Solutions to Assigned Problems Chapter 4

 10. (a)The 20.0 kg box resting on the table has the free-body diagram shown. Its F weight N is mg =(20.0 kg)( 9.80m s2 ) = 196 N . Since the box is at rest, the net force on the box must be 0, and so the normal force must also be 196 N .   mg (b)Free-body diagrams are shown for both boxes. F12 is the force on box 1 (the top box) due to box 2 (the bottom box), and is the normal force on box    FN1= F 12 1. F21 is the force on box 2 due to box 1, and has the same magnitude   Top box (#1) as F12 by Newton’s third law. FN2 is the force of the table on box 2. That is the normal force on box 2. Since both boxes are at rest, the  m g net force on each box must be 0. Write Newton’s second law in the 1 vertical direction for each box, taking the upward direction to be positive.  F= F - m g = 0 1 N1 1 FN2 2 Bottom FN1= m 1 g =(10.0 kg)( 9.80 m s) = 98.0 N = F 12 = F 21 box (#2) F= F - F - m g = 0 2 N2 21 2   m g 2 2 F21 FN2= F 21 + m 2 g =98.0 N +( 20.0 kg)( 9.80 m s) = 294 N

12. Choose up to be the positive direction. Write Newton’s second law for the  F vertical T direction, and solve for the tension force. F= F - mg = ma � F + m g a  T T ( ) mg 2 2 4 FT =(1200 kg)( 9.80m s + 0.70 m s) = 1.3 10 N

16. In both cases, a free-body diagram for the elevator would look like the  F adjacent T diagram. Choose up to be the positive direction. To find the MAXIMUM tension, assume that the acceleration is up. Write Newton’s second law for the elevator.  mg F= ma = FT - mg

2 FT = ma + mg = m( a + g) = m(0.0680 g + g ) = ( 4850 kg)( 1.0680)( 9.80 m s ) = 5.08 104 N To find the MINIMUM tension, assume that the acceleration is down. Then Newton’s second law for the elevator becomes the following.

F= ma = FT - mg � F T + ma = mg + m( a = g) - m( + 0.0680 g g ) =(4850 kg)( 0.9320)( 9.80 m s2) = 4.43 10 4 N

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18. There will be two forces on the person – their weight, and the normal force of the scales pushing up on the person. A free-body diagram for the person is shown. Choose up to be the positive direction, and use Newton’s second law to find the acceleration.   F= FN - mg = ma � 0.75 mg = mg ma mg FN a= -0.25 g = - 0.25( 9.8 m s2) = - 2.5m s 2 Due to the sign of the result, the direction of the acceleration is down . Thus the elevator must have started to move down since it had been motionless.

32. The window washer pulls down on the rope with her hands with a tension force F , so the rope pulls up on her hands with a tension force F . The T T  tension in the rope is also applied at the other end of the rope, where it FT  FT attaches to the bucket. Thus there is another force FT pulling up on the

bucket. The bucket-washer combination thus has a net force of 2FT upwards. See the adjacent free-body diagram, showing only forces on the bucket-washer combination, not forces exerted by the combination (the pull down on the rope by the person) or internal forces (normal force of bucket on person).  (a)Write Newton’s second law in the vertical direction, with up as positive. mg The net force must be zero if the bucket and washer have a constant speed.

F= FT + F T - mg = 0 � 2 F T mg

1 1 2 FT =2 mg = 2 (72 kg)( 9.80 m s) = 352.8 N 350 N

(b)Now the force is increased by 15%, so FT =358.2 N( 1.15) = 405.72 N. Again write Newton’s second law, but with a non-zero acceleration.

F= FT + F T - mg = ma

2 2F- mg 2( 405.72 N) - ( 72 kg)( 9.80 m s ) a =T = = 1.47 m s2 1.5m s 2 m 72 kg

36. Since all forces of interest in this problem are horizontal, draw the free-body diagram  showing only the horizontal forces. F is the tension in the coupling between the T1  locomotive and the first car, and it pulls to the right on the first car. F is the tension in the T2  coupling between the first car an the second car. It pulls to the right on car 2, labeled F  T2R and to the left on car 1, labeled F . Both cars have the same mass m and the same T2L  acceleration a. Note that F= F = F by Newton’s third law. T2R T2L T 2  FT2 FT2 FT1

Write a Newton’s second law expression for each car.

邋F1= FT 1 - F T 2 = ma F 2 = F T 2 = ma Substitute the expression for ma from the second expression into the first one.

FT1- F T 2 = ma = F T 2 � F T1 � 2 F T2 F T1 F T2 2 This can also be discussed in the sense that the tension between the locomotive and the first car is pulling 2 cars, while the tension between the cars is only pulling one car.

  46. (a)In the free-body diagrams below, FAB = force on block A exerted by block B, FBA = force on   block B exerted by block A, F = force on block B exerted by block C, and F = force BC   CB on block C exerted by block B. The magnitudes of F and F are equal, and the   BA AB magnitudes of FBC and FCB are equal, by Newton’s third law.  FA N   FB N F    C N  F  F AB F F CB F BA BC    mAg mBg mCg

(b)All of the vertical forces on each block add up to zero, since there is no acceleration in the

vertical direction. Thus for each block, FN = mg . For the horizontal direction, we have the following. F F= F - FAB + F BA - F BC + F CB = F =( m A + m B + m C ) a � a mA+ m B + m C (c) For each block, the net force must be ma by Newton’s second law. Each block has the same acceleration since they are in contact with each other.

mA mB mC FA net = F FB net = F F3net = F mA+ m B + m C mA+ m B + m C mA+ m B + m C

mC (d) From the free-body diagram, we see that for mC, FCB= F C net = F . And by mA+ m B + m C m   F= F = F C . Newton’s third law, BC CB Of course, F23 and F32 are in mA+ m B + m C

opposite directions. Also from the free-body diagram, we use the net force on mA.

mA m A F- FAB = F A net = F � F AB - F F mA+ m B + m C m A + m B + m C

mB+ m C FAB = F mA+ m B + m C

m2+ m 3 By Newton’s third law, FBC= F AB = F . m1+ m 2 + m 3

F 96.0 N 2 (e) Using the given values, a = = = 3.20 m s . Since all three m1+ m 2 + m 3 30.0 kg masses are the same value, the net force on each mass is 2 Fnet = ma =(10.0 kg)( 3.20 m s) = 32.0 N .

This is also the value of FCB and FBC. The value of FAB and FBA is found as follows. 2 FAB= F BA =( m 2 + m 3 ) a =(20.0 kg)( 3.20m s) = 64.0 N To summarize:

FA net= F B net = F C net =32.0 N F AB = F BA = 64.0 N F BC = F CB = 32.0 N

The values make sense in that in order of magnitude, we should have F> FBA > F CB , since

F is the net force pushing the entire set of blocks, FAB is the net force pushing the right two blocks, and FBC is the net force pushing the right block only.

54. We draw a free-body diagram for each mass. We choose UP to be  the positive direction. The tension force in the cord is found from FC analyzing the two hanging masses. Notice that the same tension force is applied to each mass. Write Newton’s second law for each of the masses. F- m g = m a F - m g = m a T 1 1 1 T 2 2 2   Since the masses are joined together by the cord, their accelerations FT FT will have the same magnitude but opposite directions. Thus a= - a . Substitute this into the force expressions and solve for the   1 2 F F tension force. T T m g- F m m F- m g = - m a � F - m g � m a a 1 T 2 1 T 1 1 2 T 1 1 2 2 m 1.2 3.2 1 kg  kg  m2g m1g 骣m1 g- F T2 m 1 m 2 g FT- m 2 g = m 2 a 2 = m 2琪 � F T 桫 m1 m 1+ m 2 Apply Newton’s second law to the stationary pulley. 4 3.2 kg 1.2 kg 9.80m s2 4m1 m 2 g ( )( )( ) FC-2 FT = 0 � F C = 2 F T = = 34 N m1+ m 2 4.4 kg

55. If m doesn’t move on the incline, it doesn’t move in the vertical  q direction, and so has no vertical component of acceleration. This FN suggests that we analyze the forces parallel and perpendicular to the floor. See the force diagram for the small block, and use Newton’s second law to find the acceleration of the small block. mg  q F= Fcosq - mg = 0 � F mg y N N cosq Fsinq mg sin q F= Fsinq = ma � a =N = g tan q x N m m cosq Since the small block doesn’t move on the incline, the combination of both masses has the same horizontal acceleration of g tanq . That can be used to find the applied force.

Fapplied =( m + M) a =( m + M) g tanq Note that this gives the correct answer for the case of q = 0, , where it would take no applied force to keep m stationary. It also gives a reasonable answer for the limiting case of q  90 , where no force would be large enough to keep the block from falling, since there would be no upward force to counteract the force of gravity.   FTC FTC

a m a 69. (a)A free-body diagram is shown for each block.  C F C C  F  N-B T  We define the positive x-direction for mA to FN-A FT y   mCg F F be up its incline, and the positive x-direction y TA TA m x for B to be down its incline. With that   x qA q definition the masses will both have the qA FTA qB FTA B same acceleration. Write Newton’s second m a m a law for each body in the x direction, and  A A  B B m g m g combine those equations to find the A  B  m g acceleration. A mBg

mA: Fx = F T - m A g sinq A = m A a

mB: Fx = m B g sinq B - F T = m B a add these two equations

mBsinq B- m A sin q A (FT- m A gsinq A) +( m B g sin q B - F T) = m A a + m B a � a g mA+ m B (b)For the system to be at rest, the acceleration must be 0.

mBsinq B- m A sin q A a= g = 0 � mB sinq B m A sin q A mA+ m B

sinqA sin 32 mB= m A =(5.0 kg) = 6.8kg sinqB sin 23 The tension can be found from one of the Newton’s second law expression from part (a). 2 mA: F T- m A g sinq A = 0 � F T = m A g sin q A �( 5.0 kg)( 9.80 m s) sin 32 26 N (c) As in part (b), the acceleration will be 0 for constant velocity in either direction.

mBsinq B- m A sin q A a= g = 0 � mB sinq B m A sin q A mA+ m B m sinq sin 23 A= B = = 0.74 mBsinq A sin 32

74. Consider the free-body diagram for the watch. Write Newton’s second law  F for both the x and y directions. Note that the net force in the y direction is  T 0 because there is no acceleration in the y direction. y mg F= Fcosq - mg = 0 � F x  y T T cosq mg mg F= Fsinq = ma � sin q ma x T cosq a= g tanq =( 9.80 m s2) tan 25 = 4.57 m s 2

Use Eq. 2-12a with v0 = 0 to find the final velocity (takeoff speed). 2 v- v0 = at � v + v 0 = at + 0 = ( 4.57 m s)( 16s) 73m s

© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 89