hibbAppC

618

APPENDIX

Review for the Fundamentals of Engineering Examination

Chapter 2—Review All SectionsC-1. Two forces act on the hook. Determine the magnitudeof the resultant force.

C-2. The force acts on the frame. Resolve thisforce into components acting along members AB and AC,and determine the magnitude of each component.

F = 450 lb

30�

40�

500 N

200 N

A

C

B

450 lb

45�

30�

Prob. C–2Prob. C–1

The Fundamentals of Engineering (FE) exam is given semiannually by the NationalCouncil of Engineering Examiners (NCEE) and is one of the requirements forobtaining a Professional Engineering License.A portion of this exam contains problemsin statics, and this appendix provides a review of the subject matter most often askedon this exam. Before solving any of the problems, you should review the sectionsindicated in each chapter in order to become familiar with the boldfaced definitions andthe procedures used to solve the various types of problems. Also, review the exampleproblems in these sections.

The following problems are arranged in the same sequence as the topics in eachchapter. Besides helping as a preparation for the FE exam, these problems also provideadditional examples for general practice of the subject matter. Solutions to all theproblems are given at the back of this appendix.

HIBBMCAPPC_0132215004_FPP 4/3/06 8:41 AM Page 618

C-3. Determine the magnitude and direction of theresultant force.

C-4. If determine themagnitude and coordinate direction angles of the force.

F = 530i + 50j - 45k6 N,

APPENDIX C REVIEW FOR THE FUNDAMENTALS OF ENGINEER ING EXAMINAT ION 619

y

x300 N

400 N

250 N

34

5

30�

F

z

y

x

Prob. C–4

Prob. C–3

C-5. The force has a component of 20 N directed along theaxis as shown. Represent the force F as a Cartesian vector.-y

C-6. The force acts on the beam as shown. Determine itscoordinate direction angles.

F

z

y

x

150�

70�

20 N

y

z

x 30�

F � 75 lb

45�

Prob. C–6

Prob. C–5


620 APPENDIX C REVIEW FOR THE FUNDAMENTALS OF ENGINEER ING EXAMINAT ION

C-7. The cables supporting the antenna are subjected to theforces shown. Represent each force as a Cartesian vector.

C-8. Determine the angle between the two cords.u

C-9. Determine the component of projection of the forceF along the pipe AB.

Chapter 3—Review Sections 3.1–3.3C-10. The crate at D has a weight of 550 lb. Determine theforce in each supporting cable.

x

y

B

A

2 ft

4 ft

F � {�20i �30j �60k} lb

3 ft

z

30�

4

35

A

B

C

D

Prob. C–10

Prob. C–9

100 ft

20 ft

x

y

z

30 ft

60 ft

10 ft

20 ft

F1 � 160 lb

F2 � 80 lb

F3 � 100 lb

Prob. C–7

y

x

z

A

B

2 m

2 m

2 m

4 m

u

2 m2 m

Prob. C–8



C

30�

P

B

A

4

3

5

Prob. C–13

C-11. The beam has a weight of 700 lb. Determine theshortest cable ABC that can be used to lift it if the maxi-mum force the cable can sustain is 1500 lb.

5 ft

0.5 ft

600 lb

20�

30�

O

Prob. C–14

45�

0.4 m

0.3 m

k � 200 N/m

Prob. C–12

10 ft

A C

B

u u

Prob. C–11

C-12. The block has a mass of 5 kg and rests on thesmooth plane. Determine the unstretched length of thespring.

C-13. The post can be removed by a vertical force of 400 lb. Determine the force P that must be applied to thecord in order to pull the post out of the ground.

Chapter 4—Review All SectionsC-14. Determine the moment of the force about point O.



C-15. Determine the moment of the force about point O.Neglect the thickness of the member.

C-16. Determine the moment of the force about point O.

C-17. Determine the moment of the force about point A.Express the result as a Cartesian vector.

2 m

6 m

5 m

1 m1 m

1 m

z

y

x

A

B

F � {30i � 40j � 50k} N

Prob. C–17

4 ft

2 ft

2 ft

14 ft

3 ft

5 ft

1 ft

z

y

x

B

A

C

F � 130 lb

Prob. C–18

50 N

60�

45�

100 mm

100 mm

200 mmO

Prob. C–15

500 N

3 m

O

45�

Prob. C–16

C-18. Determine the moment of the force about point A.Express the result as a Cartesian vector.



0.2 m

200 N

200 N

A

300 N300 N

400 N 400 N

3 m 2 m

Prob. C–19

C-19. Determine the resultant couple moment acting onthe beam.

C-20. Determine the resultant couple moment acting onthe triangular plate.

4 ft

4 ft 4 ft

300 lb

200 lb

200 lb

300 lb

150 lb

150 lb

Prob. C–20

34

5

50 N

200 N � m

30 N40 N

AB

3 m 3 m

Prob. C–21

C-21. Replace the loading shown by an equivalent resultantforce and couple-moment system at point A.

C-22. Replace the loading shown by an equivalent resultantforce and couple-moment system at point A.

A

3 ft 3 ft

4 ft

150 lb

200 lb

100 lb

Prob. C–22

HIBBMCAPPC_0132215004_FPP 4/3/06 8:41 AM 23


C-23. Replace the loading shown by an equivalent singleresultant force and specify where the force acts, measuredfrom point O.

C-24. Replace the loading shown by an equivalent singleresultant force and specify the x and y coordinates of its lineof action.

C-25. Replace the loading shown by an equivalent singleresultant force and specify the x and y coordinates of its lineof action.

500 lb 500 lb250 lb

O x

y

3 ft 3 ft 3 ft 3 ft

Prob. C–23

z

x

y

100 N

400 N

500 N

4 m

4 m3 m

Prob. C–24

C-26. Determine the resultant force and specify where itacts on the beam measured from A.

3 m2 m

3 m

3 m

1 m

1 m

z

y

x

2 m

200 N

200 N

100 N

100 N

Prob. C–25

A B

6 ft 8 ft

150 lb/ft

Prob. C–26





Chapter 5—Review Sections 5.1–5.6C-29. Determine the horizontal and vertical componentsof reaction at the supports. Neglect the thickness of thebeam.

4 m

w � 2.5x3

160 N/m

w

Ax

Prob. C–27

BA

6 ft 3 ft 3 ft

500 lb200 lb/ft

150 lb/ft

Prob. C–28

C-30. Determine the horizontal and vertical componentsof reaction at the supports.

BA

5 ft 5 ft 5 ft

500 lb

600 lb � ft43

5

Prob. C–29

A B

2 m 2 m

0.5 m400 N

500 N

Prob. C–30



C-31. Determine the components of reaction at the fixedsupport A. Neglect the thickness of the beam.

C-32. Determine the tension in the cable and the horizon-tal and vertical components of reaction at the pin A.Neglect the size of the pulley.

Chapter 6—Review Sections 6.1–6.4, 6.6C-34. Determine the force in each member of the truss.State if the members are in tension or compression.

60�

30�

3 m1 m 1 m 1 m 400 N

200 N 200 N 200 N

A

Prob. C–31

600 lb � ft

300 lb

4 ft

6 ft

B

D

C

4 ft 4 ft

A

Prob. C–32

zA

B C

y

x

200 lb

3 ft

2 ft

2 ft

Prob. C–33

D

A

C

B

2 ft 2 ft

300 lb

3 ft

Prob. C–34

C-33. The uniform plate has a weight of 500 lb. Determinethe tension in each of the supporting cables.


C-35. Determine the force in members AE and DC. Stateif the members are in tension or compression.

C-36. Determine the force in members BC, CF, and FE.State if the members are in tension or compression.


A CB

F E D

4 ft 4 ft

3 ft

800 lb

Prob. C–35

A DCB

G F E

4 ft

4 ft 4 ft 4 ft

600 lb 600 lb800 lb

Prob. C–36

C-37. Determine the force in members GF, FC, and CD.State if the members are in tension or compression.

C-38. Determine the force P needed to hold the 60-lbweight in equilibrium.

P

Prob. C–38

D

C

BA

G

F

E

500 lb

700 lb

1000 lb

8 ft

6 ft

6 ft

6 ft

Prob. C–37



C-39. Determine the horizontal and vertical componentsof reaction at pin C.

C-40. Determine the horizontal and vertical componentsof reaction at pin C.

C-41. Determine the normal force that the 100-lb plate Aexerts on the 30-lb plate B.

C-42. Determine the force P needed to lift the load. Also,determine the proper placement x of the hook forequilibrium. Neglect the weight of the beam.

3 ft3 ft

400 lb500 lb

3 ft3 ft

4 ft

B

A

C

Prob. C–39

A

B

C

400 N

800 N � m2 m1 m

1 m

1 m

1 m

Prob. C–40

4 ft

B

A

1 ft 1 ft

Prob. C–41

PB

C

A

0.9 m

100 mm 100 mm

100 mm

6 kN

x

Prob. C–42



Chapter 7—Review Section 7.1C-43. Determine the internal normal force, shear force,and moment acting in the beam at point B.

C-44. Determine the internal normal force, shear force,and moment acting in the beam at point B, which is locatedjust to the left of the 800-lb force.

1.5 m 1.5 m 1.5 m 1.5 m

8 kN3 kN/m

AB

C

Prob. C–43

CBA

3 ft3 ft

800 lb

6 ft 2 ft

400 lb

300 lb � ft

Prob. C–44

C-45. Determine the internal normal force, shear force,and moment acting in the beam at point B.

Chapter 8—Review Sections 8.1–8.2C-46. Determine the force P needed to move the 100-lbblock. The coefficient of static friction is and thecoefficient of kinetic friction is Neglect tipping.mk = 0.25.

ms = 0.3,

6 m 3 m

3 kN/m

A BC

30�

30�

P

Prob. C–46

Prob. C–45

C-47. Determine the vertical force P needed to rotate the200-lb spool. The coefficient of static friction at all contact-ing surfaces is ms = 0.4.

P

A

B

6 in.

12 in.

Prob. C–47



C-48. Block A has a weight of 30 lb and block B weighs 50 lb.If the coefficient of static friction is between all con-tacting surfaces, determine the frictional force at each surface.

ms = 0.4

C-49. Determine the force P necessary to move the 250-lbcrate which has a center of gravity at G. The coefficient ofstatic friction at the floor is ms = 0.4.

C-50. The filing cabinet A has a mass of 60 kg and centerof mass at G. It rests on a 10-kg plank. Determine thesmallest force P needed to move it. The coefficient of staticfriction between the cabinet A and the plank B is and between the plank and the floor ms = 0.3.

ms = 0.4,

Chapter 9—Review Sections 9.1–9.3(Integration is covered in the mathematics portion of theexam.)C-51. Determine the location ( ) of the centroid of thearea.

yx,

30�

10 lb

20 lb

B

A

Prob. C–48

1.5 ft 1.5 ft

2.5 ft

3.5 ft4.5 ft

P

A

G

0.2 m0.2 m

P

G

A

BC

1.3 m

0.8 m

1 m

2 ft

2 ft 3 ft 3 ft

y

x

3 ft

Prob. C–51

Prob. C–50

Prob. C–49



C-52. Determine the location ( ) of the centroid of thearea.

yx,

3 in. 3 in.

2 in.

8 in.

1 in.

0.5 in.

x

y

Prob. C–52

Chapter 10—Review Sections 10.1–10.5(Integration is covered in the mathematics portion of theexam.)C-53. Determine the moment of inertia of the cross-sectional area of the channel with respect to the y axis.

C-54. Determine the moment of inertia of the area withrespect to the x axis.

y

x

8 in.

12 in.

4 in.

8 in.

6 in.

Prob. C–54

150 mm150 mm

y

x

20 mm

20 mm100 mm

20 mm

Prob. C–53

C-55. Determine the moment of inertia of the cross-sectional area of the T-beam with respect to the axispassing through the centroid of the cross section.

x¿

x2 in.

8 in.

2 in.

x¿

8 in.

C

Prob. C–55



C–1.Ans.

C–2.

Ans.

Ans.

C–3.

Ans.

Ans.

C–4. Ans.

Ans.

Ans.

Ans.

C–5.

Ans.

C–6.

Ans.

Ans.

Ans.

C–7.

Ans.

Ans.

Ans.

C–8.

Ans.

C–9.

Ans.

C–10.

Ans.Ans.

C–11.

LABC = 2a 5 ftcos 13.5°

b = 10.3 ft

u = 13.5°-2115002 sin u + 700 = 0+ c ©Fy = 0;

FAC = 518 lbFAB = 478 lb

35

FAC + FAB sin 30° - 550 = 0+ c ©Fy = 0;

45

FAC - FAB cos 30° = 0:+ ©Fx = 0;

= 1-20i - 30j + 60k2 # a- 35

i -

45

jb = 36 lb

ƒ FAB ƒ = F # uAB

u = 90°

1-2i + 2j + 2k2 # 12i + 4j - 2k2212224

= 0

cos u =

rOA# rOB

ƒ rOA ƒ ƒ rOB ƒ

rOB = 52i + 4j - 2k6 m rOA = 5-2i + 2j + 2k6 m = 549.8i + 24.9j - 83.0k6 lb

F3 = 100 lb a 60120.4

i +

30120.4

j -

100120.4

kb = 57.81i - 15.6j - 78.1k6 lb

F2 = 80 lb a 10102.5

i -

20102.5

j -

100102.5

kb = 5-31.4i - 157k6 lb

F1 = 160 lb a - 20

102.0 i -

100102.0

kb

g = cos-1a -37.575b = 120°

b = cos-1a45.9375b = 52.2°

a = cos-1a45.9375b = 52.2°

Fz = -75 sin 30° = -37.5 Fy = 75 cos 30° cos 45° = 45.93 Fx = 75 cos 30° sin 45° = 45.93

= 57.90i - 20j + 8.42k6 N + 23.09 cos 68.61°k

F = 23.09 cos 70°i + 23.09 cos 150°j g = 68.61° 1From Fig. g 6 90°2

cos g = 41 - cos2 70° - cos2 150°

ƒ F ƒ = ` -20cos 150°

` = 23.09 N

Fy

ƒ F ƒ

= cos b

Fy = -20

g = cos-1a -4573.7b = 128°

b = cos-1a 5073.7b = 47.2°

a = cos-1a 3073.7b = 66.0°

F = 4302+ 502

+ 1-4522 = 73.7 N

u = tan-1

350446.4

= 38.1° a

FR = 41446.422 + 3502= 567 N

FRy = 400 sin 30° + 250a35b = 350 N

FRx = 300 + 400 cos 30° - 250a45b = 446.4 N

FAC = 636 lb

FAC

sin 45°=

450sin 30°

= 869 lb

FAB

sin 105°=

450sin 30°

= 666 N FR = 42002

+ 5002- 21200215002 cos 140°

Partial Solutions and Answers


PARTIAL SOLUTIONS AND ANSWERS 633

C–12.

Ans.

C–13. At A:

Ans.Ans.

C–14.Ans.

C–15.

Ans.

C–16.

Ans.

C–17.

Ans.

C–18.

Ans.

C–19.Ans.

Also,

Ans. = 740 N # m e+MCR

= 300152 - 400122 + 20010.22

+ 20010.22 = 740 N # m e+MCR

= ©MA = 400132 - 400152 + 300152 = 5160i - 780j - 300k6 lb # ft

MA = rAB * F = 3 i j k-3 -6 14-30 40 -120

3 = 5-30i + 40j - 120k6 lb

F = 130 lb a -

313

i +

413

j -

1213

kb

= 5-500i + 200j - 140k6 N # m

MA = rAB * F = 3 i j k1 6 530 40 -50

3 = 1.06 kN # m - 500 cos 45° 13 sin 45°2

d+MO = 500 sin 45° 13 + 3 cos 45°2 = 11.2 N # m - 50 cos 60° 10.2 sin 45°2

e+MO = 50 sin 60° 10.1 + 0.2 cos 45° + 0.12 = 2.49 kip # ft

d+MO = 600 sin 50° 152 + 600 cos 50° 10.52 TAC = 242 lb

P = 349 lb

45

P + TAC sin 30° - 400 = 0+ c ©Fy = 0;

35

P - TAC cos 30° = 0;+ ©Fx = 0;

l0 = 0.283 m43.35 = 20010.5 - l02 Fsp = k1l - l02;

Fsp = 43.35 N

45

1Fsp2 - 519.812 sin 45° = 0+Q©Fx = 0;C–20.

Ans.

C–21.

Ans.

Ans.

Ans.

C–22.

Ans.

Ans.

Ans.

C–23.

Ans.

Ans.

C–24.

Ans.

Ans.

Ans. x = 2.125 m

800x = 500142 - 100132 MRy = ©My ;

y = 4.50 m

-800y = -400142 - 500142 MRx = ©Mx ;

= 800 N

FR = 400 + 500 - 100 + TFR = ©Fz ;

x = 6 ft

12501x2 = 500132 + 250162 + 500192 +bFRx = ©MO ;

= 1250 lb

FR = 500 + 250 + 500+ TFR = ©Fy ;

MRA= 210 lb # ft

MAR=

35

11002142 -

45

11002162 + 150132 +bMAR

= ©MA ;

u = tan-1a 70140b = 26.6° d

FR = 41402+ 702

= 157 lb

FRy = 150 -

45

11002 = 70 lb+ TFRy = ©Fy ;

FRx = 200 -

35

11002 = 140 lb;+ FRx = ©Fx ;

= 470 N # m

MAR= 30132 +

35

1502162 + 200

+bMAR= ©MA ;

u = tan-1a10040b = 68.2° c

FR = 414022 + 110022 = 108 N

= 100 N

FRy = 40 + 30 +

35

1502+ TFRy = ©Fy ;

FRx =

45

1502 = 40 N:+ FRx = ©Fx ;

= 2600 lb # ft d+MCR

= 300142 + 200142 + 150142


C–25.Ans.

Ans.

Ans.

C–26. Ans.

Ans.

C–27. Ans.

Ans.

C–28.

Ans.

Ans.

C–29.

Ans.

Ans.

Ans. Ay = 140 lb

Ay + 260 - 500a45b = 0+ c ©Fy = 0;

By = 260 lb

By1102 - 500a45b152 - 600 = 0+g©MA = 0;

Ax = 300 lb

-Ax + 500a35b = 0:+ ©Fx = 0;

d = 5.03 ft

1550d = c12

1502162 d142 + [150162]132 + 500192+bMAR

= ©MA ; = 1550 lb

FR =

12

1502162 + 150162 + 500+ TFR = ©Fy ;

x =

Lxw1x2 dx

Lw1x2 dx

=

L

4

02.5x4 dx

160= 3.20 m

+bMAR= ©MA ;

FR =

Lw1x2 dx =

L

4

02.5x3 dx = 160 N

d = 8.36 ft

1650d = c12

16211502 d142 + [811502]1102 +bMAR

= ©MA ;

FR =

12

16211502 + 811502 = 1650 lb

x = 0.667 m+ 200122 - 200132

600x = 100132 + 100132MRy = ©My ;y = -0.667 m

+ 100132 - 100132-600y = 200112 + 200112 MRx = ©Mx ;

= 600 NFR = 200 + 200 + 100 + 100 + TFR = ©Fy ; C–30. Ans.

Ans.

Ans.

C–31.Ans.

Ans.

Ans.

C–32.

Ans.

Ans.

Ans.

C–33.

Ans.

C–34. Joint D:

Ans.

Ans.FAD = 400 lb 1C2-FAD +

45

15002 = 0;:+ ©Fx = 0;

FCD = 500 lb 1T235

FCD - 300 = 0;+ c ©Fy = 0;

TC = 100 lbTB = 250 lb,TA = 350 lb,

-TB142 - TC142 + 500122 + 200122 = 0©My = 0;

TA132 + TC132 - 50011.52 - 200132 = 0©Mx = 0;

TA + TB + TC - 200 - 500 = 0©Fz = 0;

Ay = -129 lb

Ay + 267.9 + a35b1267.92 - 300 = 0+ c ©Fy = 0;

Ax = 214 lb

Ax - a 45b1267.92 = 0:+ ©Fx = 0;

T = 267.9 = 268 lb

T142 +

35

T1122 - 300182 - 600 = 0+g©MA = 0;

MA = 3.90 kN # m

-400 sin 30°14.52 - 400 cos 30°13 sin 60°2 = 0

MA - 20012.52 - 20013.52 - 20014.52d+ ©MA = 0;

Ay = 800 N

Ay - 200 - 200 - 200 - 400 sin 30° = 0

+ c ©Fy = 0;

Ax = 346 N-Ax + 400 cos 30° = 0:+ ©Fx = 0;

Ay = 200 N

Ay + 300 - 500 = 0+ c ©Fy = 0;

By = 300 N

By142 - 40010.52 - 500122 = 0d+ ©MA = 0;

Ax = 400 N-Ax + 400 = 0;:+ ©Fx = 0;



Joint C:

Ans.

Ans.

Joint A:

Ans.

C–35.

Joint A:

Ans.

Joint C:

Ans.

C–36. Section truss through FE, FC, BC. Use the right segment.

Ans.

Ans.

Ans.

C–37. Section truss through GF, FC, DC. Use the top segment.

Ans.

Ans.

Ans.FCD = 750 lb 1C2FCD182 - 1000162 = 0d+ ©MF = 0;

FFC = 2125 lb 1C2-

45

FFC + 700 + 1000 = 0:+ ©Fx = 0;

FGF = 2025 lb 1T2FGF182 - 700162 - 10001122 = 0+g©MC = 0;

FBC = 2200 lb 1C2FBC142 - 600142 - 800182 = 0T + ©MF = 0;

FFE = 800 lb 1T2FFE142 - 800142 = 0+g©MC = 0;

FCF = 1980 lb 1T2FCF sin 45° - 600 - 800 = 0+ c ©Fy = 0;

FDC = 400 lb 1C2-FDC + 400 = 0;+ c ©Fy = 0;

FAE = 667 lb 1C2-

35

FAE + 400 = 0;+ c ©Fy = 0;

Ay = Cy = 400 lbAx = 0,

FAB = 0+ c ©Fy = 0;

FCB = 500 lb 1T2FCB - 500 = 0;+Q©Fx = 0;

FCA = 0+R©Fy = 0;

C–38.

Ans.

C–39.

Ans.

Ans.

C–40.

Ans.

Ans.

C–41. Plate A:

Plate B:

Ans.

C–42. Pulley C:

Beam:

Ans.

Ans.x = 0.333 m

2112 - 61x2 = 0+g©MA = 0;

P = 2 kN

2P + P - 6 = 0+ c ©Fy = 0;

T = 2PT - 2P = 0;+ c ©Fy = 0;

NAB = 35 lbT = 32.5 lb,

2T - NAB - 30 = 0+ c ©Fy = 0;

2T + NAB - 100 = 0+ c ©Fy = 0;

Cy = 400 N

-Cy + 1131.37 sin 45° - 400 = 0+ c ©Fy = 0;

Cx = 800 N

-Cx + 1131.37 cos 45° = 0:+ ©Fx = 0;

FAB = 1131.37 N

+ 400122 = 0

FAB cos 45°112 - FAB sin 45°132 + 800

+g©MC = 0;

Cy = 467 lb

Cy +

45

1541.672 - 400 - 500 = 0+ c ©Fy = 0;

Cx = 325 lb

-Cx +

35

1541.672 = 0:+ ©Fx = 0;

FAB = 541.67 lb

- a45b1FAB2192 + 400162 + 500132 = 0

+g©MC = 0;

P = 20 lb

3P - 60 = 0+ c ©Fy = 0;



C–43. Use segment AB.

Ans.

Ans.

Ans.

C–44. Use segment AB.

Ans.

Ans.

Ans.

C–45.Use segment AB.

Ans.

Ans.

Ans.

C–46.

Ans.

C–47.

Ans. P = 98.2 lb

P + 0.4NA + NB - 200 = 0+g©Fy = 0;

0.4NA1122 + NA1122 - P162 = 0+g©MB = 0;

0.4NB - NA = 0:+ ©Fx = 0;

P = 41.9 lb-P cos 30° + 0.3Nb = 0:+ ©Fx = 0;

Nb - P sin 30° - 100 = 0+ c ©Fy = 0;

MB = 15 kN # m

MB + c12

162122 d122 - 4.5162 = 0+g©MB = 0;

VB = 1.5 kN

4.5 -

12

162122 + VB = 0+ c ©Fy = 0;

NB = 0:+ ©Fx = 0;

wB = 2 kN>m.Ay = 4.5 kN,Ax = 0,

MB = 600 lb # ft

MB - 100162 = 0+g©MB = 0;

VB = 100 lb

100 - VB = 0+ c ©Fy = 0;

NB = 0:+ ©Fx = 0;

Ay = 100 lb.Ax = 0,

MB = 9.75 kN # m

MB + 311.5210.752 - 8.7511.52 = 0+g©MB = 0;

VB = 4.25 kN

8.75 - 311.52 - VB = 0+ c ©Fy = 0;

NB = 0:+ ©Fx = 0;

Ay = 8.75 kN. C–48. Block B:

Ans.

Blocks A and B:

Ans.

C–49. If slipping occurs:

If tipping occurs:

Ans.

C–50. P for A to slip on B:

P for B to slip:

P to tip A:

Ans.P = 90.6 N

P11.32 - 6019.81210.22 = 0d+ ©MC = 0;

P = 206 N

0.31686.72 - P = 0:+ ©Fx = 0;

NB = 686.7 N

NB - 6019.812 - 1019.812 = 0+ c ©Fy = 0;

P = 235 N

0.41588.62 - P = 0:+ ©Fx = 0;

NA = 588.6 N

NA - 6019.812 = 0+ c ©Fy = 0;

P = 83.3 lb

-P14.52 + 25011.52 = 0d+ ©MA = 0;

P = 100 lb

P - 0.412502 = 0:+ ©Fx = 0;

NC = 250 lb

NC - 250 lb = 0+ c ©Fy = 0;

FA = 27.3 lb 160.4190 lb22 FA - 20 cos 30° - 10 = 0:+ ©Fx = 0;

NA = 90 lb

NA - 30 - 50 - 20 sin 30° = 0+ c ©Fy = 0;

FB = 17.3 lb 160.4160 lb22 FB - 20 cos 30° = 0:+ ©Fx = 0;

NB = 60 lb

NB - 20 sin 30° - 50 = 0+ c ©Fy = 0;



C–51.

Ans.

Ans.

C–52. Ans.

Ans.

C–53.

Ans.

C–54.

Ans.

C–55.

Ans. + c 112

1821223 + 218219 - 6.522 d = 291 in4

Ix¿= ©1I + Ad22 = c 1

12 1221823 + 18212216.5 - 422 d

x =

©x'

A

©A=

4182122 + 91221828122 + 2182 = 6.5 in.

+ c 112

16211223 + 16211221-222 d = 5760 in4

I = ©1I + Ad22 = c 112

18211223 + 18211221622 d

= 124 11062 mm4

Iy =

112

11202130023 -

112

11002126023

y =

©y'

A

©A=

4111822 + 91621221182 + 6122 = 7 in.

x = 0 1symmetry2

1122122 + 1.5132132 + 1a12b132132

2122 + 3132 +

12

132132= 1.26 ft

y =

©y'

A

©A=

1-12122122 + 1.5132132 + 4a12b132132

2122 + 3132 +

12

132132= 1.57 ft

x =

©x'

A

©A=



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