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Solutions to Assigned Problems Chapter 5 2. A free-body diagram for the box is shown. Since the box does not accelerate vertically, FN mg . (a) To start the box moving, the pulling force must just overcome the force of static friction, and that means the force of static friction will reach its maximum value of Ffr s FN . Thus we have for the starting motion, Fx FP Ffr 0 FP Ffr s FN s mg s FP 35.0 N mg 6.0 kg 9.80 m s2 FN Ffr FP mg 0.60 (b) The same force diagram applies, but now the friction is kinetic friction, and the pulling force is NOT equal to the frictional force, since the box is accelerating to the right. F FP Ffr ma FP k FN ma FP k mg ma k 4. FP ma mg 35.0 N 6.0 kg 0.60 m s 2 6.0 kg 9.80 m s 2 0.53 See the included free-body diagram. To find the maximum angle, assume that the car is just ready to slide, so that the force of static friction is a maximum. Write Newton’s second law for both directions. Note that for both directions, the net force must be zero since the car is not accelerating. F y FN mg cos 0 FN mg cos F s x mg sin Ffr 0 mg sin Ffr s FN s mg cos mg sin mg cos tan 0.90 tan 1 0.90 42 11. A free-body diagram for the box is shown, assuming that it is moving to the right. The “push” is not shown on the free-body diagram because as Ffr soon as the box moves away from the source of the pushing force, the push is no longer applied to the box. It is apparent from the diagram that FN mg for the vertical direction. We write Newton’s second law for the horizontal direction, with positive to the right, to find the acceleration of the box. Fx Ffr ma ma k FN k mg FN mg a k g 0.15 9.80 m s 2 1.47 m s 2 Eq. 2-12c can be used to find the distance that the box moves before stopping. The initial speed is 4.0 m/s, and the final speed will be 0. v 2 v02 2a x x0 x x0 v 2 v02 2a 0 3.5 m s 2 2 1.47 m s 2 4.17 m 4.2 m © 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. 118 Physics for Scientists & Engineers, 4th Edition Giancoli 17. (a) Since the two blocks are in contact, they can be FP treated as a m1 + single object as long as no information is needed about m2 Ffr internal forces (like the force of one block pushing on the m1 m2 g FN other block). Since there is no motion in the vertical direction, it is apparent that FN m1 m2 g , and so Ffr k FN k m1 m2 g. Write Newton’s second law for the horizontal direction. Fx FP Ffr m1 m2 a a FP Ffr m1 m2 FP k m1 m2 g m1 m2 650 N 0.18190 kg 9.80 m s 2 190 kg 1.657 m s2 1.7 m s 2 (b) To solve for the contact forces between the blocks, an individual block must be analyzed. Look at the free-body diagram for the second block. F21 is the force of the first block pushing on the second F21 m2 Ffr2 m2 g FN2 block. Again, it is apparent that FN2 m2 g and so Ffr2 k FN2 k m2 g. Write Newton’s second law for the horizontal direction. F x F21 Ffr2 m2 a F21 k m2 g m2 a 0.18125kg 9.80 m s 2 125kg 1.657 m s 2 430 N By Newton’s third law, there will also be a 430 N force to the left on block # 1 due to block # 2. (c) If the crates are reversed, the acceleration of the system will F12 remain the same – the analysis from part (a) still applies. We can also repeat the analysis from part (b) to find the force of one block on the other, if we simply change m1 to m2 in the free-body diagram and the resulting equations. a 1.7 m s2 ; F x m1 Ffr1 FN1 m1g F12 Ffr1 m1a F12 k m1 g m1a 0.18 65 kg 9.80 m s 2 65 kg 1.657 m s 2 220 N 23. (a) For mB to not move, the tension must be equal to mB g , and so mB g FT . For mA to not move, the tension must be equal to the force of static friction, and so FS FT . Note that the normal force on mA is equal to its weight. Use these relationships to solve for mA . © 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. 119 Physics for Scientists & Engineers, 4th Edition Giancoli mB g FT Fs s mA g mA mB s 2.0 kg 0.40 5.0 kg mA 5.0 kg (b) For mB to move with constant velocity, the tension must be equal to mB g . For mA to move with constant velocity, the tension must be equal to the force of kinetic friction. Note that the normal force on mA is equal to its weight. Use these relationships to solve for mA . mB g Fk k mA g mA mB k 2.0 kg 0.30 6.7 kg 24. We define f to be the fraction of the cord y that is handing down, between mB and FN x FT the pulley. Thus the mass of that piece of FT Ffr cord is fmC . We assume that the system x is moving to the right as well. We take mAg the tension in the cord to be FT at the 1 f mCg pulley. We treat the hanging mass and mBg fmCg hanging fraction of the cord as one mass, and the sliding mass and horizontal part of the cord as another mass. See the free-body diagrams. We write Newton’s second law for each object. FyA FN mA 1 f mC g 0 xA FT Ffr FT k FN mA 1 f mC a xB mB fmC g FT mB fmC a F F Combine the relationships to solve for the acceleration. In particular, add the two equations for the x-direction, and then substitute the normal force. mB fmC k mA 1 f mC g mA mB mC a 33. To find the limiting value, we assume that the blocks are NOT slipping, but that the force of static friction on the smaller block is at its maximum value, so that Ffr FN . For the two-block system, there is no friction on the system, and so F M m a FN Ffr y x describes the horizontal motion of the system. Thus the upper block has a vertical acceleration of 0 and a horizontal mg F acceleration of . Write Newton’s second law for the M m upper block, using the force diagram, and solve for the applied force F. Note that the static friction force will be DOWN the plane, since the block is on the verge of sliding UP the plane. © 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. 120 Physics for Scientists & Engineers, 4th Edition Giancoli F FN cos Ffr sin mg FN cos sin mg 0 FN F FN sin Ffr cos FN sin cos ma m y x F FN sin cos M m mg cos sin sin cos M m g cos sin m F cos sin M m sin cos mg M m m 34. A free-body diagram for the car at one instant of time is shown. In the FN diagram, the car is coming out of the paper at the reader, and the center of Ffr the circular path is to the right of the car, in the plane of the paper. If the car has its maximum speed, it would be on the verge of slipping, and the mg force of static friction would be at its maximum value. The vertical forces (gravity and normal force) are of the same magnitude, because the car is not accelerating vertically. We assume that the force of friction is the force causing the circular motion. FR Ffr m v 2 r s FN s mg v s rg 0.6580.0 m 9.80 m s 2 22.57 m s 23m s Notice that the result is independent of the car’s mass . 40. At the top of a circle, a free-body diagram for the passengers would be as shown, assuming the passengers are upside down. Then the car’s normal force would be pushing DOWN on the passengers, as shown in the diagram. We assume no safety devices are present. Choose the positive direction to be down, and write Newton’s second law for the passengers. F FN mg ma m v 2 r FN m v 2 r g We see from this expression that for a high speed, the normal force is positive, meaning the passengers are in contact with the car. But as the speed decreases, the normal force also decreases. If the normal force becomes 0, the passengers are no longer in contact with the car – they are in free fall. The limiting condition is as follows. 2 vmin r g 0 vmin rg 9.80 m s 7.6 m 8.6 m s 2 41. A free-body diagram for the car is shown. Write Newton’s second law for the car in the vertical direction, assuming that up is positive. The normal force is twice the weight. F FN mg ma 2mg mg m v 2 r v rg 95 m 9.80 m FN mg s 2 30.51m s 31m s 51. The force of static friction is causing the circular motion – it is the centripetal force. The coin slides off when the static frictional force is not large enough to move the coin in a circle. The maximum static frictional force is the coefficient of static friction times the normal force, and the normal FN Ffr mg © 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. 121 Physics for Scientists & Engineers, 4th Edition Giancoli force is equal to the weight of the coin as seen in the free-body diagram, since there is no vertical acceleration. In the free-body diagram, the coin is coming out of the paper and the center of the circle is to the right of the coin, in the plane of the paper. The rotational speed must be changed into a linear speed. rev 1 min 2 0.120 m v 35.0 0.4398 m s min 60 s 1 rev FR Ffr m v 2 r s FN s mg s 66. (a) v2 rg 0.4398 m s 2 0.120 m 9.80 m s2 0.164 The terminal velocity is given by Eq. 5-9. This can be used to find the value of b. vT mg b mg 3 10 5 kg 9.80 m s2 3.27 10 5 kg s 3 105 kg s b vT 9 m s (b) From Example 5-17, the time required for the velocity to reach 63% of terminal velocity is the time constant, m b . m b 3 105 kg 3.27 105 kg s 0.917 s 1s 84. The car moves in a horizontal circle, and so there must be a net horizontal centripetal force. The car is not accelerating vertically. Write Newton’s second law for both the x and y directions. mg Fy FN cos mg 0 FN cos Fx FR FN sin max y x FN mg The amount of centripetal force needed for the car to round the curve is as follows. 2 1.0 m s 85km h 3.6 km h 9.679 103 N FR m v 2 r 1250 kg 72 m The actual horizontal force available from the normal force is as follows. mg FN sin sin mg tan 1250 kg 9.80 m s 2 tan14 3.054 103 N cos Thus more force is necessary for the car to round the curve than y can be supplied by the normal force. That extra force will have to x have a horizontal component to the right in order to provide the extra centripetal force. Accordingly, we add a frictional force pointed down the plane. That corresponds to the car not being able to make the curve without friction. mg Again write Newton’s second law for both directions, and again the y acceleration is zero. FN Ffr © 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. 122 Physics for Scientists & Engineers, 4th Edition Giancoli F FN cos mg Ffr sin 0 FN F FN sin Ffr cos m v r y x mg Ffr sin cos 2 Substitute the expression for the normal force from the y equation into the x equation, and solve for the friction force. mg Ffr sin v2 sin Ffr cos m v 2 r mg Ffr sin sin Ffr cos2 m cos cos r Ffr m v2 r cos mg sin 9.679 103 N cos14 1250 kg 9.80 m s 2 sin14 6.428 103 N So a frictional force of 6.4 103 N down the plane is needed to provide the necessary centripetal force to round the curve at the specified speed. 92. From Example 5-15 in the textbook, the no-friction banking angle is given by tan v02 1 Rg . The centripetal force in this case is provided by a component of y the normal force. Driving at a higher speed with the same radius x FN requires more centripetal force than that provided by the normal force alone. The additional centripetal force is supplied by a force of static friction, downward along the incline. See the free-body mg diagram for the car on the incline. The center of the circle of the Ffr car’s motion is to the right of the car in the diagram. Write Newton’s second law in both the x and y directions. The car will have no acceleration in the y direction, and centripetal acceleration in the x direction. Assume that the car is on the verge of skidding, so that the static frictional force has its maximum value of Ffr s FN . F y FN F x FN FN cos mg Ffr sin 0 FN cos s FN sin mg mg cos s sin FR FN sin Ffr cos m v 2 R FN sin s FN cos m v 2 R mv 2 R sin s cos Equate the two expressions for the normal force, and solve for the speed, which is the maximum speed that the car can have. mv 2 R mg sin s cos cos s sin vmax Rg sin 1 s tan cos 1 s tan v0 1 Rg 1 v 2 s 0 s Rg v02 © 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. 123 Physics for Scientists & Engineers, 4th Edition Giancoli Driving at a slower speed with the same radius requires less centripetal force than that provided by the normal force alone. The decrease in centripetal force is supplied by a force of static friction, upward along the incline. See the free-body diagram for the car on the incline. Write Newton’s second law in both the x and y directions. The car will have no acceleration in the y direction, and centripetal acceleration in the x direction. Assume that the car is on the verge of skidding, so that the static frictional force is given by Ffr s FN . F FN y x mg FN cos mg Ffr sin 0 y FN cos s FN sin mg F FN mg cos s sin FR FN sin Ffr cos m v 2 R FN sin s FN cos m v 2 R x FN Ffr mv 2 R sin s cos Equate the two expressions for the normal force, and solve for the speed. mv 2 R mg sin s cos cos s sin vmin Rg Thus vmin v0 sin 1 s tan cos 1 s tan 1 Rg v 1 v Rg s 2 s 0 2 0 v0 1 Rg v 1 v Rg 2 0 s 2 s 0 and vmax v0 1 Rg 1 v 2 s 0 s Rg v02 . © 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. 124