Vector Mechanisms for engineers

CCHHAAPPTTEERR 1133

PROPRIETARY MATERIAL. 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

507

PROBLEM 13.CQ1

Block A is traveling with a speed v0 on a smooth surface when the surface suddenly becomes rough with a coefficient of friction of causing the block to stop after a distance d. If block A were traveling twice as fast, that is, at a speed 2v0, how far will it travel on the rough surface before stopping?

(a) d/2

(b) d

(c) d2

(d) 2d

(e) 4d

SOLUTION

Answer: (e)


508

PROBLEM 13.1

A 400-kg satellite was placed in a circular orbit 1500 km above the surface of the earth. At this elevation the acceleration of gravity is 6.43 m/s2. Determine the kinetic energy of the satellite, knowing that its orbital speed is 325.6 10 km/h.

SOLUTION

Mass of satellite: 400 kgm =

Velocity: 3 325.6 10 km/h 7.111 10 m/sv = =

Kinetic energy: 2 3 21 1

(400 kg)(7.111 10 m/s)2 2

T mv= =

910.113 10 JT = 10.11 GJT =

Note: Acceleration of gravity has no effect on the mass of the satellite.


509

PROBLEM 13.2

A 1-lb stone is dropped down the bottomless pit at Carlsbad Caverns and strikes the ground with a speed of 95ft/s. Neglecting air resistance, determine (a) the kinetic energy of the stone as it strikes the ground and the height h from which it was dropped, (b) Solve Part a assuming that the same stone is dropped down a hole on the moon. (Acceleration of gravity on the moon = 5.31 ft/s2.)

SOLUTION

Mass of stone: 22

lb 1 lb0.031056 lb s /ft

32.2 ft/s

Wm

g= = =

Initial kinetic energy: 1 0 (rest)T =

(a) Kinetic energy at ground strike:

2 22 21 1

(0.031056)(95) 140.14 ft lb2 2

T mv= = =

2 140.1 ft lbT =

Use work and energy: 1 1 2 2T U T+ =

where 1 2

22

10

2

U wh mgh

mgh mv

= =

+ =

2 22 (95)

2 (2)(32.2)

vh

g= = 140.1 fth =

(b) On the moon: 25.31 ft/sg =

T1 and T2 will be the same, hence 2 140.1 ft lbT =

2 22 (95)

2 (2)(5.31)

vh

g= = 850 fth =


510

PROBLEM 13.3

A baseball player hits a 5.1-oz baseball with an initial velocity of 140 ft/s at an angle of 40 with the horizontal as shown. Determine (a) the kinetic energy of the ball immediately after it is hit, (b) the kinetic energy of the ball when it reaches its maximum height, (c) the maximum height above the ground reached by the ball.

SOLUTION

Mass of baseball: 1 lb

(5.1 oz) 0.31875 lb16 oz

W = =

22

0.31875 lb0.009899 lb s /ft

32.2 ft/s

Wm

g= = =

(a) Kinetic energy immediately after hit.

0 140 ft/sv v= =

2 211 1

(0.009899)(140)2 2

T mv= = 1 97.0 ft lbT =

(b) Kinetic energy at maximum height:

0 cos 40 140cos 40 107.246 ft/sv v= = =

2 221 1

(0.009899)(107.246)2 2

T mv= = 2 56.9 ft lbT =

Principle of work and energy: 1 1 2 2T U T+ =

1 2 2 1 40.082 ft lbU T T = =

Work of weight: 1 2U Wd =

Maximum height above impact point.

2 140.082 ft lb

125.7 ft0.31875 lb

T Td

W

= = =

125.7 ft

(c) Maximum height above ground:

125.7 ft 2 fth = + 127.7 fth =


511

PROBLEM 13.4

A 500-kg communications satellite is in a circular geosynchronous orbit and completes one revolution about the earth in 23 h and 56 min at an altitude of 35800 km above the surface of the earth. Knowing that the radius of the earth is 6370 km, determine the kinetic energy of the satellite.

SOLUTION

Radius of earth: 6370 kmR =

Radius of orbit: 66370 35800 42170 km 42.170 10 mr R h= + = + = =

Time one revolution: 23 h 56 mint = +

3(23 h)(3600 s/h) (56 min)(60 s/min) 86.160 10 st = + =

Speed: 6

3

2 2 (42.170 10 )3075.2 m/s

86.160 10

rv

t

= = =

Kinetic energy: 21

2T mv=

2 91

(500 kg)(3075.2 m/s) 2.3643 10 J2

T = =

2.36 GJT =


512

PROBLEM 13.5

In an ore-mixing operation, a bucket full of ore is suspended from a traveling crane which moves along a stationary bridge. The bucket is to swing no more than 10 ft horizontally when the crane is brought to a sudden stop. Determine the maximum allowable speed v of the crane.

SOLUTION

Let position be the position with bucket B directly below A, and position be that of maximum swing where d = 10 ft. Let L be the length AB.

Kinetic energies: 21 21

, 02

T mv T= =

Work of the weight: 1 2U Wh mgh = =

where h is the vertical projection of position above position

From geometry (see figure),

2 2

2 2

2 230 (30) (10)

1.7157 ft

y L d

h L y

L L d

= =

=

= =


21

02

mv mgh =

2 2 2 22 (2)(32.2 ft/s )(1.7157 ft) 110.49 ft /sv gh= = = 10.51 ft/sv =


513

PROBLEM 13.6

In an ore-mixing operation, a bucket full of ore is suspended from a traveling crane which moves along a stationary bridge. The crane is traveling at a speed of 10 ft/s when it is brought to a sudden stop. Determine the maximum horizontal distance through which the bucket will swing.

SOLUTION

Let position be the position with bucket B directly below A, and position be that of maximum swing where the horizontal distance is d. Let L be the length AB.

Kinetic energies: 21 21

, 02

T mv T= =

Work of the weight: 1 2U Wh mgh = =

where h is the vertical projection of position above position .


2

2 2

2

10

2

(10 ft/s)1.5528 ft

2 (2)(32.2 ft/s )

mv mgh

vh

g

=

= = =

From geometry (see figure),

2 2

2 2

2 2

( )

(30) (30 1.5528)

9.53 ft

d L y

L L h

=

=

= =

9.53 ftd =


514

PROBLEM 13.7

Determine the maximum theoretical speed that may be achieved over a distance of 110 m by a car starting from rest assuming there is no slipping. The coefficient of static friction between the tires and pavement is 0.75, and 60 percent of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels. Assume (a) front-wheel drive, (b) rear-wheel drive.

SOLUTION

Let W be the weight and m the mass. W mg=

(a) Front wheel drive: 0.60 0.60

0.75s

N W mg

= ==

Maximum friction force without slipping:

1 2

21 2 2

(0.75)(0.60 ) 0.45

0.45

10,

2

sF N W mg

U Fd mgd

T T mv

= = == =

= =


22

2 2 2 22

10 0.45

2

(2)(0.45 ) (2)(0.45)(9.81 m/s )(110 m) 971.19 m /s

mgd mv

v gd

+ =

= = =

2 31.164 m/sv = 2 112.2 km/hv =

(b) Rear wheel drive: 0.40 0.40

0.75s

N W mg

= ==

Maximum friction force without slipping:

1 2

21 2 2

(0.75)(0.40 ) 0.30

0.30

10,

2

sF N W mg

U Fd mgd

T T mv

= = == =

= =


22

2 2 2 22

10 0.30

2

(2)(0.30) (2)(0.30)(9.81 m/s )(110 m) 647.46 m /s

mgd mv

v gd

+ =

= = =

2 25.445 m/sv = 2 91.6 km/hv =

Note: The car is treated as a particle in this problem. The weight distribution is assumed to be the same for static and dynamic conditions. Compare with sample Problem 16.1 where the vehicle is treated as a rigid body.


515

PROBLEM 13.8

Skid marks on a drag racetrack indicate that the rear (drive) wheels of a car slip for the first 20 m of the 400-m track. (a) Knowing that the coefficient of kinetic friction is 0.60, determine the speed of the car at the end of the first 20-m portion of the track if it starts from rest and the front wheels are just off the ground. (b) What is the maximum theoretical speed for the car at the finish line if, after skidding for 20 m, it is driven without the wheels slipping for the remainder of the race? Assume that while the car is rolling without slipping, 60 percent of the weight of the car is on the rear wheels and the coefficient of static friction is 0.75. Ignore air resistance and rolling resistance.

SOLUTION

(a) For the first 20 m, the normal force at the real wheels is equal to the weight of the car. Since the wheels are skidding, the friction force is

k k kF N W mg = = =


22

22

2 2 2 22

10

21

02

2 (2)(0.6)(9.81 m/s )(20 m) 235.44 m /s

k

k

Fd mv

mgd mv

v gd

+ =

+ =

= = =

2 15.34 m/sv =

(b) Assume that for the remainder of the race, sliding is impending and N = 0.6 W

(0.6 ) (0.75)(0.6 ) 0.45s sF N W mg mg = = = =


2 22 31 1

(0.45 )2 2

mv mg d mv+ =

2 23 2

2 2 2

2 2

(2)(0.45)

235.44 m /s (2)(0.45)(9.81 m/s )(400 m 20 m)

3590.5 m /s

v v gd =

= +

=

3 59.9 m/sv =


516

PROBLEM 13.9

A package is projected up a 15 incline at A with an initial velocity of 8 m/s. Knowing that the coefficient of kinetic friction between the package and the incline is 0.12, determine (a) the maximum distance d that the package will move up the incline, (b) the velocity of the package as it returns to its original position.

SOLUTION

(a) Up the plane from A to B:

2 21 1 (8 m/s) 32 0

2 2

( sin15 ) 0.12 N

A A B

A B k

W WT mv T

g g

U W F d F N

= = = =

= = =

0 cos15 0 cos15F N W N W = = =

(sin15 0.12cos15 ) (0.3747)

: 32 (0.3743) 0

A B

A A B B

U W d Wd

WT U T Wd

g

= + =

+ = =

32

(9.81)(0.3747)d = 8.71 md =

(b) Down the plane from B to A: (F reverses direction)

2

2

10 8.71 m/s

2

( sin15 )

(sin15 0.12cos15 )(8.70 m/s)

1.245

10 1.245

2

A A B

B A

B A

B B A A A

WT v T d

g

U W F d

W

U W

WT U T W v

g

= = =

=

= =

+ = + =

2 (2)(9.81)(1.245)

24.43

4.94 m/s

A

A

v

v

=== 4.94 m/sA =v 15


517

PROBLEM 13.10

A 1.4 kg model rocket is launched vertically from rest with a constant thrust of 25 N until the rocket reaches an altitude of 15 m and the thrust ends. Neglecting air resistance, determine (a) the speed of the rocket when the thrust ends, (b) the maximum height reached by the rocket, (c) the speed of the rocket when it returns to the ground.

SOLUTION

Weight: (1.4)(9.81) 13.734 NW mg= = =

(a) First stage: 1 0T =

1 2 (25 13.734)(15) 169.0 N mU = =

1 1 2 2T U T+ =

22 1 21

169.0 N m2

T mv U = = =

1 222 (2) (169.0)

1.4

Uv

m= = 2 15.54 m/sv =

(b) Unpowered flight to maximum height h:

2 169.0 N mT = 3 0T =

2 3 ( 15)U W h =

2 2 3 3

2( 15)

T U T

W h T+ =

=

2169.0

1513.734

Th

W = = 27.3 mh =

(c) Falling from maximum height:

23 4 4

3 4

23 3 4 4 4

10

2

1: 0

2

T T mv

U Wh mgh

T U T mgh mv

= =

= =

= = + =

2 2 2 24 2 (2)(9.81 m/s )(27.3 m) 535.6 m /sv gh= = = 4 23.1 m/sv =


518

PROBLEM 13.11

Packages are thrown down an incline at A with a velocity of 1 m/s. The packages slide along the surface ABC to a conveyor belt which moves with a velocity of 2 m/s. Knowing that 0.25k = between the packages and the surface ABC, determine the distance d if the packages are to arrive at C with a velocity of 2 m/s.

SOLUTION

On incline AB: cos30

0.25 cos 30

sin 30

(sin 30 cos 30 )

AB

AB k AB

A B AB

k

N mg

F N mg

U mgd F d

mgd

= = = = =

On level surface BC: 7 mBC BC

BC k

B C k BC

N mg x

F mg

U mg x

= ===

At A, 21

and 1 m/s2A A A

T mv v= =

At C, 21

and 2 m/s2C C C

T mv v= =

Assume that no energy is lost at the corner B.

Work and energy. A A B B C CT U U T + + =

2 201 1

(sin 30 cos30 ) 2 2A k k BC

mv mgd mg x mv + =

Dividing by m and solving for d,

2 2

2 2

/2 /2

(sin 30 cos30 )

(2) /(2)(9.81) (0.25)(7) (1) /(2)(9.81)

sin 30 0.25cos30

C k BC A

k

v g x v gd

+ =

+ =

6.71 md =

bumblegrrl27Highlight

bumblegrrl27Sticky NoteI beliieve, it Should be Vc


519

PROBLEM 13.12

Packages are thrown down an incline at A with a velocity of 1 m/s. The packages slide along the surface ABC to a conveyor belt which moves with a velocity of 2 m/s. Knowing that

7.5 md = and 0.25k = between the packages and all surfaces, determine (a) the speed of the package at C, (b) the distance a package will slide on the conveyor belt before it comes to rest relative to the belt.

SOLUTION

(a) On incline AB:

cos 30

0.25 cos 30

sin 30

(sin 30 cos 30 )

AB

AB k AB

A B AB

k

N mg

F N mg

U mgd F d

mgd

= = = = =

On level surface BC: 7 m

BC BC

BC k

B C k BC

N mg x

F mg

U mg x

= ===

At A, 21

and 1 m/s2A A A

T mv v= =

At C, 21

and 2 m/s2C C C

T mv v= =

Assume that no energy is lost at the corner B.

Work and energy. A A B B C CT U U T + + =

2 201 1

(sin 30 cos30 ) 2 2A k k BC

mv mgd mg x mv + =

Solving for 2Cv ,

2 2

2

2 (sin 30 cos30 ) 2

(1) (2)(9.81)(7.5)(sin 30 0.25cos30 ) (2)(0.25)(9.81)(7)

C A k k BCv v gd g x = +

= +

2 28.3811 m /s= 2.90 m/sCv =

(b) Box on belt: Let beltx be the distance moves by a package as it slides on the belt.

0y y

x k k

F ma N mg N mg

F N mg = = =

= =

At the end of sliding, belt 2 m/sv v= =


520

PROBLEM 13.12 (Continued)

Principle of work and energy:

2 2belt belt

2 2belt

belt

2

1 1

2 2

2

8.3811 (2)

(2)(0.25)(9.81)

C k

C

k

mv mg x mv

v vx

g

=

=

= belt 0.893 mx =


521

PROBLEM 13.13

Boxes are transported by a conveyor belt with a velocity v0 to a fixed incline at A where they slide and eventually fall off at B. Knowing that 0.40,k = determine the velocity of the conveyor belt if the boxes leave the incline at B with a velocity of 8 ft/s.

SOLUTION

Forces when box is on AB.

0: cos15 0

cos15

yF N W

N W

= =

=

Box is sliding on AB.

cos15f k kF N W = =

Distance 20 ftAB d= =

Work of gravity force: ( ) sin15A B gU Wd =

Work of friction force: cos15f kF d Wd =

Total work (sin15 sin15 )A B kU Wd =

Kinetic energy: 20

2

1

2

1

2

A

B B

WT v

g

WT v

g

=

=

Principle of work and energy: A A B BT U T+ =

2 201 1

(sin15 cos15 )2 2k B

W Wv Wd v

g g+ =

2 20

2

2 2

2 (sin15 cos15 )

(8) (2)(32.2)(20)[sin15 (0.40)(cos15 )]

228.29 ft /s

B kv v gd =

=

=

0 15.11 ft/s=v 15


522

PROBLEM 13.14

Boxes are transported by a conveyor belt with a velocity v0 to a fixed incline at A where they slide and eventually fall off at B. Knowing that 0.40,k = determine the velocity of the conveyor belt if the boxes are to have zero velocity at B.

SOLUTION

Forces when box is on AB.

0: cos15 0

cos15

yF N W

N W

= =

=

Box is sliding on AB.

cos15f k kF N W = =

Distance 20 ftAB d= =

Work of gravity force: ( ) sin15A B gU Wd =

Work of friction force: cos15f kF d Wd =

Total work (sin15 cos15 )A B kU Wd =

Kinetic energy: 20

2

1

2

1

2

A

B B

WT v

g

WT v

g

=

=

Principle of work and energy: A A B BT U T+ =

2 201 1

(sin15 cos15 )2 2k B

W Wv Wd v

g g+ =

2 20

2 2

2 (sin15 cos15 )

0 (2)(32.2)(20)[sin15 (0.40)(cos15 )]

164.29 ft /s

B kv v gd = =

=

0 12.81 ft/s=v 15


523

PROBLEM 13.15

A 1200-kg trailer is hitched to a 1400-kg car. The car and trailer are traveling at 72 km/h when the driver applies the brakes on both the car and the trailer. Knowing that the braking forces exerted on the car and the trailer are 5000 N and 4000 N, respectively, determine (a) the distance traveled by the car and trailer before they come to a stop, (b) the horizontal component of the force exerted by the trailer hitch on the car.

SOLUTION

Let position 1 be the initial state at velocity 1 72 km/h 20 m/sv = = and position 2 be at the end of braking

2( 0).v = The braking forces and 5000 NCF = for the car and 4000 N for the trailer.

(a) Car and trailer system. ( braking distance)d =

21 1 2

1 2

1 1 2 2

1( ) 0

2( )

C T

C T

T m m v T

U F F d

T U T

= + =

= ++ =

211

( ) ( ) 02 C T C T

m m v F F d+ + =

2 21( ) (2600)(20) 57.778

2( ) (2)(9000)C T

C T

m m vd

F F

+= = =

+ 57.8 md =

(b) Car considered separately.

Let H be the horizontal pushing force that the trailer exerts on the car through the hitch.

21 1 2

1 2

1 1 2 2

10

2( )

C

C

T m v T

U H F d

T U T

= =

= + =

211

( ) 02 C C

m v H F d+ =

2 21 (1400)(20)5000

2 (2)(57.778)C

C

m vH F

d= =

Trailer hitch force on car: 154 N=H


bumblegrrl27Sticky NoteIt should be Ft. Force of the trailer.


524

PROBLEM 13.16

A trailer truck enters a 2 percent uphill grade traveling at 72 km/h and reaches a speed of 108 km/h in 300 m. The cab has a mass of 1800 kg and the trailer 5400 kg. Determine (a) the average force at the wheels of the cab, (b) the average force in the coupling between the cab and the trailer.

SOLUTION

Initial speed: 1 72 km/h 20 m/sv = =

Final speed: 2 108 km/h 30 m/sv = =

Vertical rise: (0.02)(300) 6.00 mh = =

Distance traveled: 300 md =

(a) Traction force. Use cab and trailer as a free body.

1800 5400 7200 kgm = + = 3(7200)(9.81) 70.632 10 NW mg= = =

Work and energy: 1 1 2 2T U T+ = 2 21 2

1 1

2 2tmv Wh F d mv + =

2 2 2 3 21 11 1 1 1 1

(7200)(30) (70.632 10 )(6.00) (7200)(20)2 300 2 2

tF mv Wh mvd = + = +

37.4126 10 N= 7.41 kNtF =

(b) Coupling force .cF Use the trailer alone as a free body.

35400 kg (5400)(9.81) 52.974 10 Nm W mg= = = =

Assume that the tangential force at the trailer wheels is zero.

Work and energy: 1 1 2 2T U T+ = 2 21 2

1 1

2 2cmv Wh F d mv + =

The plus sign before cF means that we have assumed that the coupling is in tension.

2 2 2 3 22 11 1 1 1 1 1

(5400)(30) (52.974 10 )(6.00) (5400)(20)2 2 300 2 2

cF mv Wh mvd = + = +

35.5595 10 N= 5.56 kN (tension)cF =


525

PROBLEM 13.17

The subway train shown is traveling at a speed of 30 mi/h when the brakes are fully applied on the wheels of cars B and C, causing them to slide on the track, but are not applied on the wheels of car A. Knowing that the coefficient of kinetic friction is 0.35 between the wheels and the track, determine (a) the distance required to bring the train to a stop, (b) the force in each coupling.

SOLUTION

0.35 (0.35)(100 kips) 35 kips

(0.35)(80 kips) 28 kipsk B

C

F

F

= = == =

1 30 mi/h 44 ft/sv = = 2 20 0v T= =

(a) Entire train: 1 1 2 2T U T+ =

22

1 (80 kips 100 kips 80 kips)(44 ft/s) (28 kips 35 kips) 0

2 32.2 ft/sx

+ + + =

124.07 ftx = 124.1 ftx =

(b) Force in each coupling: Recall that 124.07 ftx =

Car A: Assume ABF to be in tension

1 1 2 2

21 80 kips (44) (124.07 ft) 02 32.2

19.38 kips

AB

AB

T V T

F

F

+ =

=

= +

19.38 kips (tension)ABF =

Car C: 1 1 2 2T U T+ =

21 80 kips

(44) ( 28 kips)(124.07 ft) 02 32.2

28 kips 19.38 kips

BC

BC

F

F

+ =

=

8.62 kipsBCF = + 8.62 kips (tension)BCF =


526

PROBLEM 13.18

The subway train shown is traveling at a speed of 30 mi/h when the brakes are fully applied on the wheels of cars A, causing it to slide on the track, but are not applied on the wheels of cars A or B. Knowing that the coefficient of kinetic friction is 0.35 between the wheels and the track, determine (a) the distance required to bring the train to a stop, (b) the force in each coupling.

SOLUTION

(a) Entire train:

1

(0.35)(80 kips) 28 kips

30 mi/h 44 ft/s A AF N

v

= = == = 2 20 0v T= =

1 1 2 2

22

1 (80 kips 100 kips 80 kips)(44 ft/s) (28 kips) 0

2 32.2 ft/s

T V T

x

+ =+ + =

279.1 ftx = 279 ftx =

(b) Force in each coupling:

Car A: Assume ABF to be in tension

1 1 2 2T V T+ =

22

1 80 kips(44 ft/s) (28 kips )(279.1 ft) 0

2 32.2 ft/sABF + =

28 kips 8.62 kipsABF+ = +

19.38 kipsABF = 19.38 kips (compression)ABF =

Car C: 1 1 2 2T V T+ =

22

1 80 kips(44 ft/s) (279.1 ft) 0

2 32.2 ft/sBCF+ =

8.617 kipsBCF = 8.62 kips (compression)BCF =


527

PROBLEM 13.19

Blocks A and B weigh 25 lbs and 10 lbs, respectively, and they are both at a height 6 ft above the ground when the system is released from rest. Just before hitting the ground block A is moving at a speed of 9 ft/s. Determine (a) the amount of energy dissipated in friction by the pulley, (b) the tension in each portion of the cord during the motion.

SOLUTION

By constraint of the cable block B moves up a distance h when block A moves down a distance h. ( 6 ft)h = Their speeds are equal.

Let FA and FB be the tensions on the A and B sides, respectively, of the pulley.

Masses: 2

2

250.7764 lb s /ft

32.2

100.31056 lb s /ft

32.2

AA

BB

WM

g

WM

g

= = =

= = =

Let position 1 be the initial position with both blocks a distance h above the ground and position 2 be just before block A hits the ground.

Kinetic energies: 1 1( ) 0, ( ) 0A BT T= =

2 22

2 22

1 1( ) (0.7764)(9) 31.444 ft lb

2 21 1

( ) (0.31056)(9) 12.578 ft lb2 2

A A

B B

T m v

T m v

= = =

= = =


Block A: 1 2 ( )A AU W F h =

0 (25 )(6) 31.444AF+ = 19.759 lbAF =

Block B: 1 2 ( )B BU F W h =

0 ( 10)(6) 12.578BF+ = 12.096 lbBF =

At the pulley FA moves a distance h down, and FB moves a distance h up. The work done is

1 2 ( ) (19.759 12.096)(6) 46.0 ft lbA BU F F h = = =


528


Since the pulley is assumed to be massless, it cannot acquire kinetic energy; hence,

(a) Energy dissipated by the pulley: 46.0 ft lbpE =

(b) Tension in each portion of the cord: :19.76 lbA

:12.10 lbB


529

PROBLEM 13.20

The system shown is at rest when a constant 30 lb force is applied to collar B. (a) If the force acts through the entire motion, determine the speed of collar B as it strikes the support at C. (b) After what distance d should the 30 lb force be removed if the collar is to reach support C with zero velocity?

SOLUTION

Let F be the cable tension and vB be the velocity of collar B when it strikes the support. Consider the collar B. Its movement is horizontal so only horizontal forces acting on B do work. Let d be the distance through which the 30 lb applied force moves.

1 1 2 2

2

( ) ( ) ( )

1 180 30 (2 )(2)

2 32.2

B B B

B

T U T

d F v

+ =

+ =

230 4 0.27950 Bd F v = (1)

Now consider the weight A. When the collar moves 2 ft to the left, the weight moves 4 ft up, since the cable length is constant. Also, 2 .A Bv v=

1 1 2 2

2

2

( ) ( ) ( )

10 ( )(4)

2

1 64 (6)(4) (2 )

2 32.2

A A B

AA A

B

T U T

WF W v

g

F v

+ =

+ =

=

24 24 0.37267 BF v = (2)

Add Eqs. (1) and (2) to eliminate F.

230 24 0.65217 Bd v = (3)

(a) Case a: 2 ft, ?Bd v= =

2(30)(2) (24) 0.65217 Bv =

2 2 255.2 ft /sBv = 7.43 ft/sBv =

(b) Case b: ?, 0.Bd v= =

30 24 0d = 0.800 ftd =

bumblegrrl27HighlightI believe they mean A


530

PROBLEM 13.21

Car B is towing car A at a constant speed of 10 m/s on an uphill grade when the brakes of car A are fully applied causing all four wheels to skid. The driver of car B does not change the throttle setting or change gears. The masses of the cars A and B are 1400 kg and 1200 kg, respectively, and the coefficient of kinetic friction is 0.8. Neglecting air resistance and rolling resistance, determine (a) the distance traveled by the cars before they come to a stop, (b) the tension in the cable.

SOLUTION

0.8 AF N=

Given: Car B tows car A at 10 m/s uphill.

Car A brakes so 4 wheels skid. 0.8k =

Car B continues in same gear and throttle setting.

Find: (a) Distance d, traveled to stop

(b) Tension in cable

(a) 1F = traction force (from equilibrium)

1 (1400 )sin 5 (1200 )sin 5F g g= +

2600(9.81)sin 5=

For system: A B+

1 2 1[( 1400 sin 5 1200 sin 5 ) ]U F g g F d =

2 22 11 1

0 (2600)(10)2 2A B

T T m v+= = =

Since 1( 1400 sin 5 1200 sin 5 ) 0F g g =

0.8[1400(9.81)cos5 ] 130,000 N mFd d = =

11.88 md =

(b) Cable tension, T

1 2 2 1[ 0.8 ](11.88)AU T N T T = =

21400( 0.8(1400)(9.81)cos5 )11.88 (10)2

T =

( 10945) 5892T =

5.053 kN=

5.05 kNT =


531

PROBLEM 13.22

The system shown is at rest when a constant 250-N force is applied to block A. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between block A and the horizontal surface, determine (a) the velocity of block B after block A has moved 2 m, (b) the tension in the cable.

SOLUTION

Constraint of cable:

3 constant

3 0

3 0

A B

A B

A B

x y

x y

v v

+ = + =

+ =

Let F be the tension in the cable.

Block A: 130 kg, 250 N, ( ) 0A Am P T= = =

1 1 2 2

2

2

( ) ( ) ( )

10 ( )( )

21

0 (250 )(2) (30)(3 )2

A A A

A A A

B

T U T

P F x m v

F v

+ =

+ =

+ =

2500 2 135 BF v = (1)

Block B: 25 kg, 245.25 NB B Bm W m g= = =

1 1 2 2

2

2

( ) ( ) ( )

10 (3 )( )

22 1

(3 ) 245.25) (25)3 2

B B B

B B B B

B

T U T

F W y m v

F v

+ =

+ =

=

22 163.5 12.5 BF v = (2)


532



2

2 2 2

500 163.5 147.5

2.2814 m /s

B

B

v

v

=

=

(a) Velocity of B. 1.510 m/sB =v

(b) Tension in the cable.

From Eq. (2), 2 163.5 (12.5)(2.2814)F = 96.0 NF =


533

PROBLEM 13.23

The system shown is at rest when a constant 250-N force is applied to block A. Neglecting the masses of the pulleys and the effect of friction in the pulleys and assuming that the coefficients of friction between block A and the horizontal surface are 0.25s = and 0.20,k = determine (a) the velocity of block B after block A has moved 2 m, (b) the tension in the cable.

SOLUTION

Check the equilibrium position to see if the blocks move. Let F be the tension in the cable.

Block B: 3 0

(25)(9.81)81.75 N

3 3

B

B

F m g

m gF

=

= = =

Block A: 0: 0y A AF N m g = =

(30)(9.81) 294.3 NA AN m g= = =

0: 250 0x AF F F = =

250 81.75 168.25 NAF = =

Available static friction force: (0.25)(294.3) 73.57 Ns AN = =

Since ,A s AF N> the blocks move.

The friction force, FA, during sliding is

(0.20)(294.3) 58.86 NA k AF N= = =

Constraint of cable:

3 constant

3 0

3 0

A B

A B

A B

x y

x y

v v

+ = + =

+ =


534


Block A: 1

1 1 2 2

30 kg, 250 N, ( ) 0.

( ) ( ) ( )A A

A A A

m P T

T U T

= = =+ =

2

2

10 ( )( )

21

0 (250 58.86 )(2) (30)(3 )2

A A A A

B

P F F x m v

F v

+ =

+ =

2382.28 2 135 BF v = (1)

Block B: 25 kg, 245.25 NB B BM W m g= = =

1 1 2 2

2

2

( ) ( ) ( )

10 (3 )( )

22 1

(3 245.25) (25)3 2

B B B

B B B B

B

T U T

F W y m v

F v

+ =

+ =

=

22 163.5 12.5 BF v = (2)


2

2 2 2

382.28 163.5 147.5

1.48325 m /s

B

B

v

v

=

=

(a) Velocity of B: 1.218 m/sB =v (b) Tension in the cable:

From Eq. (2), 2 163.5 (12.5)(1.48325)F = 91.0 NF =


535

PROBLEM 13.24

Two blocks A and B, of mass 4 kg and 5 kg, respectively, are connected by a cord which passes over pulleys as shown. A 3 kg collar C is placed on block A and the system is released from rest. After the blocks have moved 0.9 m, collar C is removed and blocks A and B continue to move. Determine the speed of block A just before it strikes the ground.

SOLUTION

Position to Position . 1 10 0v T= =

At before C is removed from the system

2 2 22 2 2 2

1 2

21 2

1 2

1 1 2 2

1 1( ) (12 kg) 6

2 2( ) (0.9 m)

(4 3 5)( )(0.9 m) (2 kg)(9.81 m/s )(0.9 m)

17.658 J

:

A B C

A C B

T m m m v v v

U m m m g

U g

U

T U T

= + + = =

= +

= + ==

+ =

2 22 20 17.658 6 2.943v v+ = =

At Position , collar C is removed from the system.

Position to Position . 22 21 9

( ) kg (2.943) 13.244 J2 2A B

T m m v = + = =

2 23 3 31 9

( )( )2 2A B

T m m v v= + =

22 3

2 2 3 3

2 23 3

( )( )(0.7 m) ( 1 kg)(9.81 m/s )(0.7 m) 6.867 J

13.244 6.867 4.5 1.417

A BU m m g

T U T

v v

= = = + =

= =

3 1.190 m/sAv v= = 1.190 m/sAv =


536

PROBLEM 13.25

Four packages, each weighing 6 lb, are held in place by friction on a conveyor which is disengaged from its drive motor. When the system is released from rest, package 1 leaves the belt at A just as package 4 comes onto the inclined portion of the belt at B. Determine (a) the speed of package 2 as it leaves the belt at A, (b) the speed of package 3 as it leaves the belt at A. Neglect the mass of the belt and rollers.

SOLUTION

Slope angle: sin 6 ft

23.615 ft

= =

(a) Package falls off the belt and 2, 3, 4 move down

2 2 2

2 2 2 22

62 ft.

31 3 6 lb

3 0.27952 2 32.2 ft/s

T mv v v

=

= = =

1 2 (3)( )( ) (3)(6 lb)(2 ft) 36 lb ftU W R = = =

1 1 2 2 2 22 20 36 0.2795 128.8

T U T

v v+ =

+ = =

2 11.35 ft/s=v 23.6 (b) Package 2 falls off the belt and its energy is lost to the system and 3 and 4 move down 2 ft.

22 2 2

2

2 2 23 3 3 32

1 6 lb(2) (128.8)

2 32 ft/s

24 lb ft

1 6 lb(2) ( ) 0.18634

2 32.2 ft/s

T mv

T

T mv v v

= = =

= = =

2 3

2 2 3 3

(2)( )(2) (2)(6 lb)(2 ft) 24 lb ftU W

T U T

= = = + =

2 23 324 24 0.18634 257.6v v+ = = 3 16.05 ft/s=v 23.6


537

PROBLEM 13.26

A 3-kg block rests on top of a 2-kg block supported by, but not attached to, a spring of constant 40 N/m. The upper block is suddenly removed. Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block.

SOLUTION

Call blocks A and B. 2 kg, 3 kgA Bm m= =

(a) Position 1: Block B has just been removed.

Spring force: ( )S A BF m m g k x= + =

Spring stretch: 2

1( ) (5 kg)(9.81 m/s )

1.22625 m40 N/m

A Bm m gxk

+= = =

Let position 2 be a later position while the spring still contacts block A.

Work of the force exerted by the spring: 2

1

2

1

1 2

2 2 21 2

2 2 22 2

( )

1 1 1

2 2 2

1 1(40)( 1.22625) (40) 30.074 20

2 2

x

ex

x

x

U k x dx

k x k x k x

x x

=

= =

= =

Work of the gravitational force: 1 2 2 1

2 2

( ) ( )

(2)(9.81)( 1.22625) 19.62 24.059

g AU m g x x

x x

=

= + =

Total work: 21 2 2 220 19.62 6.015U x x = + +

Kinetic energies: 1

2 2 22 2 2 2

0

1 1(2)

2 2A

T

T m v v v

=

= = =


2 22 2 20 20 19.62 6.015x x v+ + =

Speed squared: 2 22 2 220 19.62 6.015v x x= + (1)

At maximum speed, 2

2

0dv

dx=


538


Differentiating Eq. (1), and setting equal to zero,

22 2

2

2 40 19.62 0

19.620.4905 m

40

dvv x

dx

x

= = =

= =

Substituting into Eq. (1), 2 2 2 22 (20)( 0.4905) (19.62)( 0.4905) 6.015 10.827 m /sv = + =

Maximum speed: 2 3.29 m/sv =

(b) Position 3: Block A reaches maximum height. Assume that the block has separated from the spring. Spring force is zero at separation.

Work of the force exerted by the spring:

1

02 2

1 3 11 1

( ) (40)(1.22625) 30.074 J2 2e x

U kxdx kx = = = = Work of the gravitational force:

1 3( ) (2)(9.81) 19.62 g AU m gh h h = = =

Total work: 1 3 30.074 19.62 U h =

At maximum height, 3 30, 0v T= =


0 30.074 19.62 0h+ =

Maximum height: 1.533 mh =


539

PROBLEM 13.27

Solve Problem 13.26, assuming that the 2-kg block is attached to the spring.

PROBLEM 13.26 A 3-kg block rests on top of a 2-kg block supported by, but not attached to, a spring of constant 40 N/m. The upper block is suddenly removed. Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block.

SOLUTION

Call blocks A and B. 2 kg, 3 kgA Bm m= =

(a) Position 1: Block B has just been removed.

Spring force: 1( )S A BF m m g kx= + =

Spring stretch: 2

1( ) (5 kg)(9.81 m/s )

1.22625 m40 N/m

A Bm m gxk

+= = =

Let position 2 be a later position. Note that the spring remains attached to block A.

Work of the force exerted by the spring: 2

1

2

1

1 2

2 2 21 2

2 2 22 2

( )

1 1 1

2 2 2

1 1(40)( 1.22625) (40) 30.074 20

2 2

x

ex

x

x

U kxdx

kx kx kx

x x

=

= =

= =

Work of the gravitational force: 1 2 2 1

2 2

( ) ( )

(2)(9.81)( 1.22625) 19.62 24.059

g AU m g x x

x x

=

= + =

Total work: 21 2 2 220 19.62 6.015U x x = +

Kinetic energies: 1

2 2 22 2 2 2

0

1 1(2)

2 2A

T

T m v v v

=

= = =


2 22 2 20 20 19.62 6.015x x v+ + =

Speed squared: 2 22 2 220 19.62 6.015v x x= + (1)

At maximum speed, 2

2

0dv

dx=


540


Differentiating Eq. (1) and setting equal to zero,

22 2

2

2

2 40 19.62 0

19.620.4905 m

40

dvv x

dx

x

= = =

= =

Substituting into Eq. (1), 2 2 1 22 (20)( 0.4905) (19.62)( 0.4905) 6.015 10.827 m /sv = + =

Maximum speed: 2 3.29 m/sv =

(b) Maximum height occurs when v2 = 0.

Substituting into Eq. (1), 22 20 20 19.62 6.015x x= +

Solving the quadratic equation

2 1.22625 m and 0.24525 mx =

Using the larger value, 2 0.24525 mx =

Maximum height: 2 1 0.24525 1.22625h x x= = + 1.472 mh =


541

PROBLEM 13.28

An 8-lb collar C slides on a horizontal rod between springs A and B. If the collar is pushed to the right until spring B is compressed 2 in. and released, determine the distance through which the collar will travel assuming (a) no friction between the collar and the rod, (b) a coefficient of friction 0.35.k =

SOLUTION

(a) 144 lb/ft

216 lb/ftB

A

k

k

==

Since the collar C leaves the spring at B and there is no friction, it must engage the spring at A.

2/12

0 0

22

0 0

144 lb/ft 2 216 lb/ftft ( )

2 12 2

A B

y

A B B A

A B

T T

U k xdx k xdx

U y

= =

=

=

2: 0 2 108 0

0.1361 ft 1.633 in.A A B BT U T y

y+ = + =

= =

Total distance 2 16 (6 1.633)d = +

13.63 in.d =

(b) Assume that C does not reach the spring at B because of friction.

6 lb

(0.35)(8 lb) 2.80 lb

0

f

A D

N W

F

T T

= == =

= =

2/12

0144 ( ) 2 2.80A D fU dx F y y = =

0 2 2.80 0

0.714 ft 8.57 in.A A D DT U T y

y+ = + =

= =

The collar must travel 16 6 2 12 in. + = before it engages the spring at B. Since 8.57 in.,y = it stops before engaging the spring at B.

Total distance 8.57 in.d =


542

PROBLEM 13.29

A 6-lb block is attached to a cable and to a spring as shown. The constant of the spring is k 8 lb/in.= and the tension in the cable is 3 lb. If the cable is cut, determine (a) the maximum displacement of the block, (b) the maximum speed of the block.

SOLUTION

8 lb/in. 96 lb/ftk = =

1

2 21 1 2 2 2

0: ( ) 6 3 3 lb

1 6 lb0 0: 0.09317

2 32.2

y sF F C

v T T v v

= = =

= = = =

For weight: 1 2 (6 lb) 6U x x = =

For spring: 21 20

(3 96 ) 3 48x

U x dx x x = + =

2 2

1 1 2 2 2

2 22

: 0 6 3 48 0.09317

3 48 0.09317

T U T x x x v

x x v

+ = + =

= (1)

(a) For 2, 0:mx v = 23 48 0x x =

3 1

0, ft48 16m

x x= = = 0.75 in.m =x


543


(b) For mv we see maximum of 2

1 2 3 48U x x =

1 23 1

3 96 0 ft ft96 32

dUx x

dx = = = =

Eq. (1): 2

21 13 ft 48 ft 0.0931732 32 m

v =

2 0.5031 0.7093 ft/sm mv v= = 8.51 in./sm =v

Note: 1 2U for the spring may be computed using 6F x curve

1 2

2

area

13 96

2

U

x x

=

= +


544

PROBLEM 13.30

A 10-kg block is attached to spring A and connected to spring B by a cord and pulley. The block is held in the position shown with both springs unstretched when the support is removed and the block is released with no initial velocity. Knowing that the constant of each spring is 2 kN/m, determine (a) the velocity of the block after it has moved down 50 mm, (b) the maximum velocity achieved by the block.

SOLUTION

(a) W = weight of the block 10 (9.81) 98.1 N= =

1

2B Ax x=

2 21 21 1

( ) ( ) ( )2 2A A A B B

U W x k x k x =

(Gravity) (Spring A) (Spring B)

21 21

(98.1 N)(0.05 m) (2000 N/m)(0.05 m)2

U =

21

(2000 N/m)(0.025 m)2

2 21 21 1

( ) (10 kg)2 2

U m v v = =

21

4.905 2.5 0.625 (10)2

v =

0.597 m/sv =

(b) Let distance moved down by the 10 kg blockx =

2

2 21 2

1 1 1( ) ( ) ( )

2 2 2 2A Bx

U W x k x k m v = =

21

( ) 0 ( ) (2 )2 8

BA

d km v W k x x

dx = =



bumblegrrl27Sticky NoteIt should be T2


545


2000

0 98.1 2000 ( ) (2 ) 98.1 (2000 250)8

x x x= = +

0.0436 m (43.6 mm)x =

For 21

0.0436, 4.2772 1.9010 0.4752 (10)2

x U v= = =

max 0.6166 m/sv =

max 0.617 m/sv =


546

PROBLEM 13.31

A 5-kg collar A is at rest on top of, but not attached to, a spring with stiffness k1 = 400 N/m; when a constant 150-N force is applied to the cable. Knowing A has a speed of 1 m/s when the upper spring is compressed 75 mm, determine the spring stiffness k2. Ignore friction and the mass of the pulley.

SOLUTION

Use the method of work and energy applied to the collar A.

1 1 2 2T U T+ =

Since collar is initially at rest, 1 0.T =

In position 2, where the upper spring is compressed 75 mm and 2 1.00 m/s,v = the kinetic energy is

2 22 21 1

(5 kg)(1.00 m/s) 2.5 J2 2

T mv= = =

As the collar is raised from level A to level B, the work of the weight force is

1 2( )gU mgh =

where 25 kg, 9.81 m/sm g= = and 450 mm 0.450 mh = =

Thus, 1 2( ) (5)(9.81)(0.450) 22.0725 JgU = =

In position 1, the force exerted by the lower spring is equal to the weight of collar A.

1 (5 kg)(9.81 m/s) 49.05 NF mg= = =

As the collar moves up a distance x1, the spring force is

1 1 2F F k x=

until the collar separates from the spring at

1

1

49.05 N0.122625 m 122.625 mm

400 N/mfF

xk

= = = =


547


Work of the force exerted by the lower spring:

1 2 1 1 10

2 2 2 21 1 1 1

2

( ) ( )

1 1 1

2 2 21

(400 N/m)(0.122625) 3.0074 J2

fx

f f f f f

U F k x dx

F x kx k x k x k x

=

= = =

= =

In position 2, the upper spring is compressed by 75 mm 0.075 m.y = = The work of the force exerted

by this spring is

2 21 2 2 2 2 21 1

( ) (0.075) 0.0028125 k2 2

U k y k = = =

Finally, we must calculate the work of the 150 N force applied to the cable. In position 1, the length AB is

2 21( ) (450) (400) 602.08 mmABl = + =

In position 2, the length AB is 2( ) 400 mm.ABl =

The displacement d of the 150 N force is

1 2( ) ( ) 202.08 mm 0.20208 mAB ABd l l= = =

The work of the 150 N force P is

1 2( ) (150 N)(0.20208 m) 30.312 JPU Pd = = =

Total work: 1 2 2

2

22.0725 3.0074 0.0028125 30.312

11.247 0.0028125

U k

k = + +

=


20 11.247 0.0028125 2.5k+ =

2 3110 N/mk = 2 3110 N/mk =


548

PROBLEM 13.32

A piston of mass m and cross-sectional area A is equilibrium under the pressure p at the center of a cylinder closed at both ends. Assuming that the piston is moved to the left a distance a/2 and released, and knowing that the pressure on each side of the piston varies inversely with the volume, determine the velocity of the piston as it again reaches the center of the cylinder. Neglect friction between the piston and the cylinder and express your answer in terms of m, a, p, and A.

SOLUTION

Pressures vary inversely as the volume

(2 ) (2 )

LL

RR

p Aa pap

P Ax xp Aa pa

pP A a x a x

= =

= =

Initially at ,

1

02

0

av x

T

= =

=

At , 221

, 2

x a T mv= =

1 2/2 /2

1 2 /2

1 2

22

1 2

21 1 2 2

1 1( )

2

[ln ln (2 )]

3ln ln ln ln

2 2

3 4ln ln ln

4 3

4 10 ln

3 2

a a

L Ra a

aa

U p p Adx paA dxx a x

U paA x a x

a aU paA a a

aU paA a paA

T U T paA mv

= = = +

= +

= =

+ = + =

( )432 2 ln 0.5754paA paAv

m m= = 0.759 paAv

m=


549

PROBLEM 13.33

An uncontrolled automobile traveling at 65 mph strikes squarely a highway crash cushion of the type shown in which the automobile is brought to rest by successively crushing steel barrels. The magnitude F of the force required to crush the barrels is shown as a function of the distance x the automobile has moved into the cushion. Knowing that the weight of the automobile is 2250 lb and neglecting the effect of friction, determine (a) the distance the automobile will move into the cushion before it comes to rest, (b) the maximum deceleration of the automobile.

SOLUTION

(a) 65 mi/h 95.3 ft/s=

Assume auto stops in 5 14 ft.d

1

2 21 1 2

1

2

2

1 2

1 1 2 2

95.33 ft/s

1 1 2250 lb(95.3 ft/s)

2 2 32.2 ft/s

317,530 lb ft

317.63 k ft

0

0

(18 k)(5 ft) (27 k)( 5)

90 27 135

27 45 k ft

317.53 27 45

v

T mv

T

v

T

U d

d

d

T U T

d

=

=

= = === + = + =

+ ==

13.43 ftd =

Assumption that 14 ftd is ok.

(b) Maximum deceleration occurs when F is largest. For 13.43 ft, 27 k.d F= = Thus, DF ma=

2

2250 lb(27,000 lb) ( )

32.2 ft/sDa

=

2386 ft/sDa =


550

PROBLEM 13.34

Two types of energy-absorbing fenders designed to be used on a pier are statically loaded. The force-deflection curve for each type of fender is given in the graph. Determine the maximum deflection of each fender when a 90-ton ship moving at 1 mi/h strikes the fender and is brought to rest.

SOLUTION

Weight: 31 (90 ton)(2000 lb/ton) 180 10 lbW = =

Mass: 3

2180 10 5590 lb s /ft32.2

Wm

g

= = =

Speed: 15280 ft

1 mi/h 1.4667 ft/s3600 s

v = = =

Kinetic energy: 2 21 1

2

1 1(5590)(1.4667)

2 26012 ft lb

0 (rest)

T mv

T

= =

= =


1 2

1 2

6012 0

6012 ft lb 72.15 kip in.

U

U

+ == =

The area under the force-deflection curve up to the maximum deflection is equal to 72.15 kip in.

Fender A: From the force-deflection curve max

max

605 kip/in.

12

FF kx k

x= = = =

20 0

1Area

2

x xfdx kx dx kx= = =

21

(5) 72.512

x =

2 228.86 in.x = 5.37 in.x = Fender B: We divide area under curve B into trapezoids

Partial area Total Area

From 0x = to 2 in.:x = 1

(2 in.)(4 kips) 4 kip in.2

= 4 kip in.

From 2 in.x = to 4 in.:x = 1

(2 in.)(4 10) 14 kip in.2

+ = 18 kip in.

From 4 in.x = to 6 in.:x = 1

(2 in.)(10 18) 28 kip in.2

+ = 46 kip in.


551


We still need 72.15 46 26.15 kip in.U = =

Equation of straight line approximating curve B from 6 in.x = to 8 in.x = is

18

18 62 30 18

x FF x

= = +

2

118 (6 ) 26.15 kip in.

2

( ) 6 8.716 0

1.209 in.

U x x x

x x

x

= + =

+ = =

Thus: 6 in. 1.209 in. 7.209 in.x = + = 7.21 in.x =


552

PROBLEM 13.35

Nonlinear springs are classified as hard or soft, depending upon the curvature of their force-deflection curve (see figure). If a delicate instrument having a mass of 5 kg is placed on a spring of length l so that its base is just touching the undeformed spring and then inadvertently released from that position, determine the maximum deflection mx of the spring and the maximum force mF exerted by the spring, assuming (a) a linear spring of constant k = 3 kN/m, (b) a hard, nonlinear spring, for which 2(3 kN/m)( 160 )F x x= + .

SOLUTION

(5 kg)

49.05 N

W mg g

W

= ==

Since 1 2 1 1 2 2 1 20, yields 0T T T U T U = = + = =

1 20 0

49.05 0m mx x

m mU Wx Fdx x Fdx = = = (1) (a) For (300 N/m)F kx x= =

Eq. (1): 0

49.05 3000 0mx

mx x dx =

249.05 1500 0m mx x = 332.7 10 m 32.7 mmmx

= =

33000 3000(32.7 10 )m mF x= = 98.1 Nm =F

(b) For 2(3000 N/m) (1 160 )F x x= +

Eq. (1) 30

49.05 3000( 160 ) 0mx

mx x x dx + =

2 41

49.05 3000 40 02m m m

x x x + =

(2)

Solve by trial: 330.44 10 mmx= 30.4 mmmx =

3 3 2(3000)(30.44 10 )[1 160(30.44 10 ) ]mF = + 104.9 Nm =F


553

PROBLEM 13.36

A rocket is fired vertically from the surface of the moon with a speed 0.v Derive a formula for the ratio /n uh h of heights reached with a speed v, if Newtons law of gravitation is used to calculate nh and a uniform gravitational field is used to calculate .uh Express your answer in terms of the acceleration of gravity mg on the surface of the moon, the radius mR of the moon, and the speeds v and 0.v

SOLUTION

Newtons law of gravitation

2 21 0 2

2

1 2 2

21 2 2

21 2

1 1 2 2

2 20

1 1

2 2

( )

1 1

1 1

2 2

m n

m

m n

m

R hm m

n nR

R h

m mR

m mm m n

mm m

m n

T mv T mv

mg RU F dr F

rdr

U mg Rr

U mg RR R h

T U T

Rmv mg R mv

R h

+

+

= =

= =

=

= +

+ =

+ = +

( )

( )2 20

2 20

22

gm

m

v vnm m

Rv vh

g R

=

(1)

Uniform gravitational field

2 21 0 2

1 2

2 21 1 2 2 0

1

2

( ) ( )

1 1

2 2

m n

m

R h

u m m u m uR

m u

T mv T mv

U F dr mg R h R mgh

T U T mv mg h mv

+

= =

= = + =

+ = =

( )2 20

2u m

v vh

g

= (2)

Dividing (1) by (2) ( )2 20(2 )

1

1m m

n

v vu

g R

h

h =


554

PROBLEM 13.37

Express the acceleration of gravity hg at an altitude h above the surface of the earth in terms of the acceleration of gravity 0g at the surface of the earth, the altitude h and the radius R of the earth. Determine the percent error if the weight that an object has on the surface of earth is used as its weight at an altitude of (a) 1 km, (b) 1000 km.

SOLUTION

( )

2

2 2

/

( ) 1

E Eh

hR

GM m GM m RF mg

h R= =

+ +

At earths surface, (h = 0) 02EGM m mg

R=

2

02 21

GME

E Rh

h

R

GMg g

R +

= =

Thus, 02

1

6370 km

hh

R

gg

R

+

=

=

At altitude h, true weight h TF mg W= =

Assumed weight 0 0W mg=

( )( )

( )

0 0 0

0 0 0

00 2

022

0

Error

1 11

11

T h h

g

hR

h hhRR

W W mg mg g gE

W mg g

g

gg E

g

= = = =

+ = = =

+ +

(a) h = 1 km: ( )2163701

1100 1001

P E = =

+ P = 0.0314%

(b) h = 1000 km: ( )2100063701

1100 1001

P E = =

+ P = 25.3%


555

PROBLEM 13.38

A golf ball struck on earth rises to a maximum height of 60 m and hits the ground 230 m away. How high will the same golf ball travel on the moon if the magnitude and direction of its velocity are the same as they were on earth immediately after the ball was hit? Assume that the ball is hit and lands at the same elevation in both cases and that the effect of the atmosphere on the earth is neglected, so that the trajectory in both cases is a parabola. The acceleration of gravity on the moon is 0.165 times that on earth.

SOLUTION

Solve for .mh

At maximum height, the total velocity is the horizontal component of the velocity, which is constant and the same in both cases.

2 21 2

1 2

1 2

1 1

2 2Earth

U Moon

H

e e

m m

T mv T mv

U mg h

mg h

= =

= =

Earth 2 21 1

2 2e e Hmv mg h mv =

Moon 2 21 1

2 2m m Hmv mg h mv =

Subtracting

0

(60 m)0.165

m ee e m m

e m

em

e

h gg h g h

h g

gh

g

+ = =

=

364 mmh =


556

PROBLEM 13.39

The sphere at A is given a downward velocity 0v of magnitude 5 m/s and swings in a vertical plane at the end of a rope of length 2 ml = attached to a support at O. Determine the angle at which the rope will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere.

SOLUTION

2 21 0

1

22

1 2

1 1(5)

2 212.5 m

1

2( )sin

T mv m

T

T mv

U mg l

= =

=

=

=

21 1 2 21

12.5 2 sin2

T U T m mg mv+ = + =

225 4 sing v+ = (1)

Newtons law at .

2 2

2

2 sin2

4 2 sin

v vmg mg m m

v g g

= =

=

(2)

Substitute for 2v from Eq. (2) into Eq. (1)

25 4 sin 4 2 sin

(4)(9.81) 25sin 0.2419

(6)(9.81)

g g g

+ = = = 14.00 =


557

PROBLEM 13.40

The sphere at A is given a downward velocity 0v and swings in a vertical circle of radius l and center O. Determine the smallest velocity 0v for which the sphere will reach Point B as it swings about Point O (a) if AO is a rope, (b) if AO is a slender rod of negligible mass.

SOLUTION

21 0

22

1 2

2 21 1 2 2 0

2 20

1

21

2

1 1

2 2

2

T mv

T mv

U mgl

T U T mv mgl mv

v v gl

=

=

=

+ = =

= +

Newtons law at

(a) For minimum v, tension in the cord must be zero.

Thus, 2v gl=

2 20 2 3v v gl gl= + = 0 3v gl=

(b) Force in the rod can support the weight so that v can be zero.

Thus, 20 0 2v gl= + 0 2v gl=


558

PROBLEM 13.41

A small sphere B of weight W is released from rest in the position shown and swings freely in a vertical plane, first about O and then about the peg A after the cord comes in contact with the peg. Determine the tension in the cord (a) just before the sphere comes in contact with the peg, (b) just after it comes in contact with the peg.

SOLUTION

Velocity of the sphere as the cord contacts A

2

0 0

1

2( )(1)

B B

C C

B C

B B C C

v T

T mv

U mg

T U T

= =

=

=+ =

21

0 12 C

mg mv+ =

2 (2)( )Cv g=

Newtons law

(a) Cord rotates about Point ( )O R L=

2

(cos60 ) Cv

T mg mL

=

(2)

(cos 60 )2

m gT mg= +

3

2T mg= 1.5 T W=


559


(b) Cord rotates about 2

LA R =

2

2

(cos 60 ) CL

mvT mg =

(2)( )

2 1

mg m gT = +

1 5

22 2

T mg mg = + =

2.5T W=


560

PROBLEM 13.42

A roller coaster starts from rest at A, rolls down the track to B, describes a circular loop of 40-ft diameter, and moves up and down past Point E. Knowing that h = 60 ft and assuming no energy loss due to friction, determine (a) the force exerted by his seat on a 160-lb rider at B and D, (b) the minimum value of the radius of curvature at E if the roller coaster is not to leave the track at that point.

SOLUTION

Let yp be the vertical distance from Point A to any Point P on the track. Let position 1 be at A and position

2 be at P. Apply the principle of work and energy.

1 0T = 2

21

2 PT mv=

1 2 PU mgy =

2

1 1 2 2

2

1: 0

2

2

P P

P P

T U T mgy mv

v gy

+ = + =

=

Magnitude of normal acceleration at P:

2 2

( ) P PP nP P

v gya

= =

(a) Rider at Point B.

60 ft

20 ft

(2 )(60)6

20

B

B

n

y h

r

ga g

= == =

= =

:F ma =

(6 )

7 7 (7)(160 lb)B

B

N mg m g

N mg W

== = =

1120 lbB =N


561


Rider at Point D.

2 20 ft

20 ft

(2 )(20)2

20

D

D

n

y h r

ga g

= ==

= =

:F ma =

(2 )

160 lbD

D

N mg m g

N mg W

+ == = =

160 lbD =N

(b) Car at Point E. 40 ft

0E

E

y h r

N

= ==

:

2n

E

E

F ma

gymg m

=

=

2E Ey = 80.0 ft =


562

PROBLEM 13.43

In Problem 13.42, determine the range of values of h for which the roller coaster will not leave the track at D or E, knowing that the radius of curvature at E is 75 ft. = Assume no energy loss due to friction.

PROBLEM 13.42 A roller coaster starts from rest at A, rolls down the track to B, describes a circular loop of 40-ft diameter, and moves up and down past Point E. Knowing that h = 60 ft and assuming no energy loss due to friction, determine (a) the force exerted by his seat on a 160-lb rider at B and D, (b) the minimum value of the radius of curvature at E if the roller coaster is not to leave the track at that point.

SOLUTION

Let yp be the vertical distance from Point A to any Point P on the track. Let position 1 be at A and position

2 be at P. Apply the principle of work and energy.

1 0T = 2

21

2 PT mv=

1 2 PU mg y =

2

1 1 2 2

2

1: 0

2

2

p P

P P

T U T mgy mv

v gy

+ = + =

=

Magnitude of normal acceleration of P:

2 2

( ) P PP nP P

v gya

= =

The condition of loss of contact with the track at P is that the curvature of the path is equal to p and the normal contact force 0.PN =

Car at Point D.

20 ft

2

2 ( 2 )( )

D

D

D n

r

y h r

g h ra

r

= ==

=

F ma =

2 ( 2 )

2 5

D

D

g h rN mg m

rh r

N mgr

+ =

=


563


For 0DN >

2 5 0

550 ft

2

h r

h r

>

> =

Car at Point E.

75 ft

20 ft

2 ( 20)( )

75

E

E

E n

y h r h

g ha

= == =

=

F ma =

2 ( 20)

75115 2

75

E

E

mg hN mg

hN mg

=

=

For 0, 115 2 0

57.5 ftEN h

h

> > 0a =

315 10 N; 25 m/sP RF F v= = =

Power: 3 3(15 10 )(25 m/s) 375 10PF v W= =

( 50 s)t > Power 375 kW=


576

PROBLEM 13.54

The elevator E has a weight of 6600 lbs when fully loaded and is connected as shown to a counterweight W of weight of 2200 lb. Determine the power in hp delivered by the motor (a) when the elevator is moving down at a constant speed of 1 ft/s, (b) when it has an upward velocity of 1 ft/s and a deceleration of 20.18 ft/s .

SOLUTION

(a) Acceleration 0=

Counterweight

Elevator

Motor

0: 0y W WF T W = = 0: 2 6600 0C WF T T = + =

2200 lbWT = 2200 lbCT = Kinematics: 2 , 2 , 2 2 ft/sE C E C C Ex x x x v v= = = =

(2200 lb)(2 ft/s) 4400 lb ft/s 8.00 hpC CP T v= = = =

8.00 hpP =

(b) 20.18 ft/sEa = , 1 ft/sEv = Counterweight

Elevator

Counterweight:

Vector Mechanisms for engineers

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Transcript of Vector Mechanisms for engineers