 The wrecker truck of Fig. P6-31 has a weight of 15,000 lb and a center of gravity at G. The force exerted on the rear (drive) wheels by the ground consists of both a normal component By and a tangential component Bx, while the force exerted on the front wheels consists of a normal force Ay only. Determine the maximum pull P that the wrecker can exert when θ = 30° if Bx cannot exceed 0.8By (because of friction considerations) and the wrecker does not tip over backward (the front wheels remain in contact with the ground).

SOLUTION From a free-body diagram of the truck, the equilibrium equations are

0 :xF→ Σ = sin 30 0xP B° − =

0 :yF↑ Σ = 15,000 cos30 0y yA B P+ − − ° =

0 :AMΣ =� ( ) ( ) ( )( ) ( ) ( )15,000 8 14.5 cos30 5 sin 30 10 0yA P P− − ° − ° =

The fourth equation needed to solve for the four unknowns is either 0yA = (the front wheels are on the verge of

lifting off the ground) or 0.8x yB B= (the rear wheels are on the verge of slipping). Guessing that the front wheels are on the verge of lifting off the ground gives the solution

0 lbyA =

12,861.56 lbP =

26,138.44 lbyB =

6430.78 lbxB =

Since

6430.78 lb 0.8 20,911 lbx yB B= < =

the guess that the front wheels are on the verge of lifting off the ground was the correct guess, and

max 12,860 lbP = ..................................................................................................................... Ans.

6-32 Pulleys A and B of the chain hoist shown in Fig. P6-32 are connected and rotate as a unit. The chain is continuous, and each of the pulleys contains slots that prevent the chain from slipping. Determine the force F required to hold a 450-kg block W in equilibrium if the radii of pulleys A and B are 90 mm and 100 mm, respectively.

SOLUTION From a free-body diagram of the lower pulley, vertical equilibrium equation gives the tension

0 :yF↑ Σ = ( )2 450 9.81 0T − =

2207.25 NT = Then, from a free-body diagram of the lower pulley, moment equilibrium equation gives the force F


163

0 :axleMΣ =� 100 90 100 0T T F− − =

221 NF = ................................................................................................................................Ans.

6-33 The crane and boom shown in Fig. P6-33 weigh 12,000 lb and 600 lb, respectively. When the boom is in the position shown, determine

(a) The maximum load that can be lifted by the crane. (b) The tension in the cable used to raise and lower the boom when the load being lifted is 3600 lb. (c) The pin reaction at boom support A when the load being lifted is 3600 lb. (d) The required size of pin A ( which is in double shear) if the shearing stress must not exceed 12 ksi. SOLUTION (a) From a free-body diagram of the entire crane,

the equilibrium equations are

0 :yF↑ Σ = 12,000 600 0N W− − − =

0 :CMΣ =� ( ) ( ) ( )( )12,000 9 600 12cos30 1Nd− − ° −

( )24cos30 1 1 0W− ° − + =

The maximum load that can be lifted corresponds to impending tip; at which point the normal force acts at the front corner C, and 0d = . At this point

max 4930 lbW W= = ...............................................................................................................Ans.

(b) From a free-body diagram of the pulley, the equilibrium equations give the axle forces

0 :xF→ Σ = 3600cos10 0xB − ° =

3545.31 lbxB =

0 :yF↑ Σ = 3600 3600sin10 0yB − − ° =

4225.13 lbyB =

Then, from the free-body diagram of the boom, the equilibrium equations give the forces

0 :AMΣ =�

( ) ( ) ( )( ) ( ) ( )( )

3545.31 24sin 30 4225.13 24cos30 600 12cos30

cos10 24sin 30 sin10 24cos30 0T T

° − ° − °

+ ° ° − ° ° =

6275.13 lb 6280 lbT = ≅ ....................................... Ans.

0 :xF→ Σ = cos10 3545.31 0xA T− ° − =

9725.11 lbxA =

0 :yF↑ Σ = sin10 4225.13 600 0yA T− ° − − =

5914.80 lbyA =

11,382.55 lb 11.38 kip= ≅A 31.31° ............ Ans.

(d) Finally, dividing the shear force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress (which must not exceed 12 ksi)

( )2

11,382.55 12,000 psi2 2 4A

s A

VA d

τπ

= = =

0.777 in.Ad = ......................................................................................................................... Ans.


164

6-34 A bar AB of negligible mass is supported in a horizontal position by two cables as shown in Fig. P6-34. Determine

(a) The magnitude of force P and the force in each cable. (b) The change in length of the 15-mm-diamerter cable BD, if it is initially 1 m long and is made of steel

with a modulus of elasticity of 200 GPa. SOLUTION (a) From a free-body diagram of the bar, the equilibrium equations give the forces

0 :BMΣ =� ( )( )1300 sin 60 0AL T L+ ° =

1501.11 N 1501 NAT = ≅ ........................................ Ans.

0 :xF→ Σ = cos 68 cos60 0B AT T° − ° =

2003.58 N 2004 NBT = ≅ ...................................... Ans.

0 :yF↑ Σ = sin 60 sin 68 1300 0A BT T P° + ° − − =

1857.69 N 1858 NP = ≅ .....................................................................................................Ans. (b) The stretch of wire BD is given by

( )

( ) ( )29

2003.58 10000.0567 mm

200 10 0.015 4PLEA

δπ

= = = ×

...................................................Ans.

6-35 The garage door ABCD shown in Fig. P6-35 is being raised by a cable DE. The one-piece door is a homogeneous rectangular slab weighing 225 lb. Frictionless rollers B and C run in tracks at each side of the door as shown. Determine

(a) The force in the cable and the reactions at the rollers when d = 75 in. (b) The shearing stress (single shear) in the 3/8-in.-diameter rods that connect the roller to the door. SOLUTION

( )1cos 25 48 58.612θ −= = °

1 12 6sintan 5.46675 6cos

θφθ

− −= = °−

(a) From a free-body diagram of the door, the equilibrium equations are

0 :xF→ Σ = cos 2 0DT Bφ − =

0 :yF↑ Σ = sin 225 2 0DT Cφ − − =

0 :DMΣ =� ( ) ( ) ( ) ( ) ( ) ( )225 36cos 2 48sin sin 6cos cos 6sin 0D DB T Tθ θ φ θ φ θ− + − =

The first equation gives

0.49773 DB T=

Substituting this into the moment equation gives

92.5356 lb 92.5 lbDT = ≅ ....................................................................................................Ans.

46.0574 lb 46.1 lbB = ≅ .....................................................................................................Ans.

108.0926 lb 108.1 lb C = − ≅ ↑ .........................................................................................Ans. (b) Dividing the shear force on the rods by their cross-sectional areas gives the shear stresses

( )2

46.0574 417 psi3 8 4

BB

s

VA

τπ

= = = .........................................................................................Ans.


165

( )2

108.0926 979 psi3 8 4

CC

s

VA

τπ

= = = .........................................................................................Ans.

6-36 The lever shown in Fig. P6-36 is formed in a quarter circular arc of radius 450 mm. Determine the angle θ if neither of the reactions at A or B can exceed 200 N.

SOLUTION From a free-body diagram of the lever, the equilibrium equations are

0 :xF→ Σ = sin 0xA B θ− =

0 :yF↑ Σ = 125 cos 0yA B θ− + =

0 :DMΣ =� ( )450 0yA− =

The fourth needed equation is either 2 2 200 Nx yA A A= + = or

200 NB = . In either case, the moment equation gives

0 NyA =

Now, guessing that 200 NB = gives

51.32θ = ° ................................................................................................................................Ans.

156.1 NxA =

2 2 156.1 N 200 Nx y xA A A A= + = = <

Therefore, the guess was correct.

6-37 A man is slowly raising a 20-ft-long homogeneous pole weighing 150 lb as shown in Fig. P6-37. The lower end of the pole is kept in place by smooth surfaces. Determine the force exerted by the man to hold the pole in the position shown.

SOLUTION From a free-body diagram of the pole, the equations of equilibrium give the tension

0 :AMΣ =� ( )( ) ( )sin 30 20 150 10cos60 0T ° − ° =

75.0 lbT = ............................................................................Ans.

6-38 The lever shown in Fig. P6-36 is a quarter circular arc of radius 450 mm. The 25-mm-diameter pin at A is smooth and frictionless and is in single shear.

(a) Plot A, the magnitude of the pin force at A, and B, the force on the smooth support B, as functions of θ (10° ≤ θ ≤ 80°), the angle at which the support is located.

(b) Plot τ, the average shear stress in the pin at A, as a function of θ (10° ≤ θ ≤ 80°). SOLUTION (a) From a free-body diagram of the lever, the equilibrium equations are


166

0 :DMΣ =� ( )450 0yA− =

0 NyA =

0 :yF↑ Σ = 125 cos 0yA B θ− + =

125 cos NB θ=

0 :xF→ Σ = sin 0xA B θ− =

sin 125 tan NxA B θ θ= =

Therefore

2 2 125 tan Nx y xA A A A θ= + = = ......................................................................................Ans.

( )125 cos NB θ= .................................................................................................................Ans.

(b) Dividing the force in the pin at A by its cross-sectional area gives the shear stress

( ) ( )

22 2

125 tan 500 tan N/m0.025 4 0.025

A

s

VA

θ θτπ π

= = = ....................................................................Ans.

6-39 The crane and boom shown in Fig. P6-39 weigh 12,000 lb and 600 lb, respectively. The pulleys at D and E are small and the cables attached to them remain essentially parallel. The 3-in.-diameter pin at A is smooth and frictionless and is in double shear. The distributed force exerted on the treads by the ground is equivalent to a single resultant force N acting at some distance d behind point C.

(a) Plot d, the location of the equivalent force N relative to point C, as a function of the boom angle θ (0° ≤ θ ≤ 80°) when the crane is lifting a 3600-lb load.

(b) Plot τ, the average shear stress in the pin at A, as a function of θ (0° ≤ θ ≤ 80°) when the crane is lifting a 3600-lb load.

(c) It is desired that the resultant force on the tread always be at least 1 ft behind C to ensure that the crane is never in danger of tipping over. Plot Wmax, the maximum load that may be lifted, as a function of θ (0° ≤ θ ≤ 80°).

SOLUTION (a) From a free-body diagram of the entire crane,


0 :yF↑ Σ = 12,000 600 0N W− − − =

0 :CMΣ =� ( )( ) ( )( )12,000 9 600 12cos 1Nd θ− − −

( )24cos 1 1 0W θ− − + =


167

If 3600 lbW = , then

16,200 lbN =

108,600 93,600cos ft

16,200d θ−= ...........................................................................................Ans.

(c) If the normal force must be no closer than 1 ft to the front of the tread, then

max12,600 lbN W= +

( ) ( )max max108,600 12,600 1 7200cos 24 cos 0W Wθ θ− + − − =

max96,000 7200cos lb

1 24cosW θ

θ−=

+...........................................................................................Ans.

(b) As the boom is raised, the cable angle φ (assumed to be the

same for both cables) and the boom angle θ are related by

24cosb θ=

24sinh θ=

6 24sin 6tan9 24cos 9

hb

θφθ

− −= =+ +

Then, from the free-body diagram of the pulley, the equilibrium equations give the forces

0 :xF→ Σ = 3600cos 0xB φ− =

3600cos lbxB φ=

0 :yF↑ Σ = 3600 3600sin 0yB φ− − =

( )3600 1 sin lbyB φ= +

Finally, from the free-body diagram of the boom, the equilibrium equations

0 :xF→ Σ = cos 0x x BDA B T φ− − =

0 :yF↑ Σ = 600 sin 0y y BDA B T φ− − − =

0 :AMΣ =� ( )( ) ( )cos 24sin 600 12cosx BDB T φ θ θ+ −

( )( )sin 24cos 0y BDB T φ θ− + =


168

are solved to get the forces

( )24 7200 cos 24 sin

lb24cos sin 24sin cos

y xBD

B BT

θ θφ θ φ θ

+ −=

−............Ans.

cos lbx x BDA B T φ= +

600 sin lby y BDA B T φ= + +

2 2x yA A A= +

Then, dividing the shear force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )

2 2

2psi

14.13722 3 4x yAA AVτ

π

+= =

..................................................................................... Ans.

Although these graphs appear to be okay, a check of the data reveals that for angles greater than about 66°, the cable tension TBD is negative when the crane is lifting 3600 lb. But the cable cannot push on the boom. Therefore, for angles greater than 66°, the boom would fall back on the cab of the crane and the solution for angles greater than 66° are meaningless. Further analysis would be required to find the critical angle for other weights being lifted.

6-40 An extension ladder arrangement is being raised into position by a hydraulic cylinder as shown in Fig. P6-40. The center of gravity of the 500-kg ladder is at G. The 25-mm-diameter pin at A is smooth, frictionless, and in double shear.

(a) Plot σn, the normal stress in the 40-mm-diameter piston rod of the hydraulic cylinder, as a function of the angle θ (10° ≤ θ ≤ 90°).

(b) Plot τ, the shear stress in the pin at A, as a function of θ (10° ≤ θ ≤ 90°). SOLUTION

( )500 9.81 4905 NW = =

(a) From a free-body diagram of the ladder, the equilibrium equations are

0 :xF→ Σ = sin 0x BA F φ+ =

0 :yF↑ Σ = cos 4905 0y BA F φ+ − =

0 :AMΣ =� ( ) ( ) ( ) ( )cos 3cos sin 3sinB BF Fφ θ φ θ−

( )( ) ( ) ( )4905cos 8 4905sin 1 0θ θ− + =

in which

1 3cos 2tan3sin 1.5

θφθ

− − = +

Solving the equilibrium equations gives


169

( )( ) ( ) ( )4905cos 8 4905sin 1

N3cos cos 3sin sinBF

θ θφ θ φ θ

−=

−

sin Nx BA F φ= −

4905 cos Ny BA F φ= −

Dividing the piston force by its cross-sectional area gives the normal stress

( )20.04 4

B Bn

n

F FA

σπ

= =

( )

22

4 N/m0.04

BFπ

= ...................... Ans.

(b) Dividing the force in the pin at A by twice its cross-sectional area (since it is in double shear) gives the shear stress

( ) ( )

2 2 2 22

22

2N/m

2 0.0252 0.025 4x y x yA

s

A A A AVA

τππ

+ += = =

.........................................................Ans.

6-41 The hydraulic cylinder BC is used to tip the box of the dump truck shown in Fig. P6-41. The pins at A (one on each side of the box) each have a diameter of 1.5 in. and are in double shear. If the combined weight of the box and the load is 22,000 lb and acts through the center of gravity G,

(a) Plot C, the force in the hydraulic cylinder, as a function of the angle θ (0° ≤ θ ≤ 80°). (b) Plot τ, the shear stress in the pin at A, as a function of θ (0° ≤ θ ≤ 80°). SOLUTION (a) From a free-body diagram of the truck box,


0 :xF→ Σ = 2 cos 0x BCA F φ− =

0 :yF↑ Σ = 2 sin 22,000 0y BCA F φ+ − =

0 :AMΣ =� ( ) ( ) ( ) ( )22,000cos 8.5 22,000sin 2θ θ−

( ) ( )sin 12.5 0BCF φ θ− − =

in which

1 12.5sin 0.5tan12.5cos 2.0

θφθ

− +=−

Therefore

( )187,000cos 44,000sin lb

12.5sinBCFθ θ

φ θ−=

−..........................Ans.

( )cos 2 lbx BCA F φ= ( )22,000 sin 2 lby BCA F φ= −

(b) Then, dividing the shear force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )

2 2

2psi

3.534292 1.5 4x yAA AVτ

π

+= =

..................................................................................Ans.


170

6-42 The wrecker truck of Fig. P6-42 has a mass of 6800 kg and a center of gravity at G. The force exerted on the rear (drive) wheels by the ground consists of both a normal component By and a tangential component Bx, while the force exerted on the front wheels consists of a normal force Ay only. Plot P, the maximum pull that the wrecker can exert, as a function of θ (0° ≤ θ ≤ 90°) if Bx cannot exceed 0.8By (because of friction considerations) and the wrecker does not tip over backward (the front wheels remain in contact with the ground).

SOLUTION

( )6800 9.81 66,708 NW = =

From a free-body diagram of the truck, the equilibrium equations are

0 :xF→ Σ = sin 0xP Bθ − =

0 :yF↑ Σ = 66,708 cos 0y yA B P θ+ − − =

0 :BMΣ =� ( )( ) ( ) ( )66,708 2.4 4.4 cos 1.5yA P θ− −

( )( )sin 3 0P θ− =

The fourth equation needed to solve for the four unknowns is either 0yA = (the front wheels are on the verge of

lifting off the ground) or 0.8x yB B= (the rear wheels are on the verge of slipping). Guessing that the front wheels are on the verge of lifting off the ground gives the solution

0 NyA =

( )66,708 2.4

N1.5cos 3sin

Pθ θ

=+

......................................................................................................Ans.

sin NxB P θ=

66,708 cos NyB P θ= +

The forces xB and 0.8 yB are plotted on the same

graph as the force P . Since xB is always less than

0.8 yB , the guess that the front wheels are on the verge of lifting off the ground was the correct guess, and the solution is valid for all values of θ .


171

6-43 The garage door ABCD shown in Fig. P6-43 is being raised by a cable DE. The one-piece door is a homogeneous rectangular slab weighing 225 lb. Frictionless rollers B and C run in tracks at each side of the door as shown.

(a) Plot T, the tension in the cable, as a function of d (0 ≤ d ≤ 100 in.). (b) Plot B and C, the forces on the frictionless rollers, as a function of d (0 ≤ d ≤ 100 in.). SOLUTION

(a) From a free-body diagram of the door with both rollers on the horizontal track ( )0 40 in.d< < , the equilibrium equations give

0 :xF→ Σ = cos 0T φ =

0 lbT =

0 :CMΣ =� ( ) ( ) ( ) ( )36 2 48 sin 6 0W B T φ− + =

3 8 675 8 84.4 lbB W= = =

0 :yF↑ Σ = sin 2 2 0T B C Wφ + + − =

8 225 8 28.1 lbC W= = =

From a free-body diagram of the door with roller B on the curve and roller C on the horizontal track ( )40 in. 53.5 in., 0 90d γ< < ° < < ° , the equilibrium equations are

0 :xF→ Σ = cos 2 sin 0T Bφ γ− =

0 :yF↑ Σ = sin 2 cos 2 0T B C Wφ γ+ + − =

0 :CMΣ =� ( ) ( ) ( )36cos 2 cos 48W Bθ γ θ− −

( )( ) ( )( )sin 6cos cos 6sin 0T Tφ θ φ θ+ − =

in which

12 12cos 1 cossin

48 4γ γθ − −= =

( )100 48cos 12 1 sind θ γ= − − −

12 6sintan

6cosdθφθ

−=−

Therefore

( )

6 cos lb8cos cos sin sin cos cos sin

WT θφ γ θ γ θ φ θ φ

=− + −

cos lb

2sinTB φ

γ= ( )2 cos sin 2 lbC W B Tγ φ= − −

From a free-body diagram of the door with roller B on the vertical track and roller C on the horizontal track ( )53.5 in. 88 in.d< < , the equilibrium equations are

0 :xF→ Σ = cos 2 0T Bφ − =

0 :yF↑ Σ = sin 2 0T C Wφ + − =

0 :CMΣ =� ( ) ( )( ) ( ) ( ) ( )( )36cos 2 48sin sin 6cos cos 6sin 0W B T Tθ θ φ θ φ θ− + − =


172

in which

100cos

48dθ −=

12 6sintan

6cosdθφθ

−=−

Therefore

6 cos lb

9sin cos cos sinWT θ

θ φ θ φ=

−

cos 2 lbB T φ=

( )sin 2 lbC W T φ= −

Finally, from a free-body diagram of the door with roller B on the vertical track and roller C on the curve ( )88 in. 100 in., 0 90d γ< < ° < < ° , the equilibrium equations are

0 :xF→ Σ = cos 2 2 sin 0T B Cφ γ− − =

0 :yF↑ Σ = sin 2 cos 0T C Wφ γ+ − =

0 :CMΣ =� ( ) ( )36cos 2 48sinW Bθ θ−

( )( ) ( )( )sin 6cos cos 6sin 0T Tφ θ φ θ+ − =

in which

12 12sin 1 sincos

48 4γ γθ − −= =

( )100 12 1 sind γ= − −

12 6sintan

6cosdθφθ

−=−

Therefore

6 cos 8 sin tan lb

9cos sin sin cos 8sin sin tanW WT θ θ γ

φ θ φ θ φ θ γ+=

− +

cos sin tan tan lb

2T T WB φ φ γ γ+ −=

sin lb

2cosW TC φ

γ−=


173

6-44 Determine all forces acting on member ABE of the frame of Fig. P6-44. SOLUTION First, from a free-body diagram of the complete frame, the equilibrium equations give the support forces

0 :AMΣ =� ( ) ( )300 150 300 0D − =

150 ND =

0 :xF→ Σ = 150 0xA + =

150 NxA = −

0 :yF↑ Σ = 0yA D+ =

150 NyA = −

Next, from a free-body diagram of member BCD, the equilibrium equations give the pin forces

0 :BMΣ =� ( ) ( )150 300 100 0yC+ =

450 NyC = −

0 :CMΣ =� ( ) ( )150 200 100 0yB− =

300 NyB =

Finally, from a free-body diagram of member ABE, the equilibrium equations give the pin forces

0 :EMΣ =� ( ) ( ) ( ) ( )300 100 150 100 150 200 100 0xB+ − − =

150 NxB =

0 :xF→ Σ = 150 150 0xE − − =

300 NxE =

0 :yF↑ Σ = 300 150 0yE − − =

450 NyE =

Therefore, the forces acting on member ABE are:

212 N=A 45° ..........................................................................................................Ans.

335 N=B 63.43° .....................................................................................................Ans.

541 N=E 56.31° .....................................................................................................Ans.

6-45 In the linkage of Fig. P6-45, a = 2.0 ft, b = 1.5 ft, θ = 30°, and P = 40 lb. Determine all forces acting on member BCD.

SOLUTION Note that member AC is a two-force member. Then, from a free-body diagram of the member BCD, the equilibrium equations give the forces

0 :BMΣ =� ( ) ( ) ( ) ( )cos 45 2 40cos30 3.5 0ACT ° − ° =

85.732 lbACT =

0 :xF→ Σ = 40cos30 cos 45 0x ACB T+ ° − ° =

25.981 lbxB =


174

0 :yF↑ Σ = sin 45 40sin 30 0y ACB T− ° + ° =

40.622 lbyB =

Therefore, the forces acting on member BCD are:

48.2 lb=B 57.40° ...................................................................................................Ans.

85.7 lbAC =T 45° ......................................................................................................Ans.

6-46 Forces of 5 N are applied to the handles of the paper punch of Fig. P6-46. Determine the force exerted on the paper at D and the force exerted on the pin at B by handle ABC.

SOLUTION From a free-body diagram of handle EBD, the equilibrium equations give the forces

0 :BMΣ =� ( ) ( )40 5 70 0DF − =

8.75 NDF =

0 :xF→ Σ = 0 NxB =

0 :yF↑ Σ = 5 0y DB F− + =

13.75 NyB =

13.75 N = ↓B .................................................................................................................Ans. The force exerted on the paper is equal and opposite to the force exerted on the handle

8.75 N D = ↓F .................................................................................................................Ans.

6-47 Forces of 25 lb are applied to the handles of the pipe pliers shown in Fig. P6-47. Determine the force exerted on the pipe at D and the force exerted on handle DAB by the pin at A.

SOLUTION From a free-body diagram of handle DAB, the equilibrium equations give the forces

0 :AMΣ =� ( ) ( )1.25 25 9 0DF − =

180 lbDF =

0 :xF→ Σ = sin 38 0D xF A° − =

110.819 lbxA =

0 :yF↑ Σ = 25 cos38 0y DA F− − ° =

166.842 lbyA =

200 lb=A 56.41° ....................................................................................................Ans. The force exerted on the pipe at D is equal and opposite to the force exerted on the handle

180 lbD =F 52° .........................................................................................................Ans.

6-48 The jaws and bolts of the wood clamp in Fig. P6-48 are parallel. The bolts pass through swivel mounts so that no moments act on them. The clamp exerts forces of 300 N on each side of the board. Treat the forces on the boards as uniformly distributed over the contact areas and determine the forces in each of the bolts. Show on a sketch all forces acting on the upper jaw of the clamp.

SOLUTION The resultant of the distributed force exerted on the upper jaw of the clamp is a force of 300 N at the center of the uniformly distributed load. From a free-body diagram of the upper jaw, the equilibrium equations give the forces


175

0 :DMΣ =� ( ) ( )300 225 100 0CF− =

675 N (T)CF = .............................................................Ans.

0 :CMΣ =� ( ) ( )300 125 100 0DF− =

375 N (C)DF = .............................................................Ans.

6-49 Determine all forces acting on member ABCD of the frame of Fig. P6-49. SOLUTION First, from a free-body diagram of the complete frame, the equilibrium equations give the support reactions

0 :AMΣ =� ( ) ( )12 100 24 0F − =

200 lbF =

0 :xF→ Σ = 100 0xA − =

100 lbxA =

0 :yF↑ Σ = 0yA F+ =

200 lbyA = −

Next, note that member BE is a two-force member. Then, from a free-body diagram of member ABCD, the equilibrium equations give the pin forces

0 :CMΣ =� ( ) ( ) ( ) ( )cos 45 6 100 18 100 6 0BEF ° − − =

565.685 lbBEF =

0 :xF→ Σ = 100 100 cos 45 0BE xF C− − ° + =

400 lbxC =

0 :yF↑ Σ = 200 sin 45 0y BEC F− + ° =

200 lbyC = −

Therefore, the forces acting on member ABCD are:

224 lb=A 63.43° ..................................................Ans.

566 lb=B 45° .........................................................Ans.

447 lb=C 26.57° ..................................................Ans.

6-50 In Fig. P6-50, a cable is attached to the structure at E, passes around the 0.8-m-diameter, frictionless pulley at A, and then is attached to a 1000-N weight W. Determine

(a) The support reaction at G. (b) All forces acting on member ABCD. SOLUTION (a) From a free-body diagram of the complete structure,

the equilibrium equations give the support reactions

0 :DMΣ =� ( ) ( )1000 3.4 2 0G− =

1700 NG =

0 :xF→ Σ = 0xD G− =


176

1700 NxD =

0 :yF↑ Σ = 1000 0yD − =

1000 NyD =

1700 N = ←G ...............................................................................................................Ans. (b) Next, from a free-body diagram of the pulley, the equilibrium equations give the pin forces

0 :xF→ Σ = 1000 0xA− =

1000 NxA =

0 :yF↑ Σ = 1000 0yA − =

1000 NyA =

Then, from a free-body diagram of member ECF, the equilibrium equations give the pin forces

0 :FMΣ =� ( ) ( )1 1000 1.4 0xC + =

1400 NxC = −

Finally, from a free-body diagram of member ABCD, the equilibrium equations give the pin forces

0 :BMΣ =� ( ) ( ) ( )1 1000 2 1000 1 0yC + + =

3000 NyC = −

0 :CMΣ =� ( ) ( ) ( )1000 2 1 1000 1 0yB− + =

3000 NyB =

0 :xF→ Σ = 1000 1700 0x xB C+ + + =

1300 NxB = −

Therefore the forces on member ABCD are

1414 N=A 45° ........................................................................................................Ans.

3270 N=B 66.57° ..................................................................................................Ans.

3310 N=C 64.98° ..................................................................................................Ans.

1972 N=D 30.47° ...................................................................................................Ans.

6-51 A pin-connected system of levers and bars is used as a toggle for a press as shown in Fig. P6-51. Three members are joined by pin D, as shown in the insert. Determine

(a) The force exerted on the can at A when a force of 1000 lb is applied to the lever at G. (b) All forces that act on member CD. SOLUTION (a) From a free-body diagram of the lever, the moment equilibrium

equation gives the force in the link DE

0 :FMΣ =� ( ) ( )1000 30 8 0DEF− =

3750 lbDEF =

Note that all three links connected to the pin D are two-force members. From the free-body diagram of the pin, the equilibrium equations give the forces in the links CD and BD


177

0 :xF→ Σ = cos78 cos 67 3750 0CD BDF F° + ° − =

0 :yF↑ Σ = sin 67 sin 78 0BD CDF F° − ° =

6395.05 lbBDF = 6018.19 lbCDF =

Finally, from a free-body diagram of the plunger, the vertical equilibrium equation gives the crushing force

0 :yF↑ Σ = 6395.05sin 67 0AF − ° =

5890 lb on the plungerA = ↑F

5890 lb on the canA = ↓F .......................................Ans.

(b) Member CD is a two force member in compression; therefore,

6020 lbCD =F 78 on at CD C° .............................................................................Ans.

6020 lbCD =F 78 on at CD D° .............................................................................Ans.

6-52 The front-wheel suspension of an automobile is shown in Fig. P6-52. The pavement exerts a vertical force of 2700 N on the tire. Determine the force in the spring and the forces at A, B, and D.

SOLUTION From a free-body diagram of the wheel, the equilibrium equations give the pin forces

0 :BMΣ =� ( ) ( )325 2700 150 0CDF − =

1246.154 NCDF =

0 :xF→ Σ = 0x CDB F− =

1246.154 NxB =

0 :yF↑ Σ = 2700 0yB− =

2700 NyB =

Next, from a free-body diagram of the control arm, the equilibrium equations give the pin forces

0 :AMΣ =� ( ) ( ) ( )250 500 50 0s y xF B B− − =

5649.23 NsF =

0 :xF→ Σ = 0x xA B− =

1246.154 NxA =

0 :yF↑ Σ = 5649.23 0y yA B+ − =

2949.23 NyA =

Therefore the force on the spring is

5650 N s = ↓F ................................................................................................................Ans.

and the forces at A, B, and D are 3200 N=A 67.09° ..................................................................................................Ans.

2970 N=B 65.22° ..................................................................................................Ans.

1246 N (C)CDF = ...........................................................................................................Ans.


178

6-53 The fold-down chair of Fig. P6-53 weighs 30 lb and has its center of gravity at G. Determine (a) All forces acting on member ABC. (b) The shearing stress on a cross section of the 3/8-in.-diameter pin at B, which is in single shear. SOLUTION (a) From a free-body diagram of the chair, the equilibrium equations give the support reactions

0 :AMΣ =� ( ) ( )30 3 21 0E− =

4.28571 lbE =

0 :xF→ Σ = 0xA E− =

4.28571 lbxA =

0 :yF↑ Σ = 30 0yA − =

30 lbyA =

Note that member BD is a two force member which makes an angle of

( )1tan 10 12 39.81θ −= = °

relative to the horizontal. Then, from the free-body diagram of the seat, the equilibrium equations give the pin forces

0 :CMΣ =� ( ) ( )( )30 3 sin 12 0BDT θ− =

11.7154 lbBDT =

0 :xF→ Σ = cos 4.28571 0BD xT Cθ − − =

4.7429 lbxC =

0 :yF↑ Σ = sin 30 0BD yT Cθ + − =

22.500 lbyC =

The forces acting on ABC are equal and opposite to those acting on the seat

30.3 lb=A 81.87° ...................................................................................................Ans.

11.72 lbB =F 39.81° ................................................................................................Ans.

22.99 lb=C 78.10° .................................................................................................Ans. (b) Dividing the force on the pin at B by its cross-sectional area gives the shear stress

( )2

11.7154 106.1 psi3 8 4

B

s

FA

τπ

= = = .................................................................................Ans.

6-54 A scissors jack for an automobile is shown in Fig. P6-54. The screw threads exert a force F on the blocks at joints A and B. Determine

(a) The force P exerted on the automobile if F = 800 N and θ = 15°, θ = 30°, and θ = 45°. (b) The shearing stress on a cross section of the 10-mm diameter pin at C, which is in single shear. Solve for

each angle in part (a). SOLUTION (a) Note that all three members attached to pin A are two-force members. From a free-body diagram of pin A, the

equilibrium equations give the forces in members AC and AE

0 :yF↑ Σ = sin sin 0AE ACF Fθ θ− =


179

AE ACF F=

0 :xF→ Σ = 800 cos cos 0AE ACF Fθ θ− − =

( )800 2cos NAE ACF F θ= =

Similarly

( )800 2cos NBD ACF F θ= =

Then, from the free-body diagram of the platform, the equilibrium equations give the force P

0 :yF↑ Σ = sin sin 0AC BDF F Pθ θ+ − =

2 sin 800 tan NACP F θ θ= =

If 15θ = ° 214 NP = ...................................Ans.

If 30θ = ° 462 NP = ...................................Ans.

If 45θ = ° 800 NP = ...................................Ans. (b) Finally, dividing the force on the pin at C by its cross-sectional area gives the shear stress

( )

( )6

22

800 2cos 5.09296 10 N/mcos0.010 4

AC

s

FA

θτ

θπ×= = =

If 15θ = ° 6 25.27 10 N/m 5.27 MPaτ = × = ............................................Ans.

If 30θ = ° 6 25.88 10 N/m 5.88 MPaτ = × = ............................................Ans.

If 45θ = ° 6 27.20 10 N/m 7.20 MPaτ = × = ............................................Ans.

6-55 A force of 20 lb is required to pull the stopper DE in Fig. P6-55. Determine (a) All forces acting on member BCD. (b) The shearing stress on the cross section of the 1/8-in.-diameter pin at B, which is in single shear. (c) The deformation of the 1/8×3/8-in. member AB, which is made of steel with a modulus of elasticity of

30(106) psi. SOLUTION (a) From a free-body diagram of the entire cork puller,

the equilibrium equations give the pulling forces

0 :GMΣ =� ( ) ( )1.5 1.5 0y yE D− =

y yE D=

0 :yF↑ Σ = 20 0y yE D− − =

10 lby yD E= =

Note that both links attached to the pin at A are two-force members. Then, from the free-body diagram of pin A, the equilibrium equations give the forces in the links

0 :xF→ Σ = cos 45 cos 45 0AF ABT T° − ° =

AF ABT T=

0 :yF↑ Σ = 20 sin 45 sin 45 0AB AFT T− ° − ° =

14.14214 lbAB AFT T= =

Next, from the free-body diagram of member BCD, the equilibrium equations give the pin forces


180

0 :CMΣ =� ( ) ( ) ( ) ( ) ( )( )2 1.5 cos 45 2 sin 45 1 0x y AB ABD D T T− − ° − ° =

22.500 lbxD =

0 :xF→ Σ = cos 45 0AB x xT C D° + + =

0 :yF↑ Σ = sin 45 0AB y yT C D° + − =

32.500 lbxC = − 0 lbyC =

Therefore, the forces acting on member BCD are

14.14 lbB =F 45° .................................................................. Ans.

32.5 lb = ←C ............................................................................ Ans.

24.6 lb=D 23.96° ............................................................... Ans. (b) Dividing the force on the pin at B by its cross-sectional area gives the shear stress

( )2

14.14214 1152 psi1 8 4

B

s

FA

τπ

= = = ..................................................................................Ans.

(c) The change in length of the link AB is given by

( )( )

( ) ( ) ( )5

6

14.14214 21.42 10 in.

30 10 1 8 3 8ABPLEA

δ −= = = ×× ×

..........................................Ans.

6-56 Member BD of the frame shown in Fig. P6-56 is made of structural steel (E = 200 GPa) and has a rectangular cross section 50 mm wide by 15 mm thick. All pins have 15 mm diameters. Determine

(a) The axial stress in member BD. (b) The shearing stress on a cross section of pin C if it is loaded in double shear. (c) The change in length of member BD. SOLUTION (a) From a free-body diagram of the entire structure, the equilibrium equations give the support reactions

0 :AMΣ =� ( ) ( )( ) ( ) ( )1.2 2000 1.2 0.6 2000 0.7 0E − − =

2366.67 NE =

0 :xF→ Σ = 2000 0xA− =

2000 NxA =

0 :yF↑ Σ = ( )2000 1.2 0yA E+ − =

33.333 NyA =

Next, from the free-body diagram of member CDE (in which sin 3 5θ = and cos 4 5θ = ), the equilibrium equations give the pin forces

0 :CMΣ =� ( ) ( ) ( ) ( ) ( )( )2366.667 1.2 2000 1.2 0.6 sin 0.8 0BDT θ− − =

2916.67 NBDT =

0 :xF→ Σ = 2916.67 cos 0xC θ− =

2333.33 NxC =


181

0 :yF↑ Σ = ( )2366.67 2000 1.2 2916.67sin 0yC θ+ − − =

1783.33 NyC =

2 2 2936.78 Nx yC C C= + =

Then, dividing the force in the member BD by its cross-sectional area gives the axial (normal) stress

( )

( )6 22916.67

3.89 10 N/m 3.89 MPa0.05 0.015

BDBD

n

TA

σ = = = × =×

...............................Ans.

(b) Dividing the force on the pin at C by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )( )

6 22

2936.788.31 10 N/m 8.31 MPa

2 2 0.015 4C

s

FA

τπ

= = = × =

...............................Ans.

(c) Finally, the change in length of the two-force member BD is given by

( )( )

( )( )9

2916.67 10000.01944 mm

200 10 0.05 0.015BDPLEA

δ = = =× ×

........................................Ans.

6-57 The hoist pulley structure of Fig. P6-57 is rigidly attached to the wall at C. A load of sand hangs from the cable that passes around the 1-ft-diameter, frictionless pulley at D. The weight of the sand can be treated as a triangular distributed load with a maximum intensity of 70 lb/ft. Determine

(a) All forces acting on member ABC. (b) The shearing stress on the cross section of the ½-in.-diameter pin D, which is in double shear. (c) The change in length of the ¼ × 1-in. member BE [E = 29(106) psi]. SOLUTION

( )( )1 70 2 70 lb2

T W= = =

(a) From a free-body diagram of the entire structure, the equilibrium equations give the reaction forces

0 :xF→ Σ = 0xT C− =

0 :yF↑ Σ = 0yC T− =

0 :DMΣ =� ( ) ( )5 1.5 0y x CTr Tr C C M− + − − =

245 lb ftCM = ⋅

70 lbxC = 70 lbyC =

Then, from the free-body diagram of member ABC, the equilibrium equations give the pin forces

0 :AMΣ =� ( ) ( )70 7 4 245 0BF− − =

61.25 lbBF =

0 :xF→ Σ = 70 0xA − =

0 :yF↑ Σ = 70 0y BA F− − =

70 lbxA =

8.75 lbyA =


182

Therefore, the forces acting on member ABC are

70.5 lb=A 7.13° ......................................................................................................Ans.

61.3 lb = ↓B .................................................................................................................. Ans.

99.0 lb=C 45° .........................................................................................................Ans.

245 lb ft C = ⋅M � ..........................................................................................................Ans.

(b) Next, from the free-body diagram of the pulley, the equilibrium equations give the pin forces

0 :xF→ Σ = 70 0xD− =

0 :yF↑ Σ = 70 0yD − =

70 lbxD =

70 lbyD =

2 2 98.995 lbx yD D D= + =

Dividing the force on the pin at D by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )2

98.995 252 psi2 2 0.5 4s

DA

τπ

= = = ...............................................................................Ans.

(c) The change in length of the two-force member BE is given by

( )( )

( ) ( )4

6

61.25 3 123.04 10 in.

29 10 1 4 1BEPLEA

δ −×= = = ×

× × .................................................Ans.

6-58 A pair of vice grip pliers is shown in Fig. P6-58. Determine the force exerted on the bolt by the jaws of the pliers when a force of magnitude 100 N is applied to the handle. Let d = 30 mm.

SOLUTION

( )1tan 30 40 36.870θ −= = ° sin 3 5θ = cos 4 5θ =

From a free-body diagram of the upper handle, the equilibrium equations give the forces

0 :BMΣ =� ( ) ( )( ) ( )( )100 93 sin 38 cos 5 0A AF Fθ θ− + =

494.681 NAF =

0 :xF→ Σ = cos 0A xF Bθ − =

0 :yF↑ Σ = sin 100 0A yF Bθ − − =

395.745 NxB =

196.809 NyB =

Next, from the free-body diagram of the jaw BCD, the moment equilibrium equation is

0 :CMΣ =� ( ) ( )35 12 0x yDa B B− − =

in which 2 235 15 38.079 mma = + = Therefore

426 ND = ........................................................................................................................ Ans.


183

6-59 Forces of 50 lb are applied to the handles of the bolt cutter of Fig. P6-59. Determine (a) All forces acting on the handle ABC. (b) The force exerted on the bolt at E. (c) The axial stress in the links at D (one on each side) if each has a 1/8 × 3/4-in. rectangular cross section. (d) The change in length of the links at D if they are 4 in. long and made of SAE 4340 heat-treated steel. SOLUTION (a) From a free-body diagram of the jaw CDE, horizontal equilibrium gives

0 lbxC =

Then, from a free-body diagram of the handle ABC, the equilibrium equations give the pin forces

0 :xF→ Σ = 0x xC B− =

0 :yF↑ Σ = 50 0y yB C− + =

0 :BMΣ =� ( ) ( )1 50 20 0yC − =

1000 lbyC =

0 lbxB = 1050 lbyB =

Therefore, the forces acting on the handle ABC are

1050 lb = ↓B .................................................................................................................Ans.

1000 lb = ↑C .................................................................................................................Ans. (b) Then, returning to the free-body diagram of the jaw CDE, the equilibrium equations give the force on the bolt

0 :DMΣ =� ( ) ( )3 2 0yC E− =

1500 lb on the jaw= ↓E

1500 lb on the bolt= ↑E ..........................................................................................Ans.

0 :yF↑ Σ = 1000 1500 0DT − − = 2500 lbDT =

(c) Dividing the link force by twice its cross-sectional area (since there are two links) gives the normal stress

( )( )

31500 13.33 10 psi 13.33 ksi2 2 1 8 3 4D

n

TA

σ = = = × =

.......................................Ans.

(d) Finally, the change in length of the two-force link is given by

( ) ( )

( ) ( ) ( )3

6

2500 2 41.839 10 in.

29 10 1 8 3 4DPLEA

δ −= = = ××

............................................Ans.

6-60 The tower crane of Fig. P6-60 is rigidly attached to the building at F. A cable is attached at D and passes over small frictionless pulleys at A and E. The object suspended from C has a mass of 1530 kg. Determine

(a) All forces acting on member ABCD. (b) The support reactions at F. (c) The shearing stress on a cross section of pin A if it has a diameter of 15 mm and is loaded in double

shear. SOLUTION

( )1530 9.81 15,009.30 NW = = 1tan 6 14.4 22.620φ −= = °

sin 3 5θ = cos 4 5θ =


184

(a) From a free-body diagram of member ABCD and pulley A, the equilibrium equations give the forces

0 :BMΣ =� ( )( ) ( )( ) ( ) ( )8 sin 8 sin 14.4 15,009.30 10 0T T Tθ φ− + − =

17,176.136 NT =

0 :xF→ Σ = cos cos 0xT B Tθ φ+ − =

2113.97 NxB =

0 :yF↑ Σ = sin 15,009.30 sin 0yT T B Tθ φ− + − + =

15, 273.51 NyB =

Therefore, the forces acting on the member ABCD are

15,363 N 15,360 N= ≅A 26.565° ......................................................................Ans.

15, 420 N=B 82.120° ............................................................................................Ans.

15,010 N = ↓C .............................................................................................................Ans.

17,180 N=D 22.620° ............................................................................................Ans.

(b) Next, from the free-body diagram of the entire structure, the equilibrium equations give the support reactions

0 :xF→ Σ = 0 NxF =

0 :yF↑ Σ = 17,176.136 15,009.30 0yF − − =

0 :FMΣ =� ( )( )17,176.136 8 FM+

( ) ( )15,009.30 10 0− =

32,185.4 N 32.2 kNyF = ≅

12,684 N mFM = ⋅

Therefore, the support reactions are

32.2 kN ≅ ↑F ................................. Ans.

12.68 kN m F ≅ ⋅M � ..................... Ans.

(c) Finally, dividing the force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )

6 22

15,363 43.5 10 N/m 43.5 MPa2 2 0.015 4A

s

VA

τπ

= = = × =

..............................Ans.

6-61 Three bars are connected with smooth pins to form the frame shown in Fig. P6-61. The weights of the bars are negligible. Determine

(a) The reactions at supports A and E. (b) The resultant forces at pins B, C, and D. SOLUTION

( )1tan 1 2 26.565θ −= = °

From a free-body diagram of the entire structure, the equilibrium equations give

0 :AMΣ =� ( )26 675 29.5 0yE − =


185

765.865 lbyE =

0 :yF↑ Σ = 675 0y yE A− − =

0 :xF→ Σ = 0x xA E− =

x xA E= 90.865 lbyA =

Next, from the free-body diagram of member BD, the equilibrium equations give the pin forces

0 :DMΣ =� ( )( ) ( )( )sin 13 675 10 0BF θ − =

1161.035 lbBF =

0 :xF→ Σ = cos 0B xF Dθ − =

1038.462 lbxD =

0 :yF↑ Σ = sin 675 0y BD F θ− − =

1194.230 lbyD =

Finally, from the free-body diagram of member ABC, the equilibrium equations give the pin forces

0 :CMΣ =� ( ) ( )2 13 13 6.5 5 0x y BA A F+ − =

603.606 lbx xA E= =

0 :xF→ Σ = cos 0x B xA F Cθ− + =

0 :yF↑ Σ = sin 0y B yA F Cθ− + − =

434.856 lbxC = 428.365 lbyC =

(a) Therefore, the reaction forces are

610 lb=A 8.56° ................................................. Ans.

975 lb=E 51.76° ............................................... Ans. (b) The forces acting on pins B, C, and D are

1161 lb=B 26.57 (on )ABC° ......................... Ans.

610 lb=C 44.57 (on )ABC° .................................................................................Ans.

1583 lb=D 48.99 (on )CDE° ...............................................................................Ans.

6-62 Figure P6-62 is a simplified sketch of the mechanism used to raise the bucket of a bulldozer. The bucket and its contents weigh 10 kN and have a center of gravity at H. Arm ABCD has a weight of 2 kN and a center of gravity at B; arm DEFG has a weight of 1 kN and a center of gravity at E. The weight of the hydraulic cylinders can be ignored.

(a) Calculate the force in the horizontal cylinders CJ and EI and all forces acting on arm DEFG for the position shown.

(b) Determine the required diameter of the pin at E if the shearing stress cannot exceed 120 MPa. The pin is in double shear.

SOLUTION (a) From a free-body diagram of the bucket, the equilibrium equations give the force in cylinder EI

0 :GMΣ =� ( ) ( )( )1.2cos30 10 0.3 0EIT ° − =


186

2.8868 kN 2.89 kNEIT = ≅ .............................................. Ans.

and the pin forces at G

0 :xF→ Σ = 0x EIG T− =

2.8868 kNxG =

0 :yF↑ Σ = 10 0yG − =

10 kNyG =

Next, from a free-body diagram of member DEFG , the equilibrium equations are

0 :DMΣ =� ( ) ( ) ( ) ( ) ( )0.6cos30 1 0.6sin 30 cos 1.2cos30EI BFT F φ° − ° + °

( ) ( ) ( ) ( )sin 1.2sin 30 1.8cos30 1.8sin 30 0BF x yF G Gφ+ ° − ° − ° =

0 :xF→ Σ = cos 0BF x EI xF D T Gφ − + − =

0 :yF↑ Σ = sin 1 0BF y yF D Gφ + − − =

in which

1 1.8cos30 1.2cos30tan 19.1071.8sin 30 1.2sin 30

φ − ° − ° = = ° ° + °

These equations are solved to give

10.438 kNBFF =

9.863 kNxD = 7.583 kNyD =

Therefore, the remaining forces acting on the member DEFG are

12.44 kN=D 37.55° ...............................................................................................Ans.

10.44 kN=F 19.11° ................................................................................................Ans.

10.41 kN=G 73.90° ...............................................................................................Ans. Then, from the free-body diagram of the entire bucket assembly,

the moment equilibrium equation gives the force in the cylinder CJ

0 :AMΣ =� ( ) ( )( )1.8cos30 2 0.9cos30CJT ° − °

( )( )1 2.4cos30 0.6cos30− ° + °

( ) ( )10 2.4cos30 1.8cos30 0.3 0− ° + ° + =

27.9 kNCJT = ..................................................Ans.

(b) Finally, dividing the force on pin E by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )3

22

2.8868 10 N/m2 2 4E

s

VA d

τπ

×= = .....................................................................................Ans.

Since the shear stress must be no greater than 120 MPa, the minimum diameter of the pin is

( )

( )3

36

2 2.8868 103.91 10 m 3.91 mm

120 10d

π−

×= = × =

×.................................................Ans.


187

6-63 The mechanism of Fig. P6-63 is designed to keep its load level while raising it. A pin on the rim of the 4-ft-diameter pulley fits in a slot on arm ABC. Arms ABC and DE are each 4 feet long and the package being lifted weighs 80 lb. The mechanism is raised by pulling on the rope that is wrapped around the pulley. Determine the force P applied to the rope and all forces acting on the arm ABC when the package has been lifted 4 feet, as shown.

SOLUTION

When 4 fth = ( )1sin 2 4 30θ −= = °

From a free-body diagram of the platform, the equilibrium equations give

0 :CMΣ =� ( )( ) ( )cos30 3 80 2 0DEF ° − =

61.584 lbDEF =

0 :xF→ Σ = cos30 0DE xF C° − =

0 :yF↑ Σ = sin 30 80 0y DEC F+ ° − =

53.333 lbxC = 49.208 lbyC =

Next, from the free-body diagram of member ABC, the equilibrium equations give the pin forces

0 :AMΣ =� ( ) ( )2 4sin 30 4cos30 0x yB C C− ° − ° =

138.564 lbB =

0 :xF→ Σ = sin 30 0x xC A B− − ° =

0 :yF↑ Σ = cos30 0y yB C A° − − =

15.949 lbxA = − 70.792 lbyA =

Finally, from the free-body diagram of the pulley, the moment equilibrium equation gives the rope force

0 :AMΣ =� 2 2 0P B− =

138.564 lb 138.6 lbP B= = ≅ ....................................................................................Ans. Note that the forces Ax and Ay represent the interaction between the arm ABC and the axle of the pulley while the forces FAx and FAy represent the interaction between the pulley support and the axle of the pulley. The forces acting on arm ABC are then

72.6 lb=A 77.30° ...................................................................................................Ans.

138.6 lb=B 60° .......................................................................................................Ans.

72.6 lb=C 42.70° ...................................................................................................Ans.

6-64 A drum of oil with a mass of 200 kg is supported by a frame (of which there are two) as shown in Fig. P6-64. Determine

(a) All forces acting on member ACE. (b) The elongation of the 20-mm diameter wire if it is made of a material with a modulus of elasticity of 200

GPa. SOLUTION (a) From a free-body diagram of the drum, the equilibrium equations give the forces

0 :xF→ Σ = 2 cos 45 2 cos 45 0D E° − ° =

0 :yF↑ Σ = ( )2 sin 45 2 sin 45 200 9.81 0D E° + ° − =

693.672 ND E= =


188

Next, from a free-body diagram of member BCD , the moment equilibrium equation is

0 :BMΣ =� ( ) ( ) ( )1 1 0.8 2 0y xC C D− + + =

Then, from a free-body diagram of member ACE , the equilibrium equations are

0 :AMΣ =� ( ) ( ) ( )1 1 0.8 2 0x yC C E+ − + =

0 :xF→ Σ = cos 45 0xT C E− + ° =

0 :yF↑ Σ = sin 45 0yA C E+ − ° =

Solving these last four equations gives

1535.937 NxC =

0 NyC =

1045.437 NT =

490.500 NA = Therefore, the forces acting on the member ACE are

491 N = ↑A ................................................................................................................... Ans.

1045 N = →T ................................................................................................................Ans.

1536 N = ←C ................................................................................................................Ans.

694 N=E 45° ...........................................................................................................Ans. (b) Finally, the stretch of the wire is given by

( )( )

( ) ( )29

1045.437 20000.0333 mm

200 10 0.02 4PLEA

δπ

= = = ×

...............................................Ans.

6-65 The mechanism shown in Fig. P6-65 is designed to keep its load level while raising it. A pin on the rim of the 4-ft-diameter pulley fits in a slot on arm ABC. Arms ABC and DE are each 4 ft long and the package being lifted weighs 80 lb. The mechanism is raised by pulling on the rope that is wrapped around the pulley.

(a) Plot P, the force required to hold the platform as a function of the platform height h (0 ≤ h ≤ 5.75 ft). (b) Plot A, C, and E, the magnitudes of the pin reaction forces at A, C, and E as a function of h (0 ≤ h ≤ 5.75

ft). SOLUTION

1 2sin4hθ − − =

(a) From a free-body diagram of the platform, the equilibrium equations give

0 :CMΣ =� ( )( ) ( )cos 3 80 2 0DEF θ − =

160 lb

3cosDEF θ=

0 :xF→ Σ = cos 0DE xF Cθ − =

0 :yF↑ Σ = sin 80 0y DEC F θ+ − =

53.333 lbxC = 160 tan80 lb

3yCθ = −


189

Next, from the free-body diagram of member ABC, the equilibrium equations give the pin forces

0 :AMΣ =� ( ) ( )2 4sin 4cos 0x yB C Cθ θ− − =

( )2 sin 2 cos lbx yB C Cθ θ= +

0 :xF→ Σ = sin 0x xC A B θ− − =

0 :yF↑ Σ = cos 0y yB C Aθ − − =

( )sin lbx xA C B θ= −

( )cos lby yA B Cθ= −

Finally, from the free-body diagram of the pulley, the moment equilibrium equation gives the rope force

0 :AMΣ =� 2 2 0P B− =

( )2 sin 2 cos lbx yP B C Cθ θ= = + ............................................................................Ans.

Note that the forces Ax and Ay represent the interaction between the arm ABC and the axle of the pulley while the forces FAx and FAy represent the interaction between the pulley support and the axle of the pulley.

(b) The magnitudes of the pin forces A, C, and E are

2 2 lbx yA A A= + ....................... Ans.

2 2 lbx yC C C= + ....................... Ans.

160 lb

3cosDEF θ= ........................ Ans.

6-66 Forces of P = 100 N are being applied to the handles of the vise grip pliers shown in Fig. P6-66. Plot the force applied on the bolt by the jaws as a function of the distance d (20 mm ≤ d ≤ 30 mm).

SOLUTION

( )1tan 40dφ −= 2 235 15 38.079 mma = + =

From a free-body diagram of the upper handle, the equilibrium equations give the forces

0 :BMΣ =� ( ) ( ) ( ) ( )( )100 93 sin 38 cos 35 0A AF F dφ φ− + − =

( )9300 N

38sin 35 cosAF dφ φ=

− −

0 :xF→ Σ = cos 0A xF Bφ − =

0 :yF↑ Σ = sin 100 0A yF Bφ − − =

cos Nx AB F φ= ( )sin 100 Ny AB F φ= −

Next, from the free-body diagram of the jaw BCD, the moment equilibrium equation is

0 :CMΣ =� ( ) ( )35 12 0D x yF a B B− − =


190

35 12

N38.079x y

D

B BF

+= ...................................................Ans.

6-67 A load P will be supported by a structure consisting of a rigid bar A, two aluminum alloy (E = 10,600 ksi) bars B, and a stainless steel (E = 28,000 ksi) bar C, as shown in Fig. P6-67. Each bar has a cross sectional area of 2.00 in.2 If the bars are unstressed before the load P is applied, determine the normal stresses in the bars after a 40-kip load is applied.

SOLUTION From a free-body diagram of bar A, the equilibrium equations give

0 :CMΣ =� 2 1 0B BT a T a− =

1 2B B BT T T= =

0 :yF↑ Σ = 31 2 40 10 0B B CT T T+ + − × =

4 2 40,000 lbB Cσ σ+ =

Since the force in each of the aluminum bars is the same, they will stretch the same amount and the steel bar must then stretch the same amount as the aluminum bars, B Cδ δ= , which in terms of the stresses can be written

( ) ( )

6 6

36 7210.6 10 28 10

B Cσ σ=

× × 1.32075C Bσ σ=

Combining the equilibrium equation and the deformation equation gives

6020 psiBσ = .................................................................................................................. Ans.

7950 psiCσ = .................................................................................................................. Ans.

6-68 The rigid bar CDE, shown in Fig. P6-68, is horizontal before the load P is applied. Tie rod A is a hot-rolled steel (E = 210 GPa) bar with a length of 450 mm and a cross-sectional area of 300 mm2. Post B is an oak timber (E = 12 GPa) with a length of 375 mm and a cross-sectional area of 4500 mm2. After the 225-kN load P is applied, determine

(a) The normal stresses in bar A and post B. (b) The vertical displacement of point D. SOLUTION (a) From a free-body diagram of the bar CDE, the moment

equilibrium equation gives

0 :CMΣ =� ( )3500 1350 1350 225 10 0AD BET F+ − × =

32.70 607.5 10 NAD BET F+ = ×


191

From similar triangles, the stretch of bar AD and the shrink of bar BE are related by

500 1350AD BEδ δ=

which gives

( )

( ) ( )( )

( )( )9 6 9 6

450 3751350 500

210 10 300 10 12 10 4500 10AD BET F

− −=

× × × ×

2.77714BE ADF T=


( )671,485.0 N (T) 300 10AD ADT σ −= = ×

( )6198,523.9 N (C) 4500 10BE BEF σ −= = ×

6 2238 10 N/m 238 MPa (T)ADσ = × = ......................................................................Ans.

6 244.1 10 N/m 44.1 MPa (C)BEσ = × = ....................................................................Ans.

(b) The vertical displacement of point D is the same as the stretch of the tie rod

( ) ( )

( )( )9 6

71,485.0 4500.511 mm

210 10 300 10ADδ−

= = ↓× ×

........................................................Ans.

6-69 A pin-connected structure is loaded and supported as shown in Fig. P6-69. Member CD is rigid and is horizontal before the load P is applied. Member A is an aluminum alloy bar with a modulus of elasticity of 10,600 ksi and a cross-sectional area of 2.25 in2. Member B is a stainless steel bar with a modulus of elasticity of 28,000 ksi and a cross-sectional area of 1.75 in2. After the load is applied to the structure, determine

(a) The normal stresses in bars A and B. (b) The vertical displacement of point D. SOLUTION (a) From a free-body diagram of the bar CD,

the moment equilibrium equation gives

0 :CMΣ =� ( )35.5 10 10 8 3 0A BF F× − − =

38 3 55 10 lbA BF F+ = ×

From similar triangles, the shrink of the two bars and the movement of point D are related by

8 3 10A B dδ δ= =

which gives

( )

( ) ( )( )

( )( )6 6

48 363 8

10.6 10 2.25 28 10 1.75A BF F

=× ×

1.02725B AF F=


( )4963.11 lb (C) 2.25A AF σ= =


192

( )5098.35 lb (C) 1.75B BF σ= =

32.21 10 psi 2.21 ksi (C)Aσ = × = ..............................................................................Ans.

32.91 10 psi 2.91 ksi (C)Bσ = × = ..............................................................................Ans.

(b) The vertical displacement of point D is then

( ) ( )

( )( )6

10 4963.11 4810 0.01249 in. 8 8 10.6 10 2.25Ad δ= = = ↓

×.................................................Ans.

6-70 Bar B of the pin-connected system of Fig. P6-70 is made of an aluminum alloy [Ea = 70 GPa, Aa = 300 mm2, and αa = 22.5(10-6)/°C] and bar A is made of a hardened carbon steel [Es = 210 GPa, As = 1200 mm2, and αs = 11.9(10-6)/°C]. Bar CDE is to be considered rigid. When the system is unloaded at 40°C, bars A and B are unstressed. After the load P is applied, the temperature of both bars decreases to 15°C. Determine

(a) The normal stresses in bars A and B. (b) The vertical displacement (deflection) of pin E. SOLUTION (a) From a free-body diagram of the bar CDE,


0 :CMΣ =� ( )3150 450 100 10 0AD BEF F+ − × =

33 300 10 NAD BEF F+ = ×

From similar triangles, the stretch of the two bars are related by

150 450AD BEδ δ=

which gives

( )( )( ) ( )( )( )

( )( ) ( ) ( )( ) ( )

69 6

69 6

250450 11.9 10 25 250

210 10 1200 10

500150 22.5 10 25 500

70 10 300 10

AD

BE

F

F

−−

−−

+ × −

× ×

= + × −

× ×

8 19,530 NAD BEF F= −


( )6212,885 N (T) 1200 10AD ADF σ −= = ×

( )629,048.2 N (T) 300 10BE BEF σ −= = ×

6 2177.4 10 N/m 177.4 MPa (T)ADσ = × = ...............................................................Ans.

6 296.8 10 N/m 96.8 MPa (T)BEσ = × = ....................................................................Ans.

(b) The vertical displacement of pin E is the same as the stretch of bar B

( ) ( )

( ) ( ) ( )( )( )69 6

29,048.2 50022.5 10 25 500

70 10 300 10BEδ −−

= + × −

× ×

0.410 mm BEδ = ↓ .........................................................................................................Ans.


193

6-71 A rigid bar CD is loaded and supported as shown in Fig. P6-71. Bars A and B are unstressed before the 30-kip load P is applied. Bar A is made of steel (E = 30,000 ksi) and has a cross-sectional area of 2 in2. Bar B is made of brass (E = 15,000 ksi) and has a cross-sectional area of 1.5 in2. Determine

(a) The stresses in bars A and B. (b) The vertical displacement (deflection) of pin C. (c) The shearing stress on a cross section of the ¾-in.-diameter pin at D, which is in double shear. SOLUTION (a) From a free-body diagram of the bar CD, the moment equilibrium equation gives

0 :DMΣ =� ( )310 30 10 2 6 0A BF F× − − =

33 150 10 lbA BF F+ = ×

From similar triangles, the stretch of the two bars and the movement of point C are related by

2 6 10A B cδ δ= =

which gives

( )

( )( )( )

( )( )6 6

8 156 2

30 10 2 15 10 1.5A BF F

=× ×

1.66667A BF F=


( )53,571 lb (T) 2A AF σ= =

( )32,143 lb (T) 1.5B BF σ= =

326.8 10 psi 26.8 ksi (T)Aσ = × = ..............................................................................Ans.

321.4 10 psi 21.4 ksi (T)Bσ = × = ..............................................................................Ans.

(b) The vertical displacement of point C is then

( )( )

( ) ( )6

10 53,571 810 0.0357 in. 2 2 30 10 2Ac δ= = = ↓

×.............................................................Ans.

(c) Returning to the free-body diagram of the bar CD, the force equilibrium equations give

0 :xF→Σ = 0xD =

0 :yF↑ Σ = 330 10 0A B yF F D+ − − × =

3 330 10 55.714 10 lby A BD F F= + − × = ×

2 2 355.714 10 lbx yD D D= + = ×

Dividing the pin force by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )

33

2

55.714 10 63.1 10 psi 63.1 ksi2 2 0.75 4D

Ds

VA

τπ

×= = = × =

.......................................Ans.


194

6-72 A rigid bar CD is loaded and supported as shown in Fig. P6-72. Bars A and B are unstressed before the 150-kN load P is applied. Bar A is made of stainless steel (E = 190 GPa) and has a cross-sectional area of 750 mm2. Bar B is made of an aluminum alloy (E = 73 GPa) and has a cross-sectional area of 1250 mm2. Determine

(a) The stresses in bars A and B. (b) The vertical displacement (deflection) of pin D. SOLUTION (a) From a free-body diagram of the bar CD,


0 :CMΣ =� ( )0.2 0.5 4 5B AF F+

( )30.6 150 10 0− × =

32 450 10 NB AF F+ = ×

From similar triangles, the stretch of bar B and the vertical movement of pins E and D are related by

0.2 0.5 0.6B e dδ = =

When pin E moves down a distance e, bar A stretches ( )4 5A eδ = . Therefore, 2A Bδ δ= and

( )

( )( )( )

( ) ( )9 6 9 6

1000 5002

190 10 750 10 73 10 1250 10A BF F

− −=

× × × ×

1.56164A BF F=


( )3 6170.432 10 N (T) 750 10A AF σ −= × = ×

( )3 6109.136 10 N (T) 1250 10B BF σ −= × = ×

6 2227 10 N/m 227 MPa (T)Aσ = × = .......................................................................Ans.

6 287.3 10 N/m 87.3 MPa (T)Bσ = × = ......................................................................Ans.

(b) The vertical displacement of pin D is then

( )( )

( )( )3

9 6

109.136 10 5003 3 1.794 mm

73 10 1250 10Bd δ−

×= = = ↓

× ×..............................................Ans.

6-73 A 40-kip load P will be supported by a structure consisting of a rigid bar A, two aluminum alloy [Ea = 10,600 ksi and αa = 12.5(10-6)/°F] bars B, and a stainless steel [Es = 28,000 ksi and αs = 9.6(10-6)/°F] bar C, as shown in Fig. P6-73. The bars are unstressed when the structure is assembled at 72°F. Each bar has a cross-sectional area of 2.00 in2. Determine the normal stresses in the bars after the 40-kip load is applied and the temperature is increased to 250°F.

SOLUTION From a free-body diagram of bar A, the equilibrium equations give

0 :CMΣ =� 2 1 0B BT a T a− =

1 2B B BT T T= =


195

0 :yF↑ Σ = 31 2 40 10 0B B CT T T+ + − × =

2 40,000 lbB CT T+ =

Since stresses are asked for, the equilibrium equation will be written in terms of stresses

4 2 40,000 psiB Cσ σ+ =

Since the force in each of the aluminum bars is the same (and the temperature change is the same for both bars), they will stretch the same amount and the steel bar must then stretch the same amount as the aluminum bars, B Cδ δ= , which in terms of the stresses can be written

( )

( ) ( )( ) ( ) ( )( ) ( )( )( )6 6

6 6

36 7212.5 10 168 36 9.6 10 168 72

10.6 10 28 10B Cσ σ− −+ × = + ×

× ×

31.32075 15.7584 10 psiC Bσ σ= − ×


310.77 10 psi 10.77 ksi (T)Bσ = × = ..........................................................................Ans.

31.536 10 psi 1.536 ksi (C)Cσ = − × = ......................................................................Ans.

6-74 A pin-connected structure is loaded and supported as shown in Fig. P6-74. Member CD is rigid and is horizontal before the 75-kN load P is applied. Bar A is made of structural steel (E = 200 GPa), and bar B is made of an aluminum alloy (E = 73 GPa). The cross-sectional areas of members A and B are 625 mm2 and 2570 mm2, respectively. Determine

(a) The vertical displacement of the pin used to apply the load. (b) The shearing stress on a cross section of the 25-mm diameter pin at C, which is in double shear. SOLUTION

( )1tan 3 4 36.870Aθ −= = ° 2 23 4 5 mAL = + =

( )1tan 3 2 56.310Bθ −= = ° 2 23 2 13 mBL = + =

(a) From a free-body diagram of the bar CD, the moment equilibrium equation gives

0 :CMΣ =� ( ) ( )4 sin 2 sinA A B BF Fθ θ+

( )35 75 10 0− × =

32.4 1.66410 375 10 NA BF F+ = ×

From similar triangles, the vertical movement of pins A, B, and D are related by

2 4 5b a d= =

When pin B moves down a distance b, bar B stretches sinB Bbδ θ= . Then, when pin A moves down a distance

2a b= , bar A will stretch

sin 2 sin

2 sin sin 1.44222A A A

B A B B

a bδ θ θδ θ θ δ

= == =


196

( )

( )( )( )

( )( )9 6 9 6

1351.44222

200 10 625 10 73 10 2570 10BAFF

− −=

× × × ×

1.44315B AF F=


378.0998 10 NAF = × 3112.7097 10 NBF = ×

The vertical displacement of pin D is then

( )( )( )( )

3

9 6

112.7097 10 132.52.5 2.5sin sin 56.310 73 10 2570 10

B

B

d b δθ −

× = = = ° × ×

0.00651 m 6.51 mm d = = ↓ ......................................................................................Ans. (b) Returning to the free-body diagram of the bar CD, the force equilibrium equations give

0 :xF→Σ = cos cos 0x A A B BC F Fθ θ− − =

0 :yF↑ Σ = 3sin sin 75 10 0A A B B yF F Cθ θ+ − − × =

3124.9997 10 NxC = × 365.6402 10 NyC = ×

2 2 3141.186 10 Nx yC C C= + = ×

Dividing the pin force by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )

36 2

2

141.186 10 143.8 10 N/m 143.8 MPa2 0.025 4

CC

s

VA

τπ

×= = = × =

..........................Ans.

6-75 Before the 20 kip load P is applied, the arms of the crank C shown in Fig. P6-75 are horizontal and vertical; there is a 0.009-in. gap between the horizontal arm and the brass (Eb = 15,000 ksi and Ab = 12 in2) post B; and the aluminum (Ea = 10,000 ksi and Aa = 2 in2) rod A is horizontal. If the crank C is rigid, determine

(a) The stresses in members A and B. (b) The change in length of members A and B. SOLUTION (a) From a free-body diagram of the crank C,


0 :CMΣ =� ( )35 20 10 10 6 0A BT F× − − =

35 3 50 10 lbA BT F+ = ×

From similar triangles,

10 6a b=

in which Aa δ= is the stretch of rod A, 0.009 in.Bb δ= + ,

and Bδ is the shrink of the post B . Therefore

( )5 3 5 3 0.015 in.A Bbδ δ= = +

( )

( )( )( )

( )( )6 6

50 155 0.015 in.310 10 2 15 10 12

A BT F = + × ×


197

( )18 108,000 lbB AF T= −


( )36.3390 10 lb 2A AT σ= × =

( )36.1017 10 lb 12B BF σ= × =

3170 psi (T)Aσ = ...........................................................................................................Ans.

508 psi (C)Bσ = .............................................................................................................Ans.

(b) The change in length of the two members is then

( )( )

( )( )

33

6

6.3390 10 5015.85 10 in. (stretch)

10 10 2Aδ −×

= = ××

.............................................Ans.

( )( )

( )( )

33

6

6.1017 10 150.508 10 in. (shrink)

15 10 12Bδ −×

= = ××

..............................................Ans.

6-76 The mechanism of Fig. P6-76 consists of a structural steel (E = 200 GPa) rod A with a cross-sectional area of 350 mm2, a cold-rolled brass (E = 100 GPa) rod B with a cross sectional area of 750 mm2, and a rigid bar C. The nuts at the top ends of rods A and B are initially tightened to the point where all slack is removed from the mechanism but the bars remain free of stress. If a nut advances 2.5 mm with each full turn (360°), determine

(a) The stresses in the rods when the nut at the top end of rod B rotates 180°. (b) The vertical displacement at the top end of rod A for part (a). SOLUTION (a) From a free-body diagram of the bar C, the moment equilibrium equation gives

0 :CMΣ =� 200 125 0A BT T− =

5 8B AT T=

From similar triangles, the vertical movement of points A and B are related by

200 125a b=

in which Aa δ= is the stretch of rod A,

1.25 mmBb δ= − , and Bδ is the stretch of rod B. Therefore

( )5 8 1.25 10 8A B Bδ δ δ= − = −

( )

( )( )( )

( )( )9 6 9 6

1500 15005 10 8

200 10 350 10 100 10 750 10A BT T

− −= −

× × × ×

( )393.33333 10 1.49333 NA BT T= × −


( )3 627.537 10 N (T) 350 10A AT σ −= × = ×

( )3 644.060 10 N (T) 750 10B BT σ −= × = ×


198

6 278.7 10 N/m 78.7 MPaAσ = × = ............................................................................Ans.

6 258.7 10 N/m 58.7 MPaBσ = × = ............................................................................Ans.

(b) The vertical displacement of the top of rod A is the same as the stretch of rod A

( ) ( )

( )( )3

9 6

27.537 10 15000.590 mm

200 10 350 10Aδ−

×= = ↑

× ×.........................................................Ans.

6-77 The pin connected structure shown in Fig. P6-77 consists of a cold-rolled bronze [Eb = 15,000 ksi and αb = 9.4(10-6)/°F] bar A which has a cross-sectional area of 3.00 in2 and two 0.2% C hardened steel [Es = 30,000 ksi and αs = 6.6(10-6)/°F] bars B which have cross-sectional areas of 2.50 in2. If the temperature of bar A decreases 50°F and the temperature of bars B increases 30°F after the 200-kip load is applied, determine

(a) The normal stresses in the bars. (b) The displacement of pin C. SOLUTION (a) From a free-body diagram of the pin, the vertical equilibrium equation gives

0 :yF↑Σ = ( ) 32 4 5 200 10 0B AT T+ − × =

31.6 200 10 lbB AT T+ = ×

The stretch of the bronze and the steel bars are related by cos 4 5B A Aδ δ θ δ= = . Therefore

( )

( )( ) ( )( )( ) ( )( )( ) ( )( ) ( )6 6

6 6

5 46.6 10 30 5 0.8 9.4 10 50 4

30 10 2.50 15 10 3B AT T− −

+ × = + × −

× ×

( )1.06667 37,410 lbB AT T= −


( )396.006 10 lb 3A AT σ= × =

( )364.996 10 lb 2.5B BT σ= × =

332.0 10 psi 32.0 ksi (T)Aσ = × = ..............................................................................Ans.

326.0 10 psi 26.0 ksi (T)Bσ = × = ..............................................................................Ans.

(b) The vertical displacement of pin C is the same as the stretch of bar A

( )( )

( )( ) ( )( )( )3

66

96.006 10 489.4 10 50 48 0.0798 in.

15 10 3Aδ −×

= + × − = ↓×

....................Ans.

6-78 Three bars are connected by smooth frictionless pins as shown in Fig. P6-78. Bar BCD is rigid, bar AB is aluminum (E = 73 GPa; L = 750 mm; d = 40 mm; σmax = 240 MPa), and bar DE is steel (E = 200 GPa; L = 500 mm; d = 30 mm; σmax = 400 MPa). The 50-mm-diameter pivot pin C is aluminum (τmax = 180 MPa) and is in double shear. After the unit is assembled, the nut D is slowly tightened.

(a) Plot σAB, σDE, and τC as functions of the distance that the nut advances (0 ≤ δnut ≤ 2 mm). (b) Plot δAB and δDE as function of δnut (0 ≤ δnut ≤ 2 mm). (c) What is the maximum amount δnut that the nut can be tightened? SOLUTION (a) From a free-body diagram of the bar BCD,

the equilibrium equations give


199

0 :xF→Σ = 0xC =

0 :yF↑Σ = 0y AB DEC T T− − =

y AB DEC T T= +

0 :CMΣ =� 200 125 0AB DET T− =

5 8DE ABT T=

From similar triangles, the vertical movement of points B and D are related by

200 125b d=

in which ABb δ= is the stretch of rod AB, mmnut DEd δ δ= − , DEδ is the stretch of rod DE, and nutδ is the movement of the nut along the rod. Therefore

( )5 8AB nut DEδ δ δ= −

( )

( ) ( )( )

( ) ( )2 29 9

750 5005 8 8

73 10 0.04 4 200 10 0.03 4AB DE

nut

T Tδ

π π= −

× ×

1.44478 282,743.34 NAB DE nutT T δ+ =


282,743.34 92,861.66 N

3.04478nut

AB nutT δ δ= =

1.6DE ABT T=

Dividing the axial forces by the cross-sectional areas gives the normal stresses

( )

22 N/m (T)

0.04 4AB

ABTσ

π= ......................................................................................Ans.

( )

22 N/m (T)

0.03 4DE

DETσ

π= ......................................................................................Ans.

Dividing the force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )

2 22

2N/m

2 0.05 4x yC

Cs

C CVA

τπ

+= =

...............................................................................Ans.

(b) The stretches of the two rods are

( )

( ) ( )29

750mm (stretch)

73 10 0.04 4AB

AB

Tδ

π=

×

.........................................................Ans.

( )

( ) ( )29

500mm (stretch)

200 10 0.03 4DE

DE

Tδ

π=

×

......................................................Ans.

(c) Checking the graph reveals that the normal stress in rod DE reaches its limiting value for 1.90 mmnutδ ≅ . The other stresses remain below their limiting values for the entire range graphed. Therefore,


200

( )max1.90 mmnutδ = .......................................................................................................Ans.

6-79 Three bars are connected by smooth frictionless pins as shown in Fig. P6-79. Bar ABCD is rigid, bar AE is aluminum (E = 10,600 ksi; L = 24 in.; d = ½ in.; σmax = 40 ksi), and bar CF is steel (E = 30,000 ksi; L = 18 in.; d = ¾ in.; σmax = 50 ksi). The ¾-in.-diameter pivot pin B is steel (τmax = 25 ksi) and is in single shear. The holes in bar CF are slightly overdrilled so that pin C moves down 0.06 in. before contact is made with bar CF.

(a) Plot σAE, σCF, and τB as functions of P (0 ≤ P ≤ 10 kip). (b) Plot δAE and δCF as functions of P (0 ≤ P ≤ 10 kip). (c) What is the maximum force P that the system can withstand? SOLUTION (a) From a free-body diagram of the bar ABCD,


0 :xF→Σ = 0xB =

0 :yF↑Σ = 0y AE CFB T F P− + − =

y AE CFB P T F= + −

0 :BMΣ =� 12 9 24 0AE CFT F P+ − =

4 3 8AE CFT F P+ =

From similar triangles, the vertical movement of points A, C, and D are related by

12 9 24a c d= =

in which AEa δ= is the stretch of rod AE, 0.06 in.CFd δ= + , and CFδ is the shrink of rod CF. Therefore

3 4 0.24 in.AE CFδ δ= +

( )

( ) ( )( )

( ) ( )2 26 6

24 183 4 0.24 in.

10.6 10 0.5 4 30 10 0.75 4AE CFT F

π π= +

× ×

6.36792 44,178.65 lbAE CFT F− =


132,536 8 lb

23.10377AEPT +=


201

8 4

3AE

CFP TF −=


( )2 psi (T)0.5 4AE

AETσ

π= .............................................................................................Ans.

( )2 psi (C)0.75 4

CFCF

Fσπ

= ..........................................................................................Ans.

Dividing the force on the pin by its cross-sectional area gives the shear stress

( )

2 2

2 psi0.75 4x yB

Bs

B BVA

τπ

+= = ...........................................................................................Ans.


( )

( ) ( )26

24in. (stretch)

10.6 10 0.5 4AE

AE

Tδ

π=

×

...........................................................Ans.

( )

( ) ( )26

18in. (shrink)

30 10 0.75 4CF

CF

Fδ

π=

×

............................................................Ans.

(c) Checking the graph reveals that the normal stress in rod AE reaches its limiting value for 6.2 kipP ≅ . The other stresses remain below their limiting values for the entire range graphed. Therefore,

max 6.2 kipP = ................................................................................................................... Ans.

6-80 Member ABCD of the pin-connected structure shown in Fig. P6-80 is rigid; bar BF is made of steel [EBF = 210 GPa; ABF = 1200 mm2; αBF = 11.9(10-6)/°C]; and bar CE is made of an aluminum alloy [ECE = 73 GPa; ACE = 900 mm2; αCE = 22.5(10-6)/°C]. As a result of a misalignment of the pin holes at A, B, and C, bar CE must be heated 80°C (after pins A and B are in place) to permit insertion of pin C. Compute and plot

(a) The axial stresses σBF (in the steel bar) and σCE (in the aluminum bar) as functions of the temperature decrease (as bar CE cools back down to room temperature) ∆T (0°C ≥ ∆T ≥ −80°C).

(b) The elongations δBF (of the steel bar) and δCE (of the aluminum bar) as functions of the temperature decrease ∆T (0°C ≥ ∆T ≥ −80°C).

SOLUTION When bar CE is heated 80°C, it stretches


202

( )( ) ( )622.5 10 80 600 1.08000 mminitδ −= × =

(a) From a free-body diagram of the bar ABCD, the moment equilibrium equation gives

0 :AMΣ =� 240 80 0CE BFT T− =

3BF CET T=

From similar triangles, the stretch of bar BF and the movement of pin C are related by

80 240BF cδ =

in which ( )1.08000 mmCEc δ= − and CEδ is the stretch of bar CE. Therefore

( )3 1.0800 mmBF CEδ δ= −

( )( ) ( )

( )( )( ) ( )( )( )

9 6

69 6

10003 1.08000

210 10 1200 10

600 22.5 10 80 600

73 10 900 10

BF

CE

T

TT

−

−−

=

× ×

− + × + ∆

× ×

1.30357 1478.250 NCE BFT T T+ = − ∆

in which 80 C 0 CT− ° ≤ ∆ ≤ ° . Combining the equilibrium equation and the deformation equation gives

1478.250 301.026 N

4.91071CETT T− ∆= = − ∆

3BF CET T=


26 N/m MPa (T)

1200 10 1200BF BF

BFT Tσ −= =

×..............................................................Ans.

26 N/m MPa (T)

900 10 900CE CE

CET Tσ −= =×

..................................................................Ans.


( )

( )( )9 6

1000mm (stretch)

210 10 1200 10BF

BF

Tδ

−=

× ×.........................................................Ans.

( )

( )( )9 6

60073 10 900 10

CECE

Tδ

−

=

× ×

( )( ) ( )622.5 10 80 600 mm (stretch)T− + × + ∆ ............................................Ans.


203

6-81 Three bars are connected by smooth frictionless pins as shown in Fig. P6-81. Bar BCD is rigid, bar AC is aluminum [E = 12,000 ksi; L = 36 in.; d = 1 in.; σmax = 20 ksi; α = 12.5(10-6)/°F], and bar DE is steel [E = 30,000 ksi; L = 30 in.; d = ½ in.; σmax = 10 ksi; α = 6.6(10-6)/°F]. The 5/8-in.-diameter pivot pin B is steel (τmax = 5 ksi) and is in single shear. If the temperature drops after the unit is assembled,

(a) Plot σAC, σDE, and τB as functions of the temperature drop ∆T (0 ≥ ∆T ≥ −60°F). (b) Plot δAC and δDE as functions of ∆T (0 ≥ ∆T ≥ −60°F). (c) What is the maximum temperature drop that the system can withstand? SOLUTION (a) From a free-body diagram of the bar BCD,


0 :xF→Σ = 0xB =

0 :yF↑Σ = 0y AC DEB T T+ − =

y DE ACB T T= −

0 :BMΣ =� 20 30 0AC DET T− =

1.5AC DET T=

From similar triangles, the shrink of bar AC and the stretch of bar DE are related by

20 30AC DEδ δ=

3 2AC DEδ δ=

( )( ) ( )

( )( ) ( )

( )( ) ( )

( )( ) ( )

626

626

363 12.5 10 36

12 10 1 4

30 2 6.6 10 30

30 10 0.5 4

AC

DE

TT

TT

π

π

−

−

− + × ∆ ×

= + × ∆ ×

1.12500 171.41315 lbDE ACT T T+ = − ∆

in which 60 F 0 FT− ° ≤ ∆ ≤ ° . Combining the equilibrium equation and the deformation equation gives

63.78164 lbDET T= − ∆ 1.5AC DET T=


204


( )2 psi (T)1 4AC

ACTσ

π= .................................................................................................Ans.

( )2 psi (T)0.5 4DE

DETσ

π= .............................................................................................Ans.

Dividing the force on the pin by its cross-sectional area gives the shear stress

( )

2 2

2 psi5 8 4x yB

Bs

B BVA

τπ

+= = .............................................................................................Ans.

(b) The shrink of bar AC and the stretch of bar DE are

( )

( ) ( )( )( ) ( )6

26

3612.5 10 36 in. (shrink)

12 10 1 4AC

ACT

Tδπ

− = − + × ∆ ×

.............Ans.

( )

( ) ( )( ) ( )( )6

26

306.6 10 30 in. (stretch)

30 10 0.5 4DE

DET

Tδπ

− = + × ∆ ×

.............Ans.

(c) Checking the graph reveals that the normal stress in rod DE reaches its limiting value for 31 FT∆ ≅ − ° . The other stresses remain below their limiting values for the entire range graphed. Therefore,

max 31 FT∆ = − ° ................................................................................................................Ans.

6-82 Use the method of joints and determine the force in each member of the truss shown in Fig. P6-82, if a = 2 m and P = 5 kN.

SOLUTION 2 tan 30 1.15470 mb = ° =

tan 30 0.66667 md b= ° = From a free-body diagram of the entire truss, the equilibrium equations give

0 :xF→Σ = 0xA =

0 :AMΣ =� ( )2.66667 2 5 0C − =

0 :yF↑Σ = 5 0yA C+ − =


205

3.7500 kNC =

1.2500 kNyA =

From a free-body diagram for joint A, the equilibrium equations give

0 :yF↑Σ = 1.25 sin 30 0ABT+ ° =

0 :xF→Σ = cos30 0AD ABT T+ ° =

2.500 kN 2.50 kN (C)ABT = − = ...............................Ans.

2.1651 kN 2.17 kN (T)ADT = ≅ ...............................Ans.

From a free-body diagram for joint D, the equilibrium equations give

0 :xF→Σ = 0CD ADT T− =

0 :yF↑Σ = 5 0BDT − =

2.1651 kN 2.17 kN (T)CDT = ≅ .............................................. Ans.

5.000 kN 5.00 kN (T)BDT = = ................................................ Ans.

Finally, from a free-body diagram for joint C, the equilibrium equations give

0 :yF↑Σ = 3.75 sin 60 0BCT+ ° =

4.3301 kN 4.33 kN (C)BCT = − = ........................................... Ans.

6-83 A 4000-lb crate is attached by light, inextensible cables to the truss of Fig. P6-83. Determine the force in each member of the truss using the method of joints.

SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give

0 :AMΣ =� ( ) ( )6 4 2000 8 2000 0E − − =

0 :xF→Σ = 0xE A− =

0 :yF↑Σ = 4000 0yA − =

4000 lbE =

4000 lbxA =

4000 lbyA =


0 :xF→Σ = ( )4 5 4000 0ABT − =

0 :yF↑Σ = ( )4000 3 5 0AE ABT T− − =

5000 lb (T)ABT = ................................................................. Ans.

1000 lb (T)AET = ................................................................. Ans.

From a free-body diagram for joint E, the equilibrium equations give

0 :xF→Σ = ( )4 5 4000 0DE BET T+ + =

0 :yF↑Σ = ( )3 5 0AE BET T+ =

1666.67 lb 1667 lb (C)BET = − ≅ .........................................Ans.


206

2666.67 lb 2670 lb (C)DET = − ≅ ........................................Ans.


0 :xF→Σ = 0CD DET T− =

0 :yF↑Σ = 2000 0BDT − =

2666.67 lb 2670 lb (C)CDT = − ≅ ........................................... Ans.

2000 lb (T)BDT = ....................................................................... Ans.


0 :yF↑Σ = ( )3 5 2000 0BCT − =

3333.33 lb 3330 lb (T)BCT = = .............................................. Ans.

6-84 Use the method of joints and determine the force in each member of the truss shown in Fig. P6-84. All members are 3 m long.


0 :xF→Σ = 0xA =

0 :FMΣ =� ( )3 3 3 0yA − = 3 kNyA =

0 :yF↑Σ = 5 3 0yF A− − − = 11 kNF =


0 :yF↑Σ = 3 sin 60 0DET− − ° =

0 :xF→Σ = cos 60 0CD DET T− − ° =

3.46410 kN 3.46 kN (C)DET = − ≅ ........................................ Ans.

1.73205 kN 1.732 kN (T)CDT = ≅ .......................................... Ans.

From a free-body diagram for joint C, the equilibrium equations give

0 :xF→Σ = cos 60 cos 60 0CD CE BCT T T+ ° − ° =

0 :yF↑Σ = sin 60 sin 60 5 0CE BCT T− ° − ° − =

1.15470 kN 1.155 kN (C)BCT = − ≅ ...................................... Ans.

4.61880 kN 4.62 kN (C)CET = − = ........................................ Ans.


0 :yF↑Σ = sin 60 3 0ABT ° − =

0 :xF→Σ = cos60 0AF ABT T+ ° =

3.46410 kN 3.46 kN (T)ABT = ≅ ........................................... Ans.

1.73205 kN 1.732 kN (C)AFT = − ≅ ...................................... Ans.

From a free-body diagram for joint B, the equilibrium equations give

0 :yF↑Σ = sin 60 sin 60 sin 60 0BC BF ABT T T° − ° − ° =


207

4.61880 kN 4.62 kN (C)BFT = − ≅ ........................................ Ans.

0 :xF→Σ = cos 60 cos 60 cos 60 0BE BC BF ABT T T T+ ° + ° − ° =

4.61880 kN 4.62 kN (T)BET = ≅ .................................... Ans.

Finally, from a free-body diagram for joint F, the equilibrium equations give

0 :yF↑Σ = 11 sin 60 sin 60 0BF EFT T+ ° + ° =

8.08291 kN 8.08 kN (C)EFT = − = .................................. Ans.

6-85 Each truss member in Fig. P6-85 is 5 ft long. Find the forces in members CD and EF using the method of sections.


0 :xF→Σ = 0xA =

0 :BMΣ =� ( ) ( ) ( )800 5 600 10 800 15+ +

( )1000 20 25 0yA+ − =

1680 lbyA =

Cut a section through CD, DE, and EF, and draw a free-body diagram of the left-hand side of the truss. The height of the truss is

5sin 60 4.33013 fth = ° = and the equilibrium equations give

0 :EMΣ =� ( ) ( ) ( )1000 5 1680 10 4.33013 0CDT− − =

2725.09 lb 2730 lb (C)CDT = − ≅ ......................Ans.

0 :DMΣ =� ( ) ( ) ( ) ( )800 2.5 1000 7.5 1680 12.5 4.33013 0EFT+ − + =

2655.81 lb 2660 lb (T)EFT = ≅ ..................................................................................Ans.

6-86 Find the forces in members CJ and KJ of the roof truss shown in Fig. P6-86 using the method of sections. SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give

0 :xF→Σ = 0xA =

0 :GMΣ =� ( ) ( ) ( ) ( ) ( )4 4 3 8 5 12 4 16 3 20 24 0yA+ + + + − = 9.33333 kNyA =

Cut a section through CD, CJ, and JK, and draw a free-body diagram of the left-hand side of the truss.


208

( )1tan 10 12 39.806CDθ −= = °

( )1tan 4 59.036CJ hθ −= = °

( )( )2 3 10 6.66667 mh = =

Then, the equilibrium equations give

0 :AMΣ =� ( ) ( ) ( )( )3 4 4 8 sin 12 0CJ CJT θ− − − =

4.28 kN 4.28 kN (C)CJT = − = ....................................................................................Ans.

0 :CMΣ =� ( ) ( ) ( )3 4 9.33333 8 6.66667 0JKT− + =

9.40 kN (T)JKT = ...........................................................................................................Ans.

6-87 Find the forces in members BC, BF, and AF, of the stairs truss of Fig. P6-87 using the method of sections. SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give

0 :xF→Σ = 0xA =

0 :DMΣ =� ( ) ( ) ( )200 4 400 8 600 12 12 0yA+ + − =

933.333 lbyA =

Cut a section through AF, BF, and BC, and draw a free-body diagram of the upper-portion of the truss. The equilibrium equations give

0 :BMΣ =� ( ) ( ) ( ) ( )600 4 933.333 4 3 5 4 0AFT− − =

555.556 lb 556 lb (T)AFT = ≅ .......................................Ans.

0 :FMΣ =� ( ) ( ) ( ) ( )600 4 933.333 4 4 5 3 0BCT− − =

555.556 lb 556 lb (C)BCT = − ≅ ....................................Ans.

0 :yF↑Σ = ( ) ( )933.333 600 3 5 3 5 0AF BC BFT T T− − − − =

333.333 lb 333 lb (T)BFT = ≅ .......................................Ans.

6-88 The Gambrel truss shown in Fig. P6-88 supports one side of a bridge; an identical truss supports the other side. Floor beams carry vehicle loads to the truss joints. Calculate the forces in members BC, BG, and DE when a truck having a mass of 3500 kg is stopped in the middle of the bridge as shown. The center of gravity of the truck is 1 m in front of the rear wheels.

SOLUTION

( )3500 9.81 34,335 NW = =

From a free-body diagram of the entire bridge (which is consists of two trusses), the equilibrium equations give

0 :xF→Σ = 2 0xA =

0 :EMΣ =� ( ) ( ) ( )34,335 4 2 10 0yA− =

0 :AMΣ =� ( ) ( ) ( )( )2 10 34,335 6 0E − =

6867.0 NyA = 10,300.5 kNE =


209

Half of the truck�s weight is supported by each of the two trusses. From a free-body diagram of the truck, the equilibrium equations give

0 :RMΣ =� ( ) ( ) ( )34,335 1 2 4 0FN− =

0 :FMΣ =� ( ) ( ) ( )2 4 34,335 3 0RN − =

4291.88 NFN =

12,875.6 NRN =

From a free-body diagram of the floor panel between joints H and G, the equilibrium equations give

0 :GMΣ =� ( )2 3 0FN H− =

2861.25 NH = Next, cut a section through BC, BG, and GH, and draw a free-body diagram of the left-hand side of the truss.

( )1tan 2 3 33.690θ −= = °

( )1tan 3 3 45φ −= = °


0 :GMΣ =� ( ) ( ) ( ) ( )2861.25 3 6867 5 cos 5 0BCT θ− − =

6190 N 6.19 kN (C)BCT = − = ......................................Ans.

0 :yF↑Σ = 6867 2861.25 sin sin 0BC BGT Tθ φ− + − =

809.30 N 0.809 kN (T)BGT = ≅ ....................................... Ans.

Finally, from a free-body diagram for joint E, the equilibrium equations give

0 :yF↑Σ = 10,300.5 cos 0DET θ+ =

12,380 N 12.38 kN (C)DET = − = .......................................Ans.

6-89 A truss is loaded and supported as shown in Fig. P6-89. Determine (a) The normal stress in member CD if it has a diameter of ½ in. (b) The change in length of member CF if it has a diameter of ½ in. and a modulus of elasticity of 29(106)

psi. (c) The change in length of member EF if it has the same diameter and modulus of elasticity as member CF. SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give

0 :AMΣ =� ( ) ( ) ( )18 1000 6 2000 12F − −

( ) ( )2000 18 1000 24 0− − =

5000 lbF = Cut a section through CD, CF, and FG, and draw a free-body diagram of the right-hand side of the truss. The equilibrium equations give


210

0 :FMΣ =� ( ) ( )8 1000 6 0CDT − =

( )2750 lb 0.5 4CD CDT σ π= =

(a) 3820 psi (T)CDσ = ...........................................................Ans.

0 :yF↑Σ = ( )4 5 5000 3000 0CFT + − =

2500 lbCFT = −

(b) ( ) ( )

( ) ( )26

2500 1200.0527 in.

29 10 0.5 4CF

PLEA

δπ

−= = = −

×

...............................................Ans.

Finally, from a free-body diagram for joint E, the equilibrium equations give

0 :yF↑Σ = ( )1000 4 5 0EFT− − =

1250 lbEFT = −

(c) ( )( )

( ) ( )26

1250 1200.0263 in.

29 10 0.5 4EF

PLEA

δπ

−= = = −

×

...............................................Ans.

6-90 A transmission line truss supports a 5-kN load, as shown in Fig. P6-90. All members of the truss are made of structural steel (see Appendix A for properties). Determine the change in length of members FG and CD if they have cross-sectional areas of 300 mm2 each.

SOLUTION Cut a section through CD, DG, and FG, and draw a free-body diagram of the upper-portion of the truss. The equilibrium equations give

0 :GMΣ =� ( )( ) ( )( ) ( )5cos30 9 5sin 30 4 3 0CDT° − ° + =

9.65705 kNCDT = −

0 :DMΣ =� ( )( ) ( )5cos30 6 3 0FGT° − =

8.66025 kNFGT =

( ) ( )

( ) ( )9 6

9657.05 40000.644 mm

200 10 300 10CDPLEA

δ−

−= = = −

× ×..............................................Ans.

( ) ( )

( )( )9 6

8660.25 40000.577 mm

200 10 300 10FGPLEA

δ−

= = = +× ×

..............................................Ans.

6-91 The truss shown in Fig. P6-91 supports a sign that weighs 3000 lb. The sign is connected to the truss at joints E, G, and H, and the connecting links are adjusted so that each joint carries one-third of the load. All members of the truss are made of structural steel, and each has a cross-sectional area of 0.564 in2. Determine the axial stresses and strains in members CD, CF, CG, and FG of the truss.


0 :AMΣ =� ( ) ( ) ( )48 1000 8 1000 24BT − −

( )1000 40 0− =


211

1500 lbBT =

Cut a section through CD, CF, and FG, and draw a free-body diagram of the right-hand side of the truss. The equilibrium equations give

0 :FMΣ =� ( ) ( ) ( )6 1500 16 1000 8 0CDT + − =

2666.667 lbCDT = −

0 :CMΣ =� ( ) ( ) ( )1500 24 1000 16 6 0FGT− − =

3333.333 lbFGT =

0 :yF↑Σ = ( )1500 1000 3 5 0CFT− + = 833.333 lbCFT = −

Next, from a free-body diagram for joint G, the equilibrium equations give

0 :yF↑Σ = 1000 0CGT − = 1000 lbCGT =

Finally, calculating the stresses and strains

2666.667 4728 psi 4.73 ksi (C)0.564

CDCD

TA

σ −= = = − ≅ ............................................Ans.

66

4728 163.0 10 163.0 in./in.29 10CD E

σε µ−−= = = − × = −×

..........................................Ans.

833.333 1478 psi 1.478 ksi (C)0.564CFσ −= = − ≅ ........................................................Ans.

66

1478 50.9 10 50.9 in./in.29 10CFε µ−−= = − × = −

×........................................................Ans.

1000 1773 psi 1.773 ksi (T)

0.564CGσ += = + ≅ ...............................................................Ans.

66

1773 61.1 10 61.1 in./in.29 10CFε µ−+= = + × = +

×........................................................Ans.

3333.333 5910 psi 5.91 ksi (T)0.564FGσ += = + ≅ ........................................................Ans.

66

5910 204 10 204 in./in.29 10CFε µ−+= = + × = +

×..........................................................Ans.

6-92 All members of the truss shown in Fig. P6-92 are made of structural steel (E = 200 GPa) and are 25-mm in diameter. Determine the normal stresses in and the change in length of members CD, DI, and HI.


0 :xF→Σ = 0xA =

0 :GMΣ =� ( ) ( ) ( )10 6 18 12 12 0yA+ − =

23 kNyA =

Cut a section through CD, DI, and HI, and draw a free-body diagram of the right-hand side of the truss. The equilibrium equations give


212

0 :IMΣ =� ( ) ( ) ( )18 3 23 3 3 0CDT− − =

5 kNCDT = −

0 :DMΣ =� ( ) ( ) ( )18 6 3 23 6 0HIT+ − =

10 kNHIT =

0 :yF↑Σ = 23 18 sin 45 0DIT− + ° =

7.07107 kNDIT = −

Finally, calculating the stresses and changes in length

( )

6 22

5000 10.19 10 N/m 10.19 MPa (C)0.025 4

CDσπ

−= = − × = ..............................Ans.

( )

6 22

7071.07 14.41 10 N/m 14.41 MPa (C)0.025 4

DIσπ

−= = − × = ................................Ans.

( )

6 22

10,000 20.4 10 N/m 20.4 MPa (T)0.025 4

HIσπ

+= = + × = ...................................Ans.

( ) ( )

( ) ( )29

5000 60000.306 mm

200 10 0.025 4CD

PLEA

δπ

−= = = −

×

........................................Ans.

( )( )

( ) ( )29

7071.07 3000 20.306 mm

200 10 0.025 4DIδ

π

−= = −

×

....................................................Ans.

( ) ( )

( ) ( )29

10,000 60000.611 mm

200 10 0.025 4HIδ

π= = +

×

....................................................Ans.

6-93 Determine the force in member BD of the truss shown in Fig. P6-93. SOLUTION

( )130 tan 12 16 66.870θ −= ° + = °

10 11.54701 ft

cos30b = =

° 1 12 sinsin 6.870b

bθφ − −= = °

Cut a section through members BD and CD, and draw a free-body diagram of the left-hand side of the truss. The moment equilibrium equation gives


213

0 :AMΣ =� ( ) ( ) ( ) ( ) ( )sin cos 2000 cos cos sin 0BD BDT b b T bφ θ θ φ θ− − =

907 lb 907 lb (C)BDT = − = ..........................................................................................Ans.

6-94 For the simple truss of Fig. P6-94 show that overall equilibrium of the truss is a consequence of equilibrium of all of the pins; hence the equations of overall equilibrium give no new information. (Hint: Write equations of equilibrium for each of the pins and eliminate the unknown member forces from these equations.)


0 :xF→Σ = 0xP A− = (a)

0 :yF↑Σ = 0yC A− = (b)

0 :AMΣ =� 0Ca Pb− = (c)

xA P= yA C Pb a= =

From a free-body diagram of joint A, the equilibrium equations give

0 :xF→Σ = cos60 0AC AB xT T A+ ° − = (d)

0 :yF↑Σ = sin 60 0AB yT A° − = (e)

From a free-body diagram of joint B, the equilibrium equations give

0 :xF→Σ = cos30 cos60 0BC ABP T T+ ° − ° = (f)

0 :yF↑Σ = sin 30 sin 60 0BC ABT T− ° − ° = (g)

From a free-body diagram of joint B, the equilibrium equations give

0 :xF→Σ = cos30 0BC ACT T− ° − = (h)

0 :yF↑Σ = sin 30 0BCC T+ ° = (i)

Adding Eqs. (d), (f), and (h) gives Eq. (a)

0xP A− = ..................................................................................................................(a)

Adding Eqs. (e), (g), and (i) gives Eq. (b)

0yC A− = ..................................................................................................................(b)

Then, Eqs. (e) and (i) can be rewritten

0sin 60 sin 60

yAB

A CT = = =° °

sin 30BCCT = −

°

Equation (f) can be rewritten

cos30 cos60cos30 cos60sin 30 sin 60 sin 30 sin 60

sin 60 cos30 sin 30 cos 60 sin 90sin 30 sin 60 sin 30 sin 60

C CP C

C C

− ° ° = − ° + ° = + ° ° ° °

° ° + ° ° °= =° ° ° °

But sin 90 1° = , sin 60 b d° = , sin 30 d a° = , sin 30 sin 60 b a° ° = , and therefore, Eq. (f) can be rewritten


214

CPb a

= ................................................................................................................... (c)

which is equivalent to Eq. (c).

6-95 Show that the overall equilibrium of a truss is a consequence of the equilibrium of the two separate parts generated by the method of sections. That is, section the bridge truss of Fig. P6-95 as indicated, and write the equilibrium equations for each piece. Eliminate the member forces from the resulting six equations and show that the result is equivalent to the equilibrium of the whole truss.


0 :xF→Σ = 0xA = (a)

0 :yF↑Σ = 0yA P D− + = (b)

0 :AMΣ =� ( ) ( )3 2 0D w P w− = (c)

Cut a section through BC, CF, and EF, and draw a free-body diagram of the left-hand side of the truss. The equilibrium equations give

0 :xF→Σ = cos 0x BC CF EFA T T Tθ+ + + = (d)

0 :yF↑Σ = sin 0CF yT Aθ + = (e)

0 :CMΣ =� ( ) ( ) ( )2 0EF x yT h A h A w+ − = (f)

0 :FMΣ =� ( ) ( ) 0BC yT h A w− − = (g)

Cut a section through BC, CF, and EF, and draw a free-body diagram of the right-hand side of the truss. The equilibrium equations give

0 :xF→Σ = cos 0BC CF EFT T Tθ− − − = (h)

0 :yF↑Σ = sin 0CFT P Dθ− − + = (i)

0 :CMΣ =� ( ) ( ) 0EFD w T h− = (j)

0 :FMΣ =� ( ) ( ) ( )2 0BCD w T h P w+ − = (k)

Adding Eqs. (d) and (h) gives Eq. (a)

0xA = ................................................................................................................................ (a)

Adding Eqs. (e) and (i) gives Eq. (b)

0yA P D− + = ................................................................................................................. (b)

Adding two times Eq. (g) to two times Eq. (k) gives

4 2 2 0yDw Pw A w− − =

Adding Eq. (f) to Eq. (j) gives

2 0x yA h A w Dw− + =

Subtracting this last equation from the previous (and using 0xA = ) gives Eq. (c)

3 2 0Dw Pw− = ................................................................................................................ (c)


215

6-96 The flat roof of a building is supported by a series of parallel plane trusses spaced 2 m apart (only one such truss is shown in Fig. P6-96). Calculate the forces in all the members of a typical truss when water collects to a depth of 0.2 m as shown. The density of water is 1000 kg/m3.

SOLUTION From a free-body diagram of a roof panel, the equilibrium equations give the force carried by the joints at each end of the roof panel as

( ) ( ) ( )1 2 2 1000 9.81 2 1.2 0.2 2

2354.4 NF F W= = = × ×

=

Then, from a free-body diagram of the entire truss, the equilibrium equations give

0 :xF→Σ = 0xA =

0 :yF↑Σ = 16 0y yA F F+ − =

0 :AMΣ =� ( ) ( ) ( ) ( )( )1 13.6 2 1.2 2 2.4yF F F− −

( ) ( )1 3.6 0F− =

13 7063.20 Ny yA F F= = =


0 :xF→Σ = 0BCT =

0 :yF↑Σ = 1 0ABT F− − =

2354.4 N 2.35 kN (C)ABT = − ≅ ............................................. Ans.

0 kNBCT = .................................................................................... Ans.


0 :xF→Σ = ( )3 5 0AH ACT T+ =

0 :yF↑Σ = ( )7063.2 4 5 0AB ACT T+ + =

5886.0 N 5.89 kN (C)ACT = − ≅ ............................................. Ans.

3531.6 N 3.53 kN (T)AHT = ≅ ................................................ Ans.

From a free-body diagram for joint H, the equilibrium equations give

0 :xF→Σ = 0GH AHT T− =

0 :yF↑Σ = 0CHT =

0 kNCHT = ................................................................................... Ans.

3531.6 N 3.53 kN (T)GHT = ≅ ................................................ Ans.


0 :xF→Σ = 0DET =

0 :yF↑Σ = 1 0EFT F− − =

0 kNDET = .................................................................................... Ans.

2354.4 N 2.35 kN (C)EFT = − ≅ ............................................. Ans.

From a free-body diagram for joint F, the equilibrium equations give


216

0 :xF→Σ = ( )3 5 0FG DFT T− − =

0 :yF↑Σ = ( )7063.2 4 5 0EF DFT T+ + =

5886.0 N 5.89 kN (C)DFT = − ≅ ............................................. Ans.

3531.6 N 3.53 kN (T)FGT = ≅ ................................................ Ans.


0 :xF→Σ = ( )3 5 0DE CD DFT T T− + =

0 :yF↑Σ = ( )12 4 5 0DG DFF T T− − − =

3531.6 N 3.53 kN (C)CDT = − ≅ ...................................... Ans.

0 kNDGT = ............................................................................ Ans.

Finally, from a free-body diagram for joint G, the equilibrium equations give

0 :yF↑Σ = ( )4 5 0DG CGT T+ =

0 kNCGT = ............................................................................. Ans.

6-97 Snow on a roof supported by the Howe truss of Fig. P6-97 can be approximated as a distributed load of 20 lb/ft (measured along the roof). Treat the distributed load as you would the weight of the members; that is, replace the total load on each of the upper members as a vertical force, half applied to the joint at each end of the member. Determine the forces in members BC, BG, CG, and GH.


( )( )1 2 2 20 80 2 89.4427 lbF F W= = = =


0 :xF→Σ = 0xA =

0 :yF↑Σ = 18 0yA E F+ − =

0 :AMΣ =� ( ) ( )( ) ( )( )1 132 2 8 2 16E F F− −

( )( ) ( ) ( )1 12 24 32 0F F− − =

13 357.771 lbyA E F= = =

Next, cut a section through members BC, BG, and GH, and draw a free-body diagram of the left-hand portion of the truss.

1tan 8 16 26.565φ −= = °

Then equilibrium equations give

0 :AMΣ =� ( )( ) ( )( )12 8 sin 16 0BGF T φ− − =

200 lb 200 lb (C)BGT = − = ..........................................................................................Ans.

0 :BMΣ =� ( ) ( ) ( )( )14 8 357.771 8 0GHT F+ − =

536.656 lb 537 lb (T)GHT = ≅ ....................................................................................Ans.


217

0 :GMΣ =� ( )( ) ( ) ( ) ( ) ( ) ( )1 12 8 16 357.771 16 sin 16 0BCF F T φ+ − − =

400 lb 400 lb (C)BCT = − = ..........................................................................................Ans.


0 :xF→Σ = cos cos 0CD BCT Tφ φ− =

0 :yF↑Σ = 12 sin sin 0CG CD BCF T T Tφ φ− − − − =

400 lbCDT = −

178.9 lb (T)CGT = ................................................................ Ans.

6-98 Determine the forces in members CD, DE, and DF of the roof truss shown in Fig. P6-98. Triangle CDF is an equilateral triangle and joints E and G are at the midpoints of their respective sides.

SOLUTION

2 23 4 5 mAFL = + = 2.5 m2AF

AELL = = 3 2sin 60 1.26795 mb = − ° =

sin 3 5φ = cos 4 5φ =


0 :xF→Σ = ( )5 8 10 sin 0xAφ+ + − =

0 :yF↑Σ = ( )5 8 10 cos 0yA B φ+ − + + =

0 :AMΣ =� ( ) ( ) ( )8 2.5 8 5 10 0B − − =

8.75 kNB =

13.800 kNxA = 9.650 kNyA =

Next, cut a section through members CD, DF, and EF, and draw a free-body diagram of the right-hand portion of the truss. Then the equilibrium equations give

0 :xF→Σ = 10sin cos cos 60 0EF DF CDT T Tφ φ− − ° − =

0 :yF↑Σ = 8.75 10cos sin sin 60 0EF DFT Tφ φ− − − ° =

0 :FMΣ =� ( ) ( )8.75 4 2sin 60 0CDT− ° =

32.2765 kN 32.3 kN (C)EFT = − ≅

20.2073 kN 20.2 kN (T)CDT = ≅ ...............................................................................Ans.

23.2279 kN 23.2 kN (T)DFT = ≅ ..............................................................................Ans.

Finally, from a free-body diagram for joint E, the component of the equilibrium equations normal to members AE and EF gives

1 1.5tan 13.0643 2

bθ − −= = °−

0 :nFΣ = ( )8 sin 0DET φ θ+ + =

10.4533 kN 10.45 kN (C)DET = − ≅ ..........................................................................Ans.


218

6-99 Find the forces in members DE, DJ, and JK of the truss of Fig. P6-99. SOLUTION Then, from a free-body diagram of the entire truss, the equilibrium equations give

0 :xF→Σ = 0xA =

0 :GMΣ =� ( ) ( )8000 20 8000 40+

( ) ( )6000 60 80 0yA+ − =

10,500 lbyA =

Next, cut a section through members CD, JK, and JL, and draw a free-body diagram of the left-hand portion of the truss. Then, the equilibrium equations give

0 :JMΣ =� ( ) ( )( ) ( )6000 20 10,500 40 30 0CDT− − =

10,000 lb 10,000 lb (C)CDT = − =

0 :CMΣ =� ( ) ( )( )30 10,500 20 0JKT − =

7000 lb 7000 lb (T)JKT = = ........................... Ans.

Finally, from a free-body diagram for joint D, the equilibrium equations give

0 :xF→Σ = 0DE CDT T− =

0 :yF↑Σ = 8000 0DJT− − =

10,000 lb 10,000 lb (C)DET = − = ..............................................................................Ans.

8000 lb 8000 lb (C)DJT = − = ......................................................................................Ans.

6-100 The Gambrel truss shown in Fig. P6-100 supports one side of a bridge; an identical truss supports the other side. A 3400-kg truck is stopped on the bridge at the location shown and floor beams carry the vehicle load to the truss joints. If the center of gravity of the truck is located 1.5 m in front of the rear wheels, plot the force in members BC, BG, and GH as a function of the truck�s location d (0 ≤ d ≤ 20 m)

SOLUTION

( )3400 9.81 33,354 NW = =

From a free-body diagram of the entire truss (one side of the bridge), the equilibrium equations give

0 :xF→Σ = 0xA =

0 :EMΣ =� 4 8 12 16 16 0F G H A yP P P P A+ + + − =

2 3 4

4F G H A

yP P P PA + + +=

To determine the joint forces PA, PE, PF, PG, and PH, we need to first consider a free-body diagram of the truck. The equilibrium equations give the forces exerted on the front and rear wheels by the bridge deck panels

0 :RMΣ =� ( ) ( )33,354 1.5 4 0FN− =

0 :FMΣ =� ( ) ( )4 33,354 2.5 0RN − =

12,507.75 NFN = 20,846.25 NRN =


219

Next, consider a bridge deck panel. Each of the two trusses supports the left end of the panel with a force F1 and the right end of the panel with a force F2. At the instant shown, one of the truck�s wheels exerts a force N on the panel at a distance b from the right end of the panel. Then the forces transmitted to the trusses are

0 :yF↑Σ = 1 22 2 0F F N+ − =

2 0 :MΣ =� ( ) ( )( )12 4 0N b F− =

1 8F Nb= ( )

2

42 8 8

N bN NbF−

= − =

As the truck moves across the bridge, there are five different cases to consider:

Case 1: 0 4 md≤ ≤ ; FN N= , b d=

0A H GP P P= = =

1 8 8F F

FN b N dP F= = =

( )

2

48

FE

N dP F

−= =

Case 2: 4 m 8 md≤ ≤ ; ( )4 mb d= −

0A HP P= =

( )

1

48 8

FFG

N dN bP F−

= = =

( ) ( )

1 2

4 88 8

R FF

N d N dP F F

− −= + = +

( )2

88

RE

N dP F

−= =

Case 3: 8 m 12 md≤ ≤ ; ( )8 mb d= −

0A EP P= =

( )

1

88 8

FFH

N dN bP F−

= = =

( ) ( )

1 2

8 128 8

R FG

N d N dP F F

− −= + = +

( )

2

128

RF

N dP F

−= =

Case 4: 12 m 16 md≤ ≤ ; ( )12 mb d= −

0E FP P= =

( )

1

128 8

FFA

N dN bP F−

= = =

( ) ( )

1 2

12 168 8

R FH

N d N dP F F

− −= + = +


220

( )

2

168

RG

N dP F

−= =

Case 5: 16 m 20 md≤ ≤ ; ( )16 mb d= −

0E F GP P P= = =

( )

1

168 8

RRA

N dN bP F−

= = =

( )

2

208

RH

N dP F

−= =

Next, cut a section through BC, BG, and GH, and draw a free-body diagram of the left-hand side of the truss.

( )1tan 3 4 36.870θ −= = ° ( )1tan 1 4 14.036φ −= = °


0 :GMΣ =� ( ) ( )4 8 4 cos 0H A y BCP P A T φ+ − − =

0 :BMΣ =� ( )3 4 0GH y AT A P− − =

0 :yF↑Σ = sin sin 0y A BC BGA P T Tφ θ− + − =

2 2cos

H A yBC

P P AT

φ+ −

= ..................... Ans.

( )4

3y A

GH

A PT

−= ............................. Ans.

sin

siny A BC

BG

A P TT

φθ

− += ................ Ans.

6-101 An overhead crane consists of an I-beam supported by a simple truss as shown in Fig. P6-101. If the uniform I-beam weighs 400 lb and is supporting a load of 1000 lb at a distance d from its left end, plot the force in members AB, BC, EF, and FG as a function of the position d (0 ≤ d ≤ 8 ft).


0 :xF→Σ = 0xG A− =

0 :yF↑Σ = 0y F DA P P− − =

0 :AMΣ =� 3 3 9 0F DG P P− − =

3F DG P P= +

3x F DA G P P= = + 1400 lby F DA P P= + =

in which the joint forces PF and PD are determined from equilibrium of the I-beam

0 :DMΣ =� ( ) ( ) ( )400 3 1000 7 6 0Fd P+ − − =

0 :FMΣ =� ( ) ( )6 400 3 1000 1 0DP d− − − =


221

( )500 7

200 lb3F

dP

−= +

( )500 1

200 lb3D

dP

−= +

Then, from a free-body diagram for joint A, the equilibrium equations give

0 :xF→Σ = cos 0AB xT Aφ − =

0 :yF↑Σ = 1400 sin 0AG ABT T φ− − =

1400 sinAG ABT T φ= −

cos

xAB

ATφ

= ....................................................................................................................... Ans.

in which ( )1tan 3 9 18.435φ −= = ° .

Then, from a free-body diagram for joint G, the equilibrium equations give

0 :xF→Σ = cos 0FG BGT T Gθ+ + =

0 :yF↑Σ = sin 0AG BGT T θ+ =

sin

AGBG

TTθ

−=

cosFG BGT G T θ= − − ......................................................................................................Ans.

in which ( )1tan 2 3 33.690θ −= = ° .

Finally, cut a section through members BC, CF, and EF, and draw a free-body diagram of the right-hand portion of the truss. Then, the equilibrium equations give

0 :FMΣ =� ( )6 sin 6 0BC DT Pφ − =

0 :CMΣ =� 1 3 0EF DT P− − =

sin

DBC

PTφ

= ......................................... Ans.

3EF DT P= − ......................................... Ans.

6-102 The masses of carton 1, 2, and 3, which rest on the platform shown in Fig. P6-102, are 300 kg, 100 kg, and 200 kg, respectively. The mass of the platform is 500 kg. Determine the tensions in the three cables A, B, and C that support the platform.

SOLUTION From a free-body diagram of the platform, the equilibrium equations give

:OΣ =M 0 ( ) ( ) ( ) ( )300 2 100g g+ × − + + × −i j k i j k


222

( ) ( ) ( )3 1.5 2 500BT g+ + × + + × −i j k i j k

( ) ( ) ( )3 200 4 Cg T+ + × − + + × =i j k i j k 0

:x 300 100 1000 600 4 0B Cg g T g g T− − + − − + =

:y 300 200 3 750 200 0B Cg g T g g T+ + − + + − =

0 :zFΣ = 1100 0A B CT T T g+ + − =

340.91 3340 NAT g= = ............................. Ans.

345.45 3390 NBT g= = ................................................................................................Ans.

413.64 4060 NCT g= = ................................................................................................Ans.

6-103 Determine the reaction at support A of the pipe system shown in Fig. P6-103 when the force applied to the pipe wrench is 50 lb.

SOLUTION From a free-body diagram of the platform, the equilibrium equations give

:AΣ =M 0 ( ) ( )7 23 10 50A + − + + × − =M i j k k 0

:x 1150 0AxM − =

:y 350 0AyM − =

:z 0AzM =

0 :xFΣ = 0xA =

0 :yFΣ = 0yA =

0 :zFΣ = 50 0zA − =

( )50 lb=A k ............................. ( )1150 350 lb ftA = + ⋅M i j ........................................Ans.

6-104 The rectangular plate of uniform thickness shown in Fig. P6-104 has a mass of 500 kg. Determine (a) The tensions in the three cables supporting the plate. (b) The deformation of the cable at A, which has a diameter of 10-mm, a length of 1.5 m and a modulus of

elasticity of 200 GPa. SOLUTION

( )500 500 9.81 4905 NW g= = =

(a) From a free-body diagram of the plate, the equilibrium equations give

0 :zFΣ = 4905 0A B CT T T+ + − =

:OΣ =M 0 ( ) ( ) ( ) ( )1.5 0.9 0.5 2.5B CT T− + × + − + ×i j k i j k

( ) ( )0.625 1.25 4905+ − + × − =i j k 0

:x 0.9 2.5 6131.250 0B CT T+ − =

:y 1.5 0.5 3065.625 0B CT T+ − =

1560.682 N 1561 NAT = ≅ ...........................................................................................Ans.

1393.466 N 1393 NBT = ≅ ..........................................................................................Ans.


223

1950.852 N 1951 NCT = ≅ ..........................................................................................Ans.

(b) The stretch of wire A is given by

( ) ( )

( ) ( )29

1560.862 15000.1490 mm

200 10 0.010 4A

PLEA

δπ

= = = ×

..........................................Ans.

6-105 A shaft is loaded through a pulley and a lever (Fig. P6-105) that are fixed to the shaft. Friction between the belt and pulley prevents slipping of the belt. Determine the force P required for equilibrium and the reactions at supports A and B. The support at A is a ball bearing and the support at B is a thrust bearing. The bearings exert only force reactions on the shaft.

SOLUTION From a free-body diagram of the shaft, the moment equilibrium equation gives

:BΣ =M 0 ( ) ( )14 14 200 P+ × −i j j k

( ) ( )18 x zA A+ − × +j i k

( ) ( )30 6 500+ − + × −j k i

( ) ( )30 6 150+ − − × − =j k i 0

:x 14 18 0zP A− − = 116.667 lbzA = −

:y 14 2100 0P − = 150 lbP =

:z 18 16,700 0xA − = 927.778 lbxA =

Then, the force equilibrium equation Σ =F 0 gives

0 :xFΣ = 650 0x xA B+ − = 277.778 lbxB = −

0 :yFΣ = 200 0yB + = 200 lbyB = −

0 :zFΣ = 0z zA B P+ − = 266.667 lbzB =

( )928 116.7 lb= −A i k ................................................................................................Ans.

( )278 200 267 lb= − − +B i j k ...................................................................................Ans.

150 lbP = ......................................................................................................................... Ans.

6-106 The homogeneous door shown in Fig. P6-106 has a mass of 60 kg, and is held in the position shown by the rod AB. The rod is held in place by smooth horizontal pins at A and B. The hinges at C and D are smooth, and the hinge at C can support thrust along its axis. Determine all forces that act on the door.

SOLUTION

( )60 60 9.81 588.60 NW g= = =

Note that the rod AB is a two-force member.

( )( )

cos 60 sin 60

0.5000 0.86603 NAB AB AB

AB AB

F F

F F

= − ° + °

= − +

F j k

j k

Then, from a free-body diagram of the door, the moment equilibrium equation gives

:CΣ =M 0 ( ) ( )950 y zD D− × +i j k


224

( ) ( )475 475cos 60 475sin 60 588.60+ − + ° + ° × −i j k k

( ) ( )950cos 60 950sin 60 0.5000 0.86603AB ABF F+ ° + ° × − + =j k j k 0

:x 139,792.5 822.724 0ABF− + = 169.914 NABF =

:y 950 279,585 0zD − = 294.300 NzD =

:z 950 0yD− = 0 NyD =


0 :xFΣ = 0xC = 0 NxC =

0 :yFΣ = ( )0.5000 169.914 0y yC D+ − = 84.957 NyC =

0 :zFΣ = ( )0.86603 169.914 588.60 0z zC D+ + − = 147.15 NzC =

( )85.0 147.2 N= +C j k ...............................................................................................Ans.

( )294 N=D k .................................................................................................................Ans.

( )85.0 147.2 NAB = − +F j k .........................................................................................Ans.

6-107 A beam is supported by a ball-and-socket joint and two cables as shown in Fig. P6-107. Determine (a) The tensions in the two cables. (b) The reaction at support A (the ball and socket joint). (c) The axial stress and deformation in each of the cables if they are made of steel (E = 30,000 ksi) and have

¼- in. diameters. SOLUTION

2 211 6 12.530 ftBDL = + = 2 2 27 11 3 13.379 ftCDL = + + =

( )2 2

11 6 0.87790 0.4788511 6

BD BD BD BDT T T− += = − ++

j kT j k

( )2 2 2

7 11 3 0.52320 0.82218 0.224237 11 3

CD CD CD CD CDT T T T− − += = − − ++ +

i j kT i j k

(a) From a free-body diagram of the beam, the moment equilibrium equation gives

:AΣ =M 0 ( ) ( ) ( ) ( )11 300 6 800BD CD× + + + × − =j T T i j k 0

:x 5.26735 2.46653 4800 0BD CDT T+ − =

:y 0 0=

:z 5.75520 3300 0CDT − =

642.77 lb 643 lbBDT = ≅ ..............................................................................................Ans.

573.39 lb 573 lbCDT = ≅ ..............................................................................................Ans.

(b) Then, the force equilibrium equation Σ =F 0 gives

0 :xFΣ = ( )0.52320 573.39 300 0xA − + = 0 lbxA =

0 :yFΣ = ( ) ( )0.87790 642.77 0.82218 573.39 0yA − − = 1035.72 lbyA =


225

0 :zFΣ = ( ) ( )0.47885 642.77 0.22423 573.39 800 0zA + + − = 363.64 lbzA =

( )1036 364 lb= +A j k .................................................................................................Ans.

(c) Finally, the stresses and deformation of the two cables is

( )

32

642.77 13.09 10 psi 13.09 ksi0.25 4

BDTA

σπ

= = = × = ...........................................Ans.

( )

32

573.39 11.68 10 psi 11.68 ksi0.25 4

CDσπ

= = × = ....................................................Ans.

( )( )

( ) ( )26

642.77 12.530 120.0656 in.

30 10 0.25 4BD

TLEA

δπ

×= = =

×

................................................Ans.

( )( )

( ) ( )26

573.39 13.379 120.0625 in.

30 10 0.25 4CDδ

π

×= =

×

...........................................................Ans.

6-108 The crankshaft-flywheel arrangement of a one-cylinder engine is shown in Fig. P6-108. A 250-N-m couple C is delivered to the crankshaft by the flywheel, which is rotating at a constant angular velocity. The support at A is a ball bearing and the support at B is a thrust bearing, both of which support only force reactions on the shaft. Determine the magnitudes of the force P and of the resultant bearing reactions at A and B.

SOLUTION sin 20 cos 20 0.34202 0.93969P P P P= − ° − ° = − −P j k j k

From a free-body diagram of the crankshaft, the moment equilibrium equation gives

:BΣ =M 0 ( ) ( )250 0.45 y zA A− + × +i i j k

( )0.225 0.125+ − × =i j P 0

:x 250 0.11746 0P− + =

:y 0.45 0.21149 0zA P− + =

:z 0.45 0.07695 0yA P− =

2128.38 NP =

363.954 NyA =

1000.293 NzA =


0 :xFΣ = 0xB = 0 NxB =

0 :yFΣ = ( )0.34202 2128.38 0y yA B+ − = 363.996 NyB =

0 :zFΣ = ( )0.93969 2128.38 0z zA B+ − = 999.728 NzB =

2 2364 1000 1064 NA = + = ......................................................................................Ans.

2 2364 1000 1064 NB = + = ......................................................................................Ans.

2130 NP = ...................................................................................................................... Ans.


226

6-109 A farmer is using the hand winch shown in Fig. P6-109 to slowly raise a 40-lb bucket of water from a well. In the position shown, the force P is vertical. The bearings at C and D exert only force reactions on the shaft. Bearing C can support thrust loading; bearing D cannot. Determine the magnitude of force P and the components of the bearing reactions.

SOLUTION From a free-body diagram of the winch drum and axle, the moment equilibrium equation gives

:CΣ =M 0 ( ) ( )20cos30 20sin 30 11 P− ° + ° + × −i j k j

( ) ( )2.5 20 40+ − × −i k j

( ) ( )50 x yD D+ − × + =k i j 0

:x 11 800 50 0yP D− + =

:y 50 0xD− =

:z 20 cos30 100 0P ° − =

5.7735 lb 5.77 lbP = ≅ ................................................................................................Ans.

0 lbxD = ............................................ 14.7298 lbyD = ..............................................Ans.

(b) Then, the force equilibrium equation Σ =F 0 gives

0 :xFΣ = 0x xC D+ = 0 lbxC = .............................................Ans.

0 :yFΣ = 40 0y yC D P+ − − = 31.0437 lbyC = ................................Ans.

0 :zFΣ = 0zC = 0 lbzC = .............................................Ans.

6-110 The block W shown in Fig. P6-110 has a mass of 250 kg. Bar AB rests against a smooth vertical wall at end B and is supported at end A with a ball-and-socket joint. The two cables are attached to a point on the bar midway between the ends. Determine

(a) The reactions at supports A and B and the tension in cable CD. (b) The axial stress and deformation in cable CD if it is made of steel (E = 210 GPa) and has an 8-mm

diameter. SOLUTION

( )250 250 9.81 2452.50 NW g= = = 2 2 2200 200 500 574.456 mmCDL = + + =

( )2 2 2

200 200 500200 200 500

0.34816 0.34816 0.87039

CD CD

CD CD CD

T

T T T

− + +=+ +

= − + +

i j kT

i j k

(a) From a free-body diagram of the beam, the moment equilibrium equation gives

:AΣ =M 0 ( ) ( )200 500 300 2452.50CD− + × −i j k T k

( ) ( )400 1000 600 yB+ − + × =i j k j 0

:x 539.643 1, 226, 250 600 0CD yT B− + − =

:y 278.526 490,500 0CDT− + = 1761.056 NCDT =

:z 104.448 400 0CD yT B− + = 459.847 NyB =

1761.056 N 1761 NCDT = ≅ .........................................................................................Ans.


227

( )460 N=B j .................................................................................................................. Ans.


0 :xFΣ = ( )0.34816 1761.056 0xA − = 613 NxA =

0 :yFΣ = ( )0.34816 1761.056 0y yA B+ + = 1073 NyA = −

0 :zFΣ = ( )0.87039 1761.056 2452.50 0zA + − = 920 NzA =

( )613 1073 920 N= − +A i j k ....................................................................................Ans.

(c) Finally, the stresses and deformation of the cable is

( )

6 22

1761.056 35.0 10 N/m 35.0 MPa0.008 4

CDσπ

= = × = .............................................Ans.

( )( )

( ) ( )29

1761.056 594.4560.0958 mm

210 10 0.008 4CDδ

π= =

×

...................................................Ans.

6-111 A 20-lb piece of electronic equipment is placed on a wooden skid that weighs 10 lb and that rests on a concrete floor (Fig. P6-111). The coefficient of static friction between the skid and the floor is 0.45.

(a) Determine the minimum pushing force along the handle necessary to cause the skid to start sliding across the floor (Fig. P6-111a).

(b) Determine the minimum force necessary to start the skid in motion if a pulling force instead of a pushing force is applied to the handle (Fig. P6-111b).

SOLUTION (a) Impending motion is to the right and friction acts to the left

to oppose the impending motion. From a free-body diagram of the skid, the equilibrium equations give

0 :xF→Σ = cos30 0.45 0P N° − =

0 :yF↑Σ = 30 sin 30 0N P− − ° =

40.5 lbN =

21.1 lbP = ....................................................................................................................... Ans. (b) Now, impending motion is to the left and friction acts to the

right to oppose the impending motion. From a free-body diagram of the skid, the equilibrium equations give

0 :xF→Σ = 0.45 cos30 0N P− ° =

0 :yF↑Σ = 30 sin 30 0N P− + ° =

23.8 lbN =

12.37 lbP = ..................................................................................................................... Ans.

6-112 A skier is at the point of experiencing impending motion on a ski slope as illustrated in Fig. P6-112. Determine the coefficient of static friction between a ski and the slope if the angle θ (the angle of repose) is 5°.

SOLUTION Impending motion is down the hill and friction acts up the hill to oppose the impending motion. From a free-body diagram of the skier, the equilibrium equations give


228

0 :nFΣ = cos5 0N mg− ° = cos5N mg= °

0 :tFΣ = sin 5 0sN mgµ − ° = sin 5sN mgµ = °

sin 5 tan 5 0.0875cos5s

mgmg

µ °= = ° =°

...............................................................................Ans.

6-113 The block in Fig. P6-113 weighs 500 lb and the coefficient of friction between the block and the inclined plane is 0.2. Determine

(a) Whether the system would be in equilibrium for P = 400 lb. (b) The minimum force P to prevent motion. (c) The maximum force P for which the system is in equilibrium. SOLUTION If the impending motion is down the hill, then the friction force would be up the hill. From a free-body diagram of the block, the equilibrium equations are

0 :nFΣ = 500cos30 sin 20 0N P− ° + ° =

0 :tFΣ = 500sin 30 cos 20 0F P− ° + ° =

(a) If 400 lbP = , then the equilibrium equations give

296.205 lbN =

125.877 lbF = −

max 0.2 59.24 lbF N= =

The negative sign on the friction indicates that the direction of impending motion is actually up the hill and the friction is acting down the hill to oppose the motion. However, since the amount of friction required for equilibrium is greater than the maximum available friction,

Not in equilibrium. ........................................................................................................Ans.

(b) If 0.2F N= , then the equilibrium equations give

368.872 lbN =

0.2 73.7744 lbF N= =

min187.5 lbP P= = ..........................................................................................................Ans.

(c) If the impending motion is up the hill, then the friction force would be down the hill. From a free-body diagram of the block, the equilibrium equations are

0 :nFΣ = 500cos30 sin 20 0N P− ° + ° =

0 :tFΣ = 500sin 30 cos 20 0F P− − ° + ° =

Now, if 0.2F N= , the equilibrium equations give

318.812 lbN =

63.7624 lbF =

max334 lbP P= = .............................................................................................................Ans.

6-114 A 75-kg man starts climbing a 5-m long ladder leaning against a wall (Fig. P6-114). The coefficient of friction is 0.25 at both surfaces. Neglect the weight of the ladder and determine how far up the ladder the man can climb before the ladder starts to slip.

SOLUTION Impending motion of the ladder is down and to the left, and friction on the wall and floor acts to oppose the impending motion. From a free-body diagram of the man and the ladder, the equilibrium equations give


229

0 :xF→Σ = 1 20.25 0N N− =

0 :yF↑Σ = ( )1 275 9.81 0.25 0N N− + =

1 0 :MΣ =� ( ) ( ) ( )2 25sin 60 0.25 5cos 60N N° + °

( ) ( )75 9.81 cos60 0d− ° =

1 692.471 NN =

2 173.118 NN =

2.33 md = .....................................................................Ans.

6-115 A device for lifting rectangular objects such as bricks and concrete blocks is shown in Fig. P6-115. Determine the minimum coefficient of static friction between the contacting surfaces required to make the device work.

SOLUTION Impending motion is for the block to slide downward. Friction must act upward on the block to oppose this impending motion. The friction force on the member AB must be equal and opposite to that on the block � it must act downward on the member. From a free-body diagram for member AB, the moment equilibrium equation gives

0 :BMΣ =� 4 12 0sN Nµ− =

4 12 0.333sµ = = ...........................................Ans.

6-116 A boy is pulling a sled with a box (at constant velocity) up an inclined surface as shown in Fig. P6-116. The mass of the box and sled is 50 kg; the mass of the boy is 40 kg. If the coefficient of kinetic friction between the sled runners and the icy surface is 0.05, determine the minimum coefficient of static friction needed between the boy�s shoes and the icy surface.

SOLUTION Impending motion is up the hill and friction acts down the hill to oppose the motion. From a free-body diagram of the sled, the equilibrium equations give

0 :nFΣ = ( )1 50 9.81 cos15 sin 25 0N P− ° + ° =

0 :tFΣ = ( ) 1cos 25 50 9.81 sin15 0.05 0P N− ° − =

1 405.142 NN =

162.426 NP = Impending motion of the boy�s foot is down the hill and friction acts up the hill to oppose the motion. From a free-body diagram of the boy, the equilibrium equations give

0 :nFΣ = ( )2 40 9.81 cos15 sin 25 0N P− ° − ° =

0 :tFΣ = ( )2 40 9.81 sin15 cos 25 0sN Pµ − ° − =

2 447.673 NN =

2 2248.769 NsN Fµ = =

0.556sµ = ......................................................................Ans.


230

6-117 A 120-lb girl is walking up a 48-lb uniform beam as shown in Fig P6-117. Determine how far up the beam the girl can walk before the beam starts to slip if

(a) The coefficient of friction is 0.20 at all surfaces. (b) The coefficient of friction at the bottom end of the beam is increased to 0.40 by placing a piece of rubber

between the beam and the floor. SOLUTION Impending motion of the beam is down and to the right and friction on the wall and floor acts to oppose the impending motion. From a free-body diagram of the girl and the beam, the equilibrium equations are

0 :xF→Σ = ( ) ( )1 1 23 5 4 5 0N F F− − =

0 :yF↑Σ = ( ) ( )1 1 24 5 3 5 168 0N F N+ − + =

1 0 :MΣ =� ( )( ) ( ) 148 4 5 6 120 4 5 10 0x N+ − =

(a) If 1 10.2F N= and 2 20.2F N= , then 0 :xF→Σ = gives

2 111 5N N=

and the other two equations give

1 53.846 lbN = 2 118.462 lbN =

3.21 ftx = ......................................................................................................................... Ans.

(b) If 1 10.2F N= and 2 20.4F N= , then 0 :xF→Σ = gives

2 111 10N N=


1 83.168 lbN = 2 91.485 lbN =

6.26 ftx = ........................................................................................................................ Ans.

6-118 The broom shown in Fig. P6-118 weighs 8 N and is held up by the two cylinders, which are wedged between the broom handle and the side rails. The coefficient of friction between the broom and cylinders and between the cylinders and side rails is 0.30. The side rails are at an angle of θ = 30° to the vertical. The weight of the cylinders may be neglected. Determine whether or not this system is in equilibrium. If the system is in equilibrium, determine the force exerted on the broom handle by the rollers.

SOLUTION Impending motion of the broom handle is downward and friction acts upward to oppose the motion. From a free-body diagram of the broom handle, the equilibrium equations give

0 :xF→Σ = 2 1 0N N− =

0 :yF↑Σ = 1 2 8 0F F+ − =

0 :centerMΣ =� 1 2 0Fb F b− =

1 2 4 NF F= =

1 2N N=

From a free-body diagram of the left-hand roller, the equilibrium equations give

0 :xF→Σ = 3 3 2cos60 cos30 0F N N° + ° − =

0 :yF↑Σ = 3 3 2sin 60 sin 30 0F N F− ° + ° − =


231

0 :centerMΣ =� 3 2 0F r F r− =

3 4 NF =

1 2 3 14.92820 NN N N= = =

max 0.3 4.478 NF N= =

Since the amount of friction required for equilibrium is less than the maximum amount of friction available at all surfaces, the broom handle

Is in equilibrium .............................................................................................................Ans.

14.93 NN = ..................................................................................................................... Ans.

4 NF = ............................................................................................................................. Ans.

6-119 The device shown in Fig. P6-119 is used to raise boxes and crates between floors of a factory. The frame, which slides on the 4-in.-diameter vertical post, weighs 50 lb. If the coefficient of friction between the post and the frame is 0.10, determine the force P required to raise a 150-lb box.

SOLUTION Impending motion of the platform is upward and friction with the post acts to oppose the impending motion. Guessing that the platform rotates slightly clockwise, contact with the post (and the location of the normal and friction forces) is on the top-left and bottom-right sides of the post. Then, from a free-body diagram of the platform and box, the equilibrium equations give

0 :xF→Σ = 0A BN N− =

0 :yF↑Σ = ( )0.1 200 0A BP N N− + − =

0 :AMΣ =� ( ) ( ) ( )26 4 0.1 8 4 50 20 150 0B BN N P+ + − − =

57.143 lbA BN N= =

211 lbP = ........................................................................................................................ Ans. (Note that the two normal forces are both positive indicating that the guess about how the platform rubs against the post was correct.)

6-120 The automobile shown in Fig. P6-120 has a mass of 1500 kg. The coefficient of friction between the rubber tires and the pavement is 0.70. Determine the maximum incline θ that the automobile can drive up if the automobile has

(a) A rear-wheel drive. (b) A front-wheel drive. SOLUTION

( )1500 9.81 14,715 NW = =

Impending slip between the tires and the ground is for the bottom of the tires to move backwards and the friction force acts to oppose the slip. From a free-body diagram of the car, the equilibrium equations are

0 :nFΣ = 1 2 14,715cos 0N N θ+ − =

0 :tFΣ = 1 2 14,715sin 0F F θ+ − =

0 :GMΣ =� ( )2 1 1 21.2 1.7 0.85 0N N F F− + + =

(a) If the vehicle has rear wheel drive, then 1 10.7F N= and 2 0F = and the moment equilibrium equation gives

2 10.92083N N=


232


1

1

14,715sin 0.7tan 0.3644314,715cos 1.92083

NN

θθθ

= = =

20.02θ = ° ........................................................................................................................ Ans.

(b) If the vehicle has front wheel drive, then 1 0F = and 2 20.7F N= and the moment equilibrium equation gives

1 21.05588N N=


2

2

14,715sin 0.7tan 0.3404914,715cos 2.05588

NN

θθθ

= = =

18.80θ = ° ......................................................................................................................... Ans.

6-121 When a drawer is pulled by only one of the handles, it tends to twist and rub as shown (highly exaggerated) in Fig. P6-121, which is a top view of the drawer. The weight of the drawer and its contents is 2 lb and is uniformly distributed. The coefficient of friction between the sides of the drawer and the sides of the dresser is 0.6; between the bottom of the drawer and the side rails the drawer rides on, µs = 0.1. Determine the minimum amount of force P necessary to pull the drawer out.

SOLUTION

( )3 4 0.1 2 0.1 lbF F W= = =

Impending slip of the drawer is to the front and friction acts to oppose the impending motion. From a free-body diagram of the drawer, the equilibrium equations give

0 :xF→Σ = 1 2 0N N− =

0 :yF↑Σ = 1 2 3 4 0F F F F P+ + + − =

2 0 :MΣ =� 3 427 30.375 0.375P F F− −

1 130.75 18 0F N− − =

1 2 0.57407 lbN N= =

1 2 10.6 0.34444 lbF F N= = =

0.889 lbP = ..................................................................................................................... Ans.

6-122 An ill-fitting window is about 10 mm narrower than its frame (see Fig. P6-122). The window weighs 40 N, and the coefficient of friction between the window and the frame is 0.20. Determine the amount of force P that must be applied at the lower corner to keep the window from lowering.

SOLUTION Impending slip of the window is downward and friction acts to oppose the motion. From a free-body diagram of the window, the equilibrium equations give

0 :xF→Σ = 1 2 0N N− =

0 :yF↑Σ = 1 2 40 0F F P+ − + =

2 0 :MΣ =� ( ) 1 1405 40 600 810 0N F− − =

1 2 21.260 NN N= =

1 2 10.2 4.252 NF F N= = =

31.5 NP = ....................................................................................................................... Ans.


233

6-123 Determine the minimum coefficient of static friction necessary for the pliers shown in Fig. P6-123 to grip the bolt.

SOLUTION Impending motion is for the bolt to slip out of the pliers and friction acts to oppose the motion. From a free-body diagram for bolt, the equilibrium equations give

0 :centerMΣ =� 1 2 0F r F r− =

1 2 1 2s sF F N Nµ µ= = =

1 2N N=

0 :xF→Σ = ( ) ( )1 2 1 2sin17.5 cos17.5 0N N F F+ ° − + ° =

( ) ( )1 2 1 2sin17.5 cos17.5sN N N Nµ+ ° = + °

tan17.5 0.315sµ = ° = ...................................................................................................Ans.

(Note that even though we did not use the third equation of equilibrium 0 :yF↑Σ = , it is also satisfied by the values solved above.)

6-124 A homogeneous book of mass 0.46 kg rests in a bookshelf as shown in Fig. P6-104. The thickness of the book is small compared to the other dimensions shown. The coefficient of friction at all surfaces is 0.4. Determine the minimum angle θ for which the book is in equilibrium.

SOLUTION

( )0.46 9.81 4.5126 NW = =

Impending motion of the book is down and to the right. Friction acts to oppose this motion. From a free-body diagram of the book, the equilibrium equations give

0 :xF→Σ = 1 20.4 0N N− =

0 :yF↑Σ = 2 10.4 4.5126 0N N+ − =

1 1.55607 NN = 2 3.89017 NN =

2 0 :MΣ =� ( ) ( ) ( )1 14.5126 130cos 260sin 0.4 260cos 0N Nθ θ θ− − =

tan 1.0500θ = 46.40θ = ° ......................................................................Ans.

6-125 A wedge is used to raise a 350-lb refrigerator onto a platform (Fig. P6-125). The coefficient of friction is 0.2 at all surfaces.

(a) Determine the minimum force P needed to insert the wedge. (b) Determine if the system would still be in equilibrium if P = 0. (c) If the system is not in equilibrium when P = 0, determine the force necessary to keep the wedge in place,

or if the system is in equilibrium when P = 0, determine the force necessary to remove the wedge. SOLUTION When the wedge is being pushed to the right, friction acts down and to the left to oppose the motion. From a free-body diagram of the wedge and refrigerator, the equilibrium equations are

0 :xF→Σ = sin15 cos15 0P N F− ° − ° =

0 :yF↑Σ = cos15 sin15 350 0N F° − ° − =

(a) If 0.2F N= , then the equilibrium equations give

382.864 lbN =


234

0.2 76.573 lbF N= =

173.1 lbP = ...................................................................................................................... Ans.

(b) If 0P = , then the equilibrium equations give

390.374 lbN = 104.600 lbF = However, this solution is not valid since the amount of friction required for equilibrium is greater than the

maximum amount of friction available max 0.2 78.075 lbF N= = . Therefore, the wedge will not stay in place if the force P is removed.

When the force P is reduced, the wedge wants to slip to the left and friction must act up and to the right to oppose the impending motion. Now from a free-body diagram of the wedge and refrigerator, the equilibrium equations give (using 0.2F N= )

0 :xF→Σ = sin15 cos15 0P N F− ° + ° =

0 :yF↑Σ = cos15 sin15 350 0N F° + ° − =

343.916 lbN =

0.2 68.783 lbF N= =

22.6 lbP = ....................................................................................................................... Ans.

6-126 A 50-kg uniform plank rests on rough supports at A and B (Fig. P6-126). The coefficient of friction is 0.60 at both surfaces. If a man weighing 800 N pulls on the rope with a force of P = 400 N, determine

(a) The minimum and maximum angles θmin and θmax for which the system will be in equilibrium. (b) The minimum coefficient of friction that must exist between the man�s shoes and the ground for each of

the cases in part (a). SOLUTION

( )50 9.81 490.5 NW = =

(a) Impending motion of the plank is to the left and friction acts to the right to oppose the motion. From a free-body diagram of the plank, the equilibrium equations are

0 :xF→Σ = 400cos 0A BF F θ+ − =

0 :yF↑Σ = 490.5 400sin 0A BN N θ+ − − =

0 :BMΣ =� ( ) ( ) ( ) ( )490.5 1 2 400sin 2 0AN θ− − =

If 0.6A AF N= and 0.6B BF N= (impending slip), then the equilibrium equations give

( )0.6 400cosA BN N θ+ = 490.5 400sinA BN N θ+ = +

666.667 cos 490.5 400sinA BN N θ θ+ = = +

19.920θ = °

If 0A AF N= = (impending tip), then the equilibrium equations give

( )1 490.5sin 37.816

2 400θ −= = °

316.0 NBF = 735.75 NBN = ( )max 0.6 441.45 NB BF N= =

Since the amount of friction required is less than the maximum amount of friction available, the solution is okay and the allowed range of angles is

min 19.92θ = ° ................. max 37.82θ = ° .........................................................................Ans.


235

(b) Impending motion of the man�s feet is to the right and friction acts to the left to oppose the motion. From a free-body diagram of the plank, the equilibrium equations are

0 :xF→Σ = 400cos 0Fθ − =

0 :yF↑Σ = 400sin 800 0N θ+ − =

If 19.920θ = ° , the equilibrium equations give

376.07 NF = 663.72 NN =

( )min

376.07 0.567663.72s

FN

µ = = = ....................................................................................Ans.

If 37.816θ = ° , the equilibrium equations give

315.99 NF = 554.75 NN =

( )min

315.99 0.570554.75sµ = = .............................................................................................Ans.

6-127 The plunger of a door latch is held in place by a spring as shown in Fig. P6-127. Friction on the sides of the plunger may be ignored. If a force of 2 lb is required to just start closing the door and the coefficient of friction between the plunger and the striker plate is 0.25, determine the force exerted on the plunger by the spring.

SOLUTION When the door is pushed in the direction of the 2-lb force, friction acts upward and to the left to oppose the motion. From a free-body diagram of the plunger, the equilibrium equations give

0 :xF→Σ = sin 50 0.25 cos50 0N N P° − ° − =

0 :yF↑Σ = cos50 0.25 sin 50 2 0N N° + ° − =

2.39722 lbN =

1.451 lbP = ...................................................................................................................... Ans.

6-128 The 25-kg block in Fig. P6-128 is held against the wall by the brake arm. The coefficient of friction between the wall and the block is 0.20; between the block and the brake arm, 0.50. Neglect the weight of the brake arm. Determine

(a) Whether the system would be in equilibrium for P = 230 N. (b) The minimum force P for which the system would be in equilibrium. (c) The maximum force P for which the system would be in equilibrium. (d) The shearing stress on a cross section of the pin at A when impending motion of the block is downward if

the pin has a 6-mm diameter and is in double shear. SOLUTION

( )25 9.81 245.25 NW = =

If the impending motion of the block is downward, friction would act upward on the block (and downward on the brake handle) to oppose the motion. From a free-body diagram of the brake handle, the equilibrium equations are

0 :xF→Σ = 2 140 0xN A− − =

0 :yF↑Σ = 2 0yA F− =

0 :AMΣ =� ( )( ) ( )2 2140 50 40 0N F− + =


236

From a free-body diagram of the block, the equilibrium equations are

0 :xF→Σ = 1 2 0N N− =

0 :yF↑Σ = 1 2 245.25 0P F F+ + − =

(a) If 230 NP = , the equilibrium equations give

1 2N N= 1 2 15.25 NF F+ = 2 2140 0.8N F= +

( )1 2 1 2 2 2max 0.2 0.5 0.7 98 0.56 98 NF F N N N F+ = + = = + >

Since the amount of friction required for equilibrium ( )15.25 N is less than the maximum amount of friction

available ( )at least 98 N , the system

Is in equilibrium .............................................................................................................Ans.

(b) If 1 10.2F N= and 2 20.5F N= (impending motion downward), the equilibrium equations give

( )( ) ( ) ( )2 2140 50 0.5 40 0N N− + = 1 2 233.333 NN N= =

1 116.667 NF = 2 46.667 NF =

min81.9 NP P= = ............................................................................................................Ans.

93.333 NxA = 116.667 NyA = 2 2 149.406 Nx yA A A= + =

(c) If impending motion of the block is upward, the friction forces ( 1 10.2F N= and 2 20.5F N= ) change direction and the equilibrium equations give

( )( ) ( )( )2 2140 50 0.5 40 0N N− − = 1 2 100.00 NN N= =

1 20.00 NF = 2 50.00 NF =

max315 NP P= = .............................................................................................................Ans.

(d) When the impending motion of the block is downward, the force on the pin is 149.406 NA = . Dividing the pin force by twice its cross sectional area (since it is in double shear) gives the shear stress

( )

6 22

149.406 2.64 10 N/m 2.64 MPa2 2 0.006 4A

As

VA

τπ

= = = × =

...........................Ans.

6-129 A 250-lb weight is suspended from a lightweight rope wrapped around the inner cylinder of a drum (Fig. P6-129). A brake arm is pressed against the outer cylinder of the drum by a hydraulic cylinder. The coefficient of friction between the brake arm and the drum is 0.40. Determine

(a) The smallest force in the hydraulic cylinder necessary to prevent motion. (b) The shearing stress on a cross section of the pin at A when motion is impending if the drum weighs 75 lb

and the pin has a ½-in. diameter and is in double shear. SOLUTION (a) Impending motion of the drum is a counterclockwise rotation and friction

acts to oppose the motion. From a free-body diagram of the drum, the equilibrium equations give

0 :xF→Σ = 0xF A− =

0 :yF↑Σ = 75 250 0yA N− − − =

0 :AMΣ =� ( )( ) ( )250 6 12 0F− =


237

125 lb 0.4F N= = 312.5 lbN =

125 lbxA = 637.5 lbyA =

2 2 649.639 lbx yA A A= + =

From a free-body diagram of the brake arm, the moment equilibrium equation gives

0 :CMΣ =� 9 9 27 0P F N− − =

1063 lbP = ...................................................................................................................... Ans. (b) Dividing the force on the pin by twice its cross sectional area (since it is in double shear) gives the shear stress

( )2

649.639 1654 psi2 2 0.5 4A

As

VA

τπ

= = =

......................................................................Ans.

6-130 The brake shown in Fig. P6-130 is used to control the motion of block B. If the mass of block B is 25 kg and the kinetic coefficient of friction between the brake drum and brake pad is 0.30, determine

(a) The force P required for a constant-velocity descent. (b) The shearing stress on cross sections of both pins. Each pin has a diameter of 10 mm and is in double

shear. SOLUTION

( )25 25 9.81 245.25 NW g= = =

(a) Motion of the drum is a clockwise rotation and friction acts to oppose the motion. From a free-body diagram of the drum, the equilibrium equations give

0 :xF→Σ = 0xB F− =

0 :yF↑Σ = 245.25 0yB N− − =

0 :BMΣ =� ( ) ( )180 245.25 90 0F − =

122.625 N 0.3F N= = 408.750 NN =

122.625 NxB = 654.000 NyB =

2 2 665.397 Nx yB B B= + =

From a free-body diagram of the brake arm, the equilibrium equations give

0 :xF→Σ = 0xF A− =

0 :yF↑Σ = 0yN A P− − =

0 :AMΣ =� 200 40 600 0N F P+ − =

122.625 NxA = 264.325 NyA =

144.4 NP = ..................................................................................................................... Ans.

2 2 291.384 Nx yA A A= + =

(b) Dividing the force on the pins by twice their cross sectional areas (since they are in double shear) gives the shear stresses


238

( )

6 22

291.384 1.855 10 N/m 1.855 MPa2 2 0.01 4A

As

VA

τπ

= = = × =

..........................Ans.

( )

6 22

665.397 4.24 10 N/m 4.24 MPa2 2 0.01 4B

Bs

VA

τπ

= = = × =

..............................Ans.

6-131 Three identical cylinders are stacked as shown in Fig. P6-131. The cylinders each weigh 22 lb and are 8 in. in diameter. The coefficient of friction is µs = 0.40 at all surfaces. Determine the maximum force P that the cylinders can support without moving.

SOLUTION From a free-body diagram for the upper cylinder, the equilibrium equations give

0 :xF→Σ = cos30 cos30 cos60 cos 60 0AC AB AB ACF F N N° − ° + ° − ° =

0 :yF↑Σ = ( ) ( )sin 60 sin 30 22 0AB AC AB ACN N F F P+ ° + + ° − − =

0 :centerMΣ =� 0AC ABF r F r− =

AC ABF F= AC ABN N=

2 sin 60 2 sin 30 22AB ABP N F= ° + ° +

From a free-body diagram for the lower-left cylinder, the equilibrium equations give

0 :xF→Σ = cos30 cos60 0B AB ABF F N+ ° − ° =

0 :yF↑Σ = 22 sin 30 sin 60 0B AB ABN F N− − ° − ° =

0 :centerMΣ =� 0B ABF r F r− =

0.26795B AB ABF F N= =

Since the amount of friction required for equilibrium ( )0.26795AB ABF N= is less than the maximum amount of

friction available ( )max 0.4AB ABF N= , the cylinders will not slip at the surface between the cylinders regardless of the value of P. Then,

( )22 sin 30 3.73205 sin 60 0B AB ABN F F− − ° − ° =

0.26795 5.89488B AB BF F N= = −

and again the amount of friction required for equilibrium ( )0.26795 5.89488B BF N= − is less than the

maximum amount of friction available ( )max 0.4B BF N= so the cylinders will not slip at the lower surface regardless of the value of P. Therefore, the cylinders will not slip for any value of P and

maxP = ∞ ............................................................................................................................ Ans.

6-132 A 10-kg drum rests on a thin lightweight piece of cardboard as shown in Fig. P6-132. The coefficient of friction is the same at all surfaces.

(a) Plot P, the maximum force that may be applied to the cardboard without moving it, as a function of the coefficient of friction µs (0.05 ≤ µs ≤ 0.8).

(b) On the same graph, plot (Af)actual and (Bf)actual, the actual amounts of friction force that act at points A and B, and (Af)avail and (Bf)avail, the maximum amounts of friction force available for equilibrium at points A and B.

(c) What happens to the system at µs ≅ 0.364? (d) What does the solution for µs > 0.364 mean?


239

SOLUTION

( )10 9.81 98.1 NW = =

Impending motion of the cardboard is downward and friction acts to oppose the motion. From a free-body diagram of the cardboard, the equilibrium equations are

0 :nFΣ = 0C BN N− =

0 :tFΣ = 0B CF F P+ − =

C s CF Nµ=

From a free-body diagram of the cylinder, the equilibrium equations are

0 :nFΣ = 98.1cos 20 sin 40 0B AN F− ° − ° =

0 :tFΣ = 98.1sin 20 cos 40 cos50 0B A AF F N− − ° − ° + ° =

0 :centerMΣ =� 0A BF r F r− =

Therefore

A BF F= B CN N=

If impending slip at A, then A s A BF N Fµ= = and

98.1sin 20 cos 40 cos50 0s A s A AN N Nµ µ− − ° − ° + ° =

( )98.1sin 20

cos50 1 cos 40As

Nµ

°=° − + °

98.1cos 20 sin 40 sin 50B s A A CN N N Nµ= ° + ° + ° =

( ) ( )B C A C s A C s A BP F F F F N N N Nµ µ= + = + = + = + .......................................Ans.

( )maxA B A s AF F F Nµ= = = ............................................................................................Ans.

( )maxB s BF Nµ= ...............................................................................................................Ans.

These five functions are all plotted on the graph below. Note that the force ( )maxB BF F< for the entire range graphed. This verifies the guess that impending slip occurs first at A. (c) When the coefficient of friction is 0.364, the normal

force AN goes to infinity. That is, the system locks

at 0.364sµ = , and no amount of pulling force will be able to move the cardboard.

(d) For coefficients of friction greater than 0.364, the

friction and normal forces become negative. The best interpretation of the solution is that a pushing force will be necessary to move the cardboard � therefore, the friction force must be in the opposite direction.


240

6-133 A 150-lb uniformly-loaded file cabinet sits on an inclined surface as shown in Fig. P6-133. A horizontal force P acts on the cabinet, and the coefficient of friction between the cabinet and the inclined surface is 0.4.

(a) Determine Pmin and Pmax, the minimum and maximum horizontal forces for which the file cabinet will be in equilibrium.

(b) Calculate and plot N and F, the normal and friction forces acting on the bottom of the file cabinet, as functions of P (Pmin ≤ P ≤ Pmax).

(c) Calculate and plot d, the distance between the line of action of the normal force and the corner C, as a function of P (Pmin ≤ P ≤ Pmax).

SOLUTION Assuming that impending motion is up the incline, friction will act down the incline to oppose the motion. From a free-body diagram of the cabinet, the equilibrium equations are

0 :xF→Σ = sin 25 cos 25 0P N F− ° − ° =

0 :yF↑Σ = cos 25 sin 25 150 0N F° − ° − =

0 :CMΣ =� ( ) ( ) ( )( )150cos 25 9 150sin 25 24° + °

( ) ( ) ( )( )sin 25 18 cos 25 32 0Nd P P− + ° − ° =

(a) If 0.4F N= (impending slip up the incline), then

203.456 lbN = 81.382 lb F =

159.742 lbP = 3.306 in.d = − But this solution is not possible since the normal force acts at a point

outside the body. If 0d = (impending tip), then

190.168 lbN = 52.887 lb F =

max128.300 lbP P= = .....................................................................................................Ans.

If impending slip is down the incline, then the friction force acts up the incline (changes sign in the equilibrium equations above), 0.4F N= , and the equations of equilibrium give

139.489 lbN = 55.796 lb F =

8.383 lbP = 18.393 in.d = But this solution is also not possible since the normal force again acts at a point outside the body. If

18 in.d = (impending tip), then

140.287 lbN = 54.083 lb F =

min10.272 lbP P= = .......................................................................................................Ans.

(b) For 10.272 lb 128.300 lbP< < , the normal and friction forces are

cos 25 150sin 25F P= ° − ° ..........................................................................................Ans.

sin 25 150cos 25N P= ° + ° ..........................................................................................Ans.

(c) For 10.272 lb 128.300 lbP< < , the location of the normal force is

2744.9413 21.39472Pd

N−= .......................................................................................Ans.


241

6-134 A 35-kg child is sitting on a swing suspended by a rope that passes over a tree branch (Fig. P6-134). The coefficient of friction between the rope and the branch (which can be modeled as a flat belt over a drum) is 0.5 and the weight of the rope can be ignored. Determine the minimum force the child must exert on the rope to keep suspended.

SOLUTION

( )35 9.81 343.35 NW = =

From a free-body diagram of the child, the vertical equilibrium equation gives

0 :yF↑Σ = 1 2 343.35 0T T+ − =

which combined with the belt friction equation

0.52 1 14.81048T T e Tπ= =

gives

1 59.1 NT = ....................................................................................................................... Ans.

6-135 A rope attached to a 500-lb block passes over a frictionless pulley and is wrapped for one full turn around a fixed post as shown in Fig. P6-135. If the coefficient of friction between the rope and the post is 0.25, determine

(a) The minimum force P that must be used to keep the block from falling. (b) The minimum force P that must be used to begin to raise the block. SOLUTION

( )0.25 2500 Pe π= min103.9 lbP P= = ........................................................Ans.

( )0.25 2500P e π= max2410 lbP P= = ........................................................Ans.

6-136 A rope attached to a 220-kg block passes over a fixed drum (Fig. P6-136) If the coefficient of friction between the rope and the drum is 0.30, determine the minimum force P that must be used to

(a) Keep the block from falling. (b) Begin to raise the block. SOLUTION

( ) ( )0.3 2220 9.81 Pe π= min1347 NP P= = .........................................................Ans.

( ) ( )0.3 2220 9.81P e π= max3460 NP P= = ........................................................Ans.

6-137 A rotating shaft with a belt-type brake is shown in Fig. P6-137. The static coefficient of friction between the brake drum and the brake belt is 0.20. When a 75-lb force P is applied to the brake arm, rotation of the shaft is prevented. Determine the maximum torque that can be resisted by the brake if the shaft is tending to rotate

(a) Counterclockwise. (b) Clockwise.


242

SOLUTION From a free-body diagram for the pulley, the moment equilibrium equation gives

0 :BMΣ =� 1 26 6 0T T C− + =

( )2 16C T T= −

From a free-body diagram for the brake arm, the moment equilibrium equation gives

0 :AMΣ =� ( )2 19 3 27 75 0T T− − =

2 13 675T T− =

(a) If the couple C is counterclockwise, then 2 1T T> and the belt friction equation gives

0.22 1 11.87446T T e Tπ= =

1 145.997 lbT = 2 273.666 lbT =

766 lb in. = ⋅C � ............................................................................................................Ans.

(a) If 1 2T T> , then the belt friction equation gives

0.21 2 21.87446T T e Tπ= =

1 1124.130 lbT = 2 599.710 lbT =

3150 lb in. 3150 lb in. = − ⋅ = ⋅C � .............................................................................Ans.

6-138 The scaffolding of Fig. P6-138 is raised using an electric motor that sits on the scaffolding. Frictionless wheels at the ends of the scaffold restrict horizontal motion. The 250-mm-diameter pulley at the top is jammed and will not rotate. The coefficient of friction between the rope and the pulley is 0.25 and the weight of the motor, scaffold, and supplies is 2500 N. Determine the minimum torque that must be supplied by the motor

(a) To raise the scaffold at a constant rate. (b) To lower the scaffold at a constant rate. SOLUTION From a free-body diagram for the scaffolding, the vertical equilibrium equation gives

0 :yF↑Σ = 2500 0A BT T+ − =

(a) If A BT T> , then the belt friction equation gives

0.25 2.19328A B BT T e Tπ= =

1717.106 NAT = 782.894 NBT =

( )( )Torque 1717.106 0.075AT r= =

128.8 N m (to raise the scaffolding)= ⋅ ......................................................Ans.

(b) If B AT T> , then the belt friction equation gives

0.25 2.19328B A AT T e Tπ= =

782.894 NAT = 1717.106 NBT =

( )( )Torque 782.894 0.075AT r= =

58.7 N m (to lower the scaffolding)= ⋅ ......................................................Ans.


243

6-139 A rope is wrapped one full turn around each of two posts as shown in Fig. P6-139. If the coefficient of friction between the rope and the posts is 0.5, determine

(a) The ratio of TA to TB. (b) The ratio of TA to TB if only one post is used. SOLUTION

(a) ( )0.5 2 23.14069B C CT T e Tπ= =

( )0.5 223.14069 535.492A AT e Tπ = =

535B AT T = ...................................................................................................................... Ans.

(b) ( )0.5 2 23.14069B A AT T e Tπ= =

23.1B AT T = ..................................................................................................................... Ans.

6-140 A conveyor belt is driven with the 200-mm-diameter multiple-pulley drive shown in Fig. P6-140. Couples CA and CB are applied to the system at pulleys A and B, respectively. The angle of contact between the belt and a pulley is 225° for each 6.5 kg pulley, and the coefficient of friction is 0.30. Determine

(a) The maximum force T that can be developed by the drive and the magnitudes of the input couples CA and CB.

(b) The shearing stress on a cross section of each 25-mm diameter pin, if each pin is in double shear. SOLUTION

( )6.5 9.81 63.765 lbW = = ( ) ( )225 360 2 3.92699 radβ π= =

(a) ( )0.3 3.926992 6.49638 kNCT e= =

( )0.3 3.92699 21.101 kNCT T e= = .......................................................................................Ans.

From a free-body diagram for pulley A, the equilibrium equations give

0 :xF→Σ = 21,101 6496.38cos 45 0xA+ ° − =

0 :yF↑Σ = 63.765 6496.38sin 45 0yA − + ° =

0 :AMΣ =� ( ) ( )21,101 0.1 6496.38 0.1 0AC − + =

25,695 NxA = 4530 NyA = −

2 2 26,091 Nx yA A A= + =

1460 N mAC = ⋅ ..............................................................................................................Ans.

From a free-body diagram for pulley B, the equilibrium equations give

0 :xF→Σ = 2000 6496.38cos 45 0xB − − ° =

0 :yF↑Σ = 63.765 6496.38sin 45 0yB − − ° =

0 :BMΣ =� ( ) ( )6496.38 0.1 2000 0.1 0BC− − =

6594 NxB = 4657 NyB =

2 2 8073 Nx yB B B= + =

450 N mBC = ⋅ ................................................................................................................Ans.


244

(b) The shearing stress on the pins is obtained by dividing the forces on the pins by twice their cross sectional areas (since they are in double shear),

( )

6 22

26,091 26.6 10 N/m 26.6 MPa2 2 0.025 4A

As

VA

τπ

= = = × =

............................Ans.

( )

6 22

8073 8.22 10 N/m 8.22 MPa2 2 0.025 4B

Bs

VA

τπ

= = = × =

............................Ans.

6-141 The electric motor shown in Fig. P6-141 weighs 30 lb and delivers 50 lb-in. of torque to pulley A of a furnace blower by means of a V-belt. The effective diameters of the 36° pulleys are 5 in. The coefficient of friction is 0.30. Determine the minimum distance a to prevent slipping of the belt if the rotation of the motor is clockwise.

SOLUTION

( )enh0.3 0.9708

sin 36 2sµ = =

From a free-body diagram for the pulley A, the moment equilibrium equation

0 :AMΣ =� ( )( )2 150 2.5 0T T− − =

is combined with the belt friction equation

0.97082 1 121.1137T T e Tπ= =

to get

1 0.99435 lbT =

2 20.9944 lbT =

Then, from a free-body diagram for the motor, the moment equilibrium equation gives

0 :CMΣ =� ( ) ( )20.9944 10.5 0.99435 5.5 50 0a+ − =

4.52 in.a = ....................................................................................................................... Ans.

6-142 Complete the derivation of Eq. (6-20). SOLUTION Starting from

( )cos 2 2T Fθ∆ ∆ = ∆

( ) ( ) ( )2 sin 2 2 sin 2 sin 2P T Tα θ θ∆ = ∆ + ∆ ∆

in the limit as 0θ∆ →

2T F∆ = ∆

( ) ( )2 sin 2 2P T Tα θ θ∆ = ∆ + ∆ ∆

and therefore as 0θ∆ →

2 0P→ 0F∆ → 0T∆ →

Then, at the point of impending slip, sF Pµ∆ = ∆ . Divide through by θ∆ and let 0θ∆ → to get

( )( )0

22limsin 2ss T TPT

θ

µµθ θ α∆ →

+ ∆∆∆ = =∆ ∆


245

( )sin 2sTdT

dµ

θ α=

( )2

1 0sin 2T sT

dT dT

βµ θα

=∫ ∫

( )2

2 11

ln ln lnsin 2

sTT TT

µ βα

− = =

( )sin 22

1

sT eT

µ β α= .................................................................................................................Ans.

6-143 The band wrench of Fig. P6-143 is used to unscrew an oil filter from a car. (The filter acts as if it were a wheel with a resisting torque of 40 lb-ft.) Neglect the weight of the handle and friction between the end of the metal handle and the metal filter case. Determine

(a) The minimum coefficient of friction between the band and the filter that will prevent slippage. (b) The shearing stress on a cross section of the pin at B when slipping is impending if the ¼-in.-diameter pin

is loaded in double shear. SOLUTION

( )1tan 2.5 4.5 29.055φ −= = °

270 299.055 5.2195 radβ φ= ° + = ° =

(a) From a free-body diagram for the filter, the moment equilibrium equation gives

0 :CMΣ =� ( )( )2 140 2.5 12 0T T− − =

2 1 192.00 lbT T− =

From a free-body diagram for the wrench, the moment equilibrium equation gives

0 :PMΣ =� ( ) ( )1 29.5 sin 7.5 0T T φ− =

2 12.60819T T=

1 119.3886 lbT =

2 311.3886 lbT =

Then, the belt friction equation applied to the tensions on the filter gives

5.2195311.3886 119.3886 se µ=

0.1837sµ = ...................................................................................................................... Ans.

(b) Dividing the force on the pin at B by twice its cross sectional area (since it is in double shear) gives the shear stress

( )

322

311.3886 3.17 10 psi 3.17 ksi2 2 0.25 4

Bs

TA

τπ

= = = × =

.......................................Ans.

6-144 A uniform belt 2 m long with a mass of 2 kg hangs over a small fixed peg (Fig. P6-144). If the coefficient of friction between the belt and the peg is 0.40, determine the maximum distance d between the two ends for which the belt will not slip off the peg.

SOLUTION

1 1T W= 2 2T W=


246

( )1 2 2 9.81 19.62 NW W W+ = = =

1 2 2 mL L L+ = =

0.42 1 13.51359W We Wπ= =

1 4.34688 NW = 2 15.27312 NW =

The weight of each portion of the belt is directly proportional to its length. Therefore

1

1

19.622

WL

=

1 0.44311 mL = 2 2 0.44311 1.55689 mL = − =

2 1 1.114 md L L= − = ....................................................................................................Ans.

6-145 The hand brake of Fig. P6-145 is used to control the rotation of a drum. The coefficient of friction between the belt and the drum is µs = 0.35, and the weight of the handle may be neglected. If a clockwise torque of 350 lb-ft is applied to the drum, determine the minimum force P that must be applied to the handle to prevent motion when a = 4 in., 8 in., and 12 in.

SOLUTION From a free-body diagram for the drum, the moment equilibrium equation

0 :EMΣ =� ( ) ( )2 1 6 12 350 0T T− − =

is combined with the belt friction equation

0.352 1 13.00284T T e Tπ= =

to get

1 349.5043 lbT =

2 1049.504 lbT =

From a free-body diagram for the brake, the moment equilibrium equation gives

0 :BMΣ =� ( ) 230 0P a T a− − =

4 in.a = 161.5 lbP = .........................................................................................Ans.

8 in.a = 382 lbP = ............................................................................................Ans.

12 in.a = 700 lbP = ............................................................................................Ans.

6-146 The structure shown in Fig. P6-146 consists of a circular tie rod AB and a rigid member BC. If the structure is to support a load P = 40 kN, determine the required diameters of the pins at A, B, and C, and the required diameter of the tie rod. The tie rod is made of structural steel and the pins are made of 0.2% C hardened steel. All pins are in double shear. The tie rod is adequately reinforced around the pins so that tensile failure does not occur at the pins. Failure is by yielding, and the factor of safety is 1.3. Take the yield strength in shear to be one-half the yield strength in tension.

SOLUTION

sin 3 5θ = cos 4 5θ =

structural steel: 250 MPayσ = 200 GPaE =

0.2% hardened steel: 250 MPayσ =

Note that the tie rod is a two-force member. Then from a free-body diagram of the rigid member BC, the equilibrium equations give


247

0 :xF→Σ = cos 0x ABC T θ− =

0 :yF↑Σ = 40 sin 0y ABC T θ− + =

0 :AMΣ =� ( )( ) ( )sin 4 40 3.5 0ABT θ − =

58.333 kNABT =

46.667 kNxC = 5.000 kNyC =

2 2 46.934 kNx yC C C= + =

Dividing the force in the tie rod by its cross sectional area gives the normal stress

3 6

2

58.333 10 250 104 1.3

yABAB

AB

TA d FS

σσ

π× ×= = ≤ =

0.01965 m 19.65 mmABd ≥ = ......................................................................Ans.

The force on the pins are divided by twice their cross sectional area since they are in double shear

( )63

2

430 10 2258.333 102 1.32 4

y yAA

s A

VA FS FSd

τ στ

π××= = ≤ = =

0.01498 m 14.98 mmAd ≥ = ........................................................................Ans.

( )63

2

430 10 258.333 101.32 4B

Bdτ

π××= ≤

14.98 mmBd ≥ ...................Ans.

( )63

2

430 10 246.934 101.32 4C

Cdτ

π××= ≤

13.44 mmCd ≥ ...................Ans.

6-147 A rigid handle is used to twist a 2.5-in.-diameter valve, as shown in Fig. P6-147. A pair of oppositely directed parallel forces P = 100 lb is needed to twist the valve. Force is transmitted from the handle to the valve shaft by means of a 1.0-in.-long (into the page) key with a square cross section. The key is to be made of 0.8% C hot-rolled steel. If failure is by fracture, and the factor of safety is to be 2, determine the cross-sectional dimensions of the square key to the nearest 1/64 in. The yield and ultimate strengths in shear are one-half the corresponding values in tension.

SOLUTION

0.8% hot-rolled steel: 122 ksifσ =

From a free-body diagram of the handle, the moment equilibrium equation gives

0 :centerMΣ =� ( ) ( )100 24 1.25 0V− =

( )1920 lb 1V bτ= = ×

( )3122 10 221920

2f y

b FS FSτ σ ×

≤ = =

0.06295 in.b ≥

( ) ( )4 64 0.06295 5 64< <

min 5 64 in.b = .................................................................................................................Ans.


248

6-148 Each member of the truss of Fig. P6-148 has a circular cross section and is made of structural steel. If failure is by yielding, and the factor of safety is 2.5, determine the minimum permissible diameter of member EJ.

SOLUTION

( )1tan 3 2 56.310θ −= = ° 2 23 2 3.60555 mEJL = + =

structural steel: 250 MPayσ =

From a free-body diagram of the entire truss, the moment equilibrium equation gives

0 :AMΣ =� ( ) ( ) ( ) ( ) ( )8 30 8 30 6 30 4 30 2 0G − − − − =

75 kNG = Next, divide the truss with a section through members DE, EJ, and HJ, and draw a free-body diagram of the right-hand side. From a free-body diagram of the section, the vertical equilibrium equation gives

0 :yF↑Σ = 75 30 30 sin 0EJT θ− − − =

18.0277 kNEJT =

Dividing the force in the member by its cross-sectional area gives the normal stress

3 6

2

18.0277 10 250 104 2.5

yEJEJ

TA d FS

σσ

π× ×= = ≤ =

0.01515 m 15.15 mmd ≥ = Ans.

6-149 A pin-connected system of levers and bars is used as a toggle for a press to crush cans, as shown in Fig. P6-149. The system is designed to provide a crushing force of 550 lb. The handle may be considered rigid. The pin at D is to be made of 2024-T4 wrought aluminum. If failure is by yielding, and the factor of safety is to be 1.5, determine the required diameter of pin D to the nearest 1/32 in.

SOLUTION

2024-T4 wrought aluminum: 48 ksiyσ =

From a free-body diagram of the plunger, the vertical equilibrium equation gives

0 :yF↑Σ = 550 sin 67 0BDF− ° =

597.498 lbBDF =

Note that all three members attached to the pin D are two-force members. Therefore, from a free-body diagram of the pin, the equilibrium equations give

0 :xF→Σ = cos67 cos 78 0BD CD DEF F F° + ° − =

0 :yF↑Σ = sin 67 sin 78 0BD CDF F° − ° =

562.287 lbCDF = 350.367 lbDEF =

Therefore, the maximum shear force on pin D is 597.498 lbBDF = on the shear surface between BD and CD.

( )3

max 2

48 10 22597.4984 1.5

y y

d FS FSτ σ

τπ

×= ≤ = =

0.2181 in.d ≥

6 32 0.2181 7 32< < min 7 32 in.d = .................................................Ans.


249

6-150 A pair of vise grip pliers is shown in Fig. P6-150. All members of the pliers, except for the pins and spring, may be treated as rigid. It is anticipated that the largest applied force on the handles will be P = 200 N. If each of the three pins is in double shear and made of 0.4% C hardened steel, determine the required diameters for the pins to the nearest mm. In the analysis, neglect the spring force. Failure is by yielding, and the factor of safety is 1.25. The yield strength in shear is one-half the tensile yield strength.

SOLUTION

( )1tan 30 50 30.964θ −= = ° ( )1tan 15 35 23.199φ −= = °

0.4% C hardened steel: 360 MPayσ =

From a free-body diagram of the handle, the equilibrium equations give

0 :xF→Σ = cos 0AB xF Cθ − =

0 :yF↑Σ = sin 200 0AB yF Cθ − − =

0 :CMΣ =� ( ) ( )( ) ( )( )200 93 sin 28 cos 5 0AB ABF Fθ θ− + =

1838.212 NABF =

1576.250 NxC = 745.759 NyC =

2 2 1743.766 Nx yC C C= + =

Next, from a free-body diagram of the jaw, the equilibrium equations give

0 :xF→Σ = sin 0x xC D N φ− − =

0 :yF↑Σ = cos 0y yC D N φ− + =

0 :DMΣ =� ( ) ( ) ( )2 235 15 35 12 0x yN C C+ − − =

1683.817 NN =

912.951 NxD = 2293.426 NyD =

2 2 2468.457 Nx yD D D= + =

Dividing the force on the pins by twice their cross-sectional areas (since they are in double shear) gives the shear stresses

( )6

2

360 10 221838.2122 1.252 4

y yABB

s B

FA FS FSd

τ στ

π×

= = ≤ = =

0.00285 mBd ≥ min 3 mmBd = ............................................Ans.

( )6

2

360 10 21743.7662 1.252 4C

s C

CA d

τπ

×= = ≤

0.00222 mCd ≥ min 3 mmCd = ............................................Ans.

( )6

2

360 10 22468.4572 1.252 4D

s D

DA d

τπ

×= = ≤

0.00264 mDd ≥ min 3 mmDd = ............................................Ans.


250

6-151 A device for lifting rectangular objects such as bricks and concrete blocks is shown in Fig. P6-151. The coefficients of friction at all vertical contact surfaces are µs = 0.4 and µk = 0.3. The device is to lift two blocks, each weighing 15 lb. The pin at B is to be made of structural steel with a strength in shear equal to one-half the strength in tension. For failure by yielding and a factor of safety of 3, determine the minimum permissible diameter of the pin at B, which is in double shear.

SOLUTION

structural steel: 36 ksiyσ =

From a free-body diagram of the blocks, the equilibrium equations give

0 :yF↑Σ = 30 0A CF F+ − =

0 :GMΣ =� 8 8 0A CF F− =

15 lbA CF F= =

Next, from a free-body diagram of member AB, the equilibrium equations give

0 :xF→Σ = 0A xN B− =

0 :yF↑Σ = 0y AB F− =

0 :BMΣ =� 4 12 0A AN F− =

3 45 lbA AN F= =

45 lbxB = 15 lbyB =

2 2 47.434 lbx yB B B= + =

Note that 3A AN F= independent of the weight being lifted. Therefore, as long as the coefficient of friction is

greater than 1 3A AF N = , the device will lift the weight without slipping.


( )3

2

36 10 2247.4342 32 4

y yB

s B

BA FS FSd

τ στ

π×

= = ≤ = =

0.0709 in.Bd ≥ .................................................................................................Ans.

6-152 The flat roof of a building is supported by a series of parallel plane trusses spaced 2 m apart (only one such truss is shown in Fig. P6-152). Water of density 1000 kg/m3 may collect on the roof to a depth of 0.2 m. All members of the truss are to be the same size. If the truss members are made of structural steel, failure is by yielding, and the factor of safety is 3.0, determine the smallest diameter solid circular rod, to the nearest mm, that can be used. The members are braced so that buckling does not occur.


( ) ( ) ( )1 2 2 1000 9.81 2 1.2 0.2 2

2354.4 NF F W= = = × ×

=


0 :xF→Σ = 0xA =

0 :yF↑Σ = 16 0y yA F F+ − =


251

0 :AMΣ =� ( ) ( ) ( ) ( )( ) ( ) ( )1 1 13.6 2 1.2 2 2.4 3.6 0yF F F F− − − =

13 7063.20 Ny yA F F= = =


0 :xF→Σ = 0BCT =

0 :yF↑Σ = 1 0ABT F− − =

2354.4 N 2.35 kN (C)ABT = − ≅ ............................................. Ans.

0 kNBCT = .................................................................................... Ans.


0 :xF→Σ = ( )3 5 0AH ACT T+ =

0 :yF↑Σ = ( )7063.2 4 5 0AB ACT T+ + =

5886.0 N 5.89 kN (C)ACT = − ≅ ............................................. Ans.

3531.6 N 3.53 kN (T)AHT = ≅ ................................................ Ans.

From a free-body diagram for joint H, the equilibrium equations give

0 :xF→Σ = 0GH AHT T− =

0 :yF↑Σ = 0CHT =

0 kNCHT = ................................................................................... Ans.

3531.6 N 3.53 kN (T)GHT = ≅ ................................................ Ans.


0 :xF→Σ = 0DET− =

0 :yF↑Σ = 1 0EFT F− − =

0 kNDET = .................................................................................... Ans.

2354.4 N 2.35 kN (C)EFT = − ≅ ............................................. Ans.

From a free-body diagram for joint F, the equilibrium equations give

0 :xF→Σ = ( )3 5 0FG DFT T− − =

0 :yF↑Σ = ( )7063.2 4 5 0EF DFT T+ + =

5886.0 N 5.89 kN (C)DFT = − ≅ ............................................. Ans.

3531.6 N 3.53 kN (T)FET = ≅ ................................................ Ans.


0 :xF→Σ = ( )3 5 0DE CD DFT T T− + =

0 :yF↑Σ = ( )12 4 5 0DG DFF T T− − − =

3531.6 N 3.53 kN (C)CDT = − ≅ ...................................... Ans.

0 kNDGT = ............................................................................ Ans.


252

Finally, from a free-body diagram for joint G, the equilibrium equations give

0 :yF↑Σ = ( )4 5 0DG CGT T+ =

0 kNCGT = ............................................................................. Ans.

Dividing the largest force ( )5886.0 N by the cross sectional area of the members gives the normal stress in the member. For structural steel

250 MPayσ = . Therefore

6

max 2

5886.00 250 104 3.0

yPA d FS

σσ

π×= = ≤ =

0.00948 m 9.48 mmd ≥ =

min 10 mmd = ................................................................................................................... Ans.

6-153 The brake shown in Fig. P6-153 is used to control the motion of block B. The weight of the block is 2200 lb, the coefficient of static friction between the brake arm and drum is 0.35, and the coefficient of kinetic friction is 0.30. A force P is applied to the rigid brake arm such that the block moves downward at a constant rate. Determine the minimum permissible diameter for the pin at the end of the brake arm (to the nearest 1/8-in.) if it is in double shear and is made of 0.2% C hardened steel. Failure is by yielding, and the factor of safety is 2.5. The strength in shear is one-half the strength in tension.

SOLUTION

0.2% C hardened steel: 62 ksiyσ =

From a free-body diagram of the drum, the moment equilibrium equation gives

0 :CMΣ =� ( )7 3.5 2200 0F − =

1100 lb 0.35F N= =

3142.857 lbN = Next, from a free-body diagram of the brake handle, the equilibrium equations give

0 :xF→Σ = 0xF A− =

0 :yF↑Σ = 0yN A P− − =

0 :AMΣ =� 8 1.6 24 0N F P+ − =

1120.952 lbP =

1100 lbxA =

4263.809 lbyA =

2 2 4403.416 lbx yA A A= + =


( )3

2

62 10 224403.4162 2.52 4

y yAA

s

VA FS FSd

τ στ

π×

= = ≤ = =

0.47547 in.d ≥

3 8 0.47547 4 8< < min 4 8 1 2 in.d = = .........................................Ans.


253

6-154 Solid circular bars AB and BC of Fig. P6-154 are pinned at the ends, and the structure is subjected to a load P = 26 kN. The angle θ may vary, but pin A is always directly above pin C. Both bars are made of 6061-T6 wrought aluminum alloy. Failure is by yielding, and the factor of safety is 4. In the force analysis assume that the weights of the bars are negligible with respect to the applied loads and that the pins are adequately designed. Determine

(a) The angle θ for a minimum weight structure. (b) The diameters of the bars. (c) The weights of the bars. SOLUTION

6061-T6 wrought aluminum: 270 MPayσ = 32710 kg/mρ =

Note that both AB and BC are two-force members. Then, from a free-body diagram of the pin, the equilibrium equations give

0 :xF→Σ = cos 0BC ABF T θ− =

0 :yF↑Σ = sin 26 0ABT θ − =

( )26 sin kNABT θ= ( )cos 26 tan kNBC ABF T θ θ= =

Dividing the force in the members by their cross sectional areas gives the normal stresses

3 6

2

26 10 sin 270 104 4

yABAB

AB

TA d FS

σθσπ× ×= = ≤ =

322.1457 10 msinABd θ

−×≥

3 6

2

26 10 tan 270 104 4BC

BCdθσ

π× ×= ≤

322.1457 10 mtanBCd θ

−×≥

(a) To minimize the weight of the structure, we need to minimize the volume of the bars making up the structure.

( )1.8 cos mABL θ= 1.8 mBCL =

( )

( )

2 2

23

1.81.84 4 cos

1.8 22.1457 10 cos 14 sin sin cos

BC ABd dV π πθ

π θθ θ θ

−

= +

× = +

To minimize the volume, we set 0dV dθ =

( )23 2

2 2 2

1.8 22.1457 10 cos 1 11 04 sin sin cos

dVd

π θθ θ θ θ

−× = − − − + =

If neither sin 0θ = nor cos 0θ = , we can multiply through by ( )2 2sin cosθ θ− to get

2 2 4 2 2sin cos cos cos sin 0θ θ θ θ θ+ + − =

( ) ( )2 2 2 2cos sin cos 1 sin 0θ θ θ θ+ + − =

2 22cos sin 0θ θ− =

( )2 22 1 sin sin 0θ θ− − =

22 3sin 0θ− =

54.736θ = ° ...................................................................................................................... Ans.


254

324.5 10 m 24.5 mmABd−≥ × = ..................................................................................Ans.

318.62 10 m 18.62 mmBCd−≥ × = ..............................................................................Ans.

( ) ( )20.0245 1.82710 9.81 39.1 N4 cos54.736ABw

π = = ° ..................................Ans.

( ) ( ) ( )20.01862

2710 9.81 1.8 13.03 N4BCw

π = =

...............................................Ans.

6-155 Determine the force P required to pull the 250-lb roller over the step shown in Fig. P6-155. SOLUTION

( )1cos 9 12 41.410θ −= = °

When the roller is about to be pulled over the step, it no longer touches the floor and the force exerted on the roller by the floor is zero. From a free-body diagram of the roller, the equilibrium equations give

0 :xF→Σ = cos30 sin cos 0P N Fθ θ° − − =

0 :yF↑Σ = cos sin sin 30 250 0N F Pθ θ− + ° − =

0 :centerMΣ =� 0Fr =

0 lbF = 1.30931N P=

168.7 lbP = ..................................................................................................................... Ans.

6-156 A machine is activated by force T when a pedal is depressed with a 40-N force, as shown in Fig. P6-156. Determine the magnitude of T and the resultant bearing reaction at B.

SOLUTION From a free-body diagram of the pedal, the equilibrium equations give

0 :xF→Σ = 0xB =

0 :yF↑Σ = 40 0yB T− − =

0 :BMΣ =� ( ) ( ) ( )200 125cos 45 40 0T − ° =

17.68 N T = ↓ ........................................................... Ans.

0 NxB = 57.7 NyB =

57.7 N = ↑B .................................................................................................................. Ans.

6-157 The 500-lb block shown in Fig. P6-157 is supported by a ball-and-socket joint at A, by a smooth pin at B, and by a cable at C. Determine the components of the reactions at supports A and B and the force in the cable at C.

SOLUTION

:AΣ =M 0 ( ) ( )14 8 y zB B− + × +i k j k

( ) ( )7 9 4 500+ − + + × −i j k k

( ) ( )7 18 8 CT+ − + + ×i j k k

( ) ( )14 18 8 200+ − + + × − =i j k k 0


255

:x 8 18 8100 0y CB T− + − = 450 lbCT = ...............................................Ans.

:y 14 7 6300 0z CB T+ − = 225 lbzB = ...............................................Ans.

:z 14 0yB− = 0 lbyB = ....................................................Ans.

0 :xFΣ = 0xA = 0 lbxA = ....................................................Ans.

0 :yFΣ = 350 0y yA B+ + = 350 lbyA = − ............................................Ans.

0 :zFΣ = 700 0z z CA B T+ + − = 25 lbzA = ..................................................Ans.

6-158 The plate shown in Fig. P6-158 has a mass of 80 kg. The brackets at supports A and B exert only force reactions on the plate. Each of the brackets can resist a force along the axis of the pins in one direction only. Determine

(a) The reactions at supports A and B and the tension in the cable. (b) The change in length of the cable. (Use E = 200 GPa and d = 2.5 mm.) SOLUTION

(a) ( )80 9.81 784.800 NW = =

1.4cos30 1.21244 my∆ = ° =

1.2 1.4sin 30 1.9000 mz∆ = + ° =

2 2 21 1.21244 1.9000 2.46577 mCDL = + + =

( )2 2 2

1 1.21244 1.90001 1.21244 1.9000

0.40555 0.49171 0.77055 N

C C

C C C

T

T T T

− − +=+ +

= − − +

i j kT

i j k

From a free-body diagram of the plate, the equilibrium equations give

:BΣ =M 0 ( ) ( ) ( )1 1.2 2C y zA A+ × + × +i k T i j k

( ) ( )1 0.60622 0.35 784.8+ + − × − =i j k k 0

:x 0.59005 475.761 0CT − = 806.307 NCT =

:y 1.25721 2 784.800 0C zT A− − + = 114.449 NzA = −

:z 0.49171 2 0C yT A− + = 198.235 NyA =

0 :xFΣ = ( )0.40555 806.307 0xB − = 326.998 NxB =

0 :yFΣ = ( )0.49171 806.307 0y yA B+ − = 198.234 NyB =

0 :zFΣ = ( )0.77055 806.307 784.800 0z zA B+ + − = 277.949 NzB =

( )198.2 114.4 N= −A j k ...................... 806 NCT = ................................................Ans.

( )327 198.2 278 N= + +B i j k ...................................................................................Ans.

(b) The stretch of the cable CD is given by

( ) ( )

( ) ( )29

806.307 2465.772.03 mm

200 10 0.0025 4CD

PLEA

δπ

= = = ×

...........................................Ans.


256

6-159 The door to an airplane hangar consists of two uniform sections that are hinged at the middle as shown in Fig. P6-159. The door is raised by means of a cable attached to a bar along the bottom edge of the door. Smooth rollers at the ends of the bar C run in a smooth vertical channel. If the door is 30 ft wide, 15 ft tall, and weighs 1620 lb, determine

(a) The force P when the door opening height h = 8 ft. (b) The resultant hinge forces at A and B when h = 8 ft. SOLUTION

( )1sin 3.5 7.5 27.818θ −= = °

(a) From a free-body diagram of the door AB, the equilibrium equations are

0 :xF→Σ = 0x xA B− =

0 :yF↑Σ = 810 0y yA B− − =

0 :AMΣ =� ( ) ( ) ( )810 3.75cos 7.5cos 7.5sin 0y xB Bθ θ θ+ − =

From a free-body diagram of the door AB, the equilibrium equations are

0 :xF→Σ = 0x CB N− =

0 :yF↑Σ = 810 0yB P+ − =

0 :CMΣ =� ( ) ( ) ( )810 3.75cos 7.5cos 7.5sin 0y xB Bθ θ θ− − =

Combining the two moment equations gives

767.562 lbxB = 0 lbyB =

Then the force equations give

767.562 lbxA = 810 lbyA =

767.562 lbCN =

810 lbP = ........................................................................................................................ Ans.

(b) 2 2 1116 lbx yA A A= + = ...............................................................................................Ans.

2 2 768 lbx yB B B= + = .................................................................................................Ans.

6-160 Three bars are connected by smooth frictionless pins as shown in Fig. P6-160. Bar BCDE is rigid, bar AB is aluminum (E = 73 GPa; L = 1.5 m; d = 30 mm) and bar EF is steel (E = 200 GPa; L = 1.2 m; d = 20 mm). The 25-mm-diameter pivot pin D is aluminum and is in double shear. When P = 100 kN determine

(a) The deformation of bars AB and EF. (b) The shearing stress on a cross section of the pin at D. SOLUTION (a) From a free-body diagram of member BCDE,


0 :xF→Σ = 100 0EF x ABT D F− + − =

0 :yF↑Σ = 0yD =

0 :DMΣ =� ( )600 300 300 100 0AB EFF T+ − =

in which EFT is a tensile force and ABF is a compressive force. The stretch of EF and the shrink of AB are related by


257

600 300AB EFδ δ= 2AB EFδ δ=

( )

( ) ( )( )

( ) ( )2 29 9

1.5 1.22

73 10 0.03 4 200 10 0.02 4AB EFF T

π π=

× ×

1.31400AB EFF T=

Combining the equilibrium equations and the deformation equation gives

36.2183 kNABF = 27.5634 kNEFT =

91.3451 kNxD = 0 kNyD =

The deformation of the two rods are

( )( )

( ) ( )

3

29

36.2183 10 15001.053 mm (shrink)

73 10 0.03 4ABδ

π

×= =

×

............................................Ans.

( )( )

( ) ( )

3

29

27.5634 10 12000.526 mm (stretch)

200 10 0.02 4EFδ

π

×= =

×

........................................Ans.

(b) Finally, dividing the force on the pin by twice its cross sectional are (since it is in double shear) gives the shear stress

( )

36 2

2

91.3451 10 93.0 10 N/m 93.0 MPa2 2 0.025 4

Ds

DA

τπ

×= = = × =

............................Ans.

6-161 A person is holding a 20-lb object as shown in Fig. P6-161. Determine the force T in the biceps muscle and the force F of the humerus against the ulna, in terms of the weight W of the forearm which acts through G. For the position shown, both T and F act vertically.

SOLUTION From a free-body diagram of the forearm, the equilibrium equations give

0 :yF↑Σ = 20 0T F W− − − =

0 :FMΣ =� ( )1.5 5.5 11.5 20 0T W− − =

3.667 153.33 lbT W= + ................................Ans.

2.667 133.33 lbF W= + ...............................Ans.

6-162 Bodies A and B shown in Fig. P6-162 have masses of 2200 kg and 450 kg, respectively. The coefficient of friction between B and the horizontal surface is 0.2, between A and B it is 0.2, and between A and the fixed surface is 0.5. Determine the force P required to cause impending motion of body B to the left.

SOLUTION

( )2200 9.81 21,582.0 NAW = =

( )450 9.81 4414.5 NBW = =

For impending motion of block B to the left, the friction forces must act to the right to oppose the motion and

0.2B s B BF N Nµ= =

From a free-body diagram for block B, the equilibrium equations are


258

0 :xF→Σ = 0A BF F P+ − =

0 :yF↑Σ = 4414.5 0B AN N− − =

From a free-body diagram for block A, the equilibrium equations are

0 :xF→Σ = 0C AF F− =

0 :yF↑Σ = 21,582 0C AN N+ − =

0 :AMΣ =� ( ) ( )21,582 300 1500 0CN− =

Therefore

4316.4 NCN = 17, 265.6 NAN = A CF F=

21,680.1 NBN = ( )0.2 21,680.1 4336.02 NBF = =

0.2A s A AF N Nµ≤ = 0.5C s C CF N Nµ≤ =

Guessing that impending slip occurs at C before A gives

( )0.5 4316.4 2158.2 NC AF F= = =

6490 NA BP F F= + = ...................................................................................................Ans.

Since the maximum amount of friction available at A ( )max 0.2 17, 265.6 3453.12 NAF = = is greater than

the amount of friction required ( )2158.2 NAF = , the guess that slip occurs at C first is correct.

6-163 Three bars are connected by smooth frictionless pins as shown in Fig. P6-163. Bar BCD is rigid, bar AC is aluminum [E = 12,000 ksi; L = 36 in.; d = 1 in.; α = 12.5(10-6)/°F], and bar DE is steel [E = 30,000 ksi; L = 30 in.; d = ½ in.; α = 6.6(10-6)/°F]. The 5/8-in.-diameter pivot pin B is steel and is in single shear. If the temperature drops 20°F after the unit is assembled, determine

(a) The normal stresses in bars AC and DE. (b) The change in length of bars AC and DE. (c) The shearing stress on the cross section of the pin at B. SOLUTION

(a) From a free-body diagram of bar BCD, the equilibrium equations give

0 :xF→Σ = 0xB =

0 :yF↑Σ = 0y AC DEB T T+ − =

0 :AMΣ =� 20 30 0AC DET T− =

1.5AC DET T=

0xB = y DE ACB T T= −

in which ACT and DET are both tensile forces. The stretch of bar DE and the shrink of bar AC are related by

30 20

ACDE δδ =

1.5DE ACδ δ=


259

( )

( ) ( )( ) ( )( )6

26

306.6 10 20 30

30 10 .05 4DET

π−+ × −

×

( )

( ) ( )( )( )( )6

26

361.5 12.5 10 20 36

12 10 1 4ACT

π−

− = − × − ×

5.09296 5.72958 9540.0 lbDE ACT T+ =

Combining the equilibrium equations and the deformation equation gives

1045.493 lb (T)ACT = 696.995 lb (T)DET = 348.498 lbyB = −

( )2

1045.493 1331 psi (T)1 4

ACAC

TA

σπ

= = = ......................................................................Ans.

( )2

696.995 3550 psi (T)0.5 4

DEσπ

= = ...............................................................................Ans.

(b) ( )( )

( ) ( )( )( )( )6

26

1045.493 3612.5 10 20 36

12 10 1 4AC

TL T LEA

δ απ

−−−= − ∆ = − × − ×

0.00501 in. (shrink) ..............................................................................................Ans.

( ) ( )

( ) ( )( ) ( )( )6

26

696.995 306.6 10 20 30

30 10 0.5 4DE

TL T LEA

δ απ

−= + ∆ = + × − ×

0.00751 in. (stretch) .............................................................................................Ans.

(c) ( )

2 2

2348.498 1136 psi

5 8 4x y

Bs

B BA

τπ

+= = = ..................................................................Ans.

6-164 An adjustable bracket is held in place by a force P of magnitude 175 N as shown in Fig. P6-164. Determine the minimum coefficient of friction between the bracket and vertical square member if the bracket is to stay in place.

SOLUTION

From a free-body diagram of the bracket, the equilibrium equations are

0 :xF→Σ = 0B AN N− =

0 :yF↑Σ = 175 0A BF F+ − =

0 :AMΣ =� ( )75 300 417.5 175 0B BF N+ − =

and for impending slip at both A and B

A s AF Nµ= B s BF Nµ=

Therefore

A BN N= 87.5 NA BF F= = 221.667 NA BN N= =

87.5 0.395

221.667A

sA

FN

µ = = = ........................................................................................Ans.


260

6-165 Determine the force in members CD, DE, and FG of the truss shown in Fig. P6-165. SOLUTION

sin 3 5θ = cos 4 5θ =

( )1tan 5 20 14.036φ −= = °

From a free-body diagram of the entire truss, the equilibrium equations give

0 :xF→Σ = 800sin 600sin 0xG A θ θ− − − =

0 :yF↑Σ = 800cos 600cos 500 0yA θ θ− − − =

0 :AMΣ =� ( ) ( ) ( ) ( )10 800 10 600 17.5 500 20 0G − − − =

2850 lbG = 2010 lbxA = 1620 lbyA =

2 447.673 NN =

From a free-body diagram of the joint A, the equilibrium equations give

0 :xF→Σ = cos 2010 0ABT θ − =

0 :yF↑Σ = 1620 sin 0AG ABT T θ− − =

2512.50 lb (T)ABT =

112.50 lb (T)AGT =

From a free-body diagram of the joint G, the equilibrium equations give

0 :xF→Σ = 2850 cos cos 0BG FGT Tβ φ+ + =

0 :yF↑Σ = sin sin 0AG BG FGT T Tβ φ+ − =

( )1tan 4 8 26.565β −= = °

1229.83 lb 1230 lb (C)BGT = − ≅

1803.86 lb 1804 lb (C)FGT = − ≅ ................................................... Ans.

Finally, from a free-body diagram of the joint D, the equilibrium equations give

0 :xF→Σ = cos cos 0CD DET Tθ φ− − =

0 :yF↑Σ = sin sin 500 0CD DET Tθ φ+ − =

1250 lb (T)CDT = ............................................................................... Ans.

1031 lb 1031 lb (C)DET = − = ......................................................................................Ans.

6-166 Determine the change in length of members CF and FG of the truss shown in Fig. P6-166. Each member of the truss is made of structural steel, and is circular with a diameter of 25 mm.

SOLUTION structural steel: 200 GPaE =

4 mCF EF FGL L L= = =

4cos30 3.46410 mDEL = ° =


261

From a free-body diagram of the upper portion of the truss, the equilibrium equations give

0 :EMΣ =� ( ) ( )1200 3.46410 3.46410 0CFT− =

1200 NCFT =

0 :CMΣ =� ( ) ( )1200 3.46410 300 6.92820− −

( )( )sin 30 6.92820 0FGT− ° =

1800 NFGT = −

( )( )

( ) ( )29

1200 40000.0489 mm

200 10 0.025 4CF

TLEA

δπ

= = = ×

........................................Ans.

( )( )

( ) ( )29

1800 40000.0733 mm

200 10 0.025 4FGδ

π

−= = −

×

.................................................Ans.

6-167 The band brake of Fig. P6-167 is used to control the rotation of a drum. The coefficient of friction between the belt and the drum is 0.25, and the weight of the handle is negligible. Determine the force P that must be applied to the end of the handle to resist a maximum torque of 200 lb-in. if the torque is applied

(a) Clockwise. (b) Counterclockwise. SOLUTION From a free-body diagram of the drum, the moment equilibrium equation gives

0 :CMΣ =� 4 4 0A BT T M− − =

0.25A BT T M− =

From a free-body diagram of the brake handle, the moment equilibrium equation gives

0 :CMΣ =� ( )3 cos30 15 0AT P° − =

0.2 cos30AP T= °

270 3 2 radβ π= ° =

(a) If 200 lb in.M = + ⋅ (the drum rotates clockwise), then the tension AT must be bigger than BT and the belt friction equation gives

( )0.25 3 2 3.24819A B BT T e Tπ= =

72.240 lbAT = 22.240 lbBT =

12.51 lbP = ...................................................................................................................... Ans.

(b) If 200 lb in.M = − ⋅ (the drum rotates counterclockwise), then the tension BT must be bigger than AT and the belt friction equation gives

( )0.25 3 2 3.24819B A AT T e Tπ= = 22.240 lbAT = 72.240 lbBT =

3.85 lbP = ....................................................................................................................... Ans.

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