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PHYSICS 171 AQ 2009 Homework #5 (due Friday, October 16, 2009) 1. Giancoli Chapter 5, Problem 10 A free-body diagram for the bar of soap is shown. There is no motion in the y direction and thus no acceleration in the y direction. Write Newton’s second law for both directions, and use those expressions to find the acceleration of the soap. Fx FN mg cos 0 FN mg cos FN F fr Fx mg sin Ffr ma y ma mg sin k FN mg sin k mg cos a g sin k cos Now use Eq. 2-12b, with an initial velocity of 0, to find the final velocity. x x0 v0t 12 at 2 t 2x a 2x g sin k cos x mg 2 9.0 m 9.80 m s sin 8.0 0.060 cos 8.0 2 4.8s 2. Giancoli Chapter 5, Problem 17 a) Since the two blocks are in contact, they can be treated as a single object as long as no information is needed about internal forces (like the force of one block pushing on the 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 s 2 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 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. F21 m2 Ffr2 FN2 m2 g F x F21 Ffr2 m2 a F21 k m2 g m2a 0.18125kg 9.80 m s2 125kg 1.657 m s2 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 remain F12 the same – the analysis from part (a) still applies. We can also repeat the m1 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 Ffr1 equations. m1g FN1 2 a 1.7 m s ; Fx 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 3. Giancoli Chapter 5, Problem 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 . 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 4. Giancoli Chapter 5, Problem 28 We define the positive x direction to be the direction of motion for each block. See the free-body diagrams. Write Newton’s second law in both dimensions for both objects. Add the two x-equations to find the acceleration. Block A: FyA FNA mA g cos A 0 FNA mA g cos A F xA FT mA g sin FfrA mA a Block B: FyB FNB mB g cos B 0 FNB mB g cos B FxB mB g sin FfrB FT mBa y FT FNA FfrA A A mAg x FT FNB y FfrB x B B m Bg Add the final equations together from both analyses and solve for the acceleration, noting that in both cases the friction force is found as Ffr FN . mA a FT mA g sin A A mA g cos A ; mBa mB g sin B BmB g cos B FT mA a mBa FT mA g sin A A mA g cos A mB g sin B BmB g cos B FT mA sin A A cos A mB sin B cos mA m B 2.0 kg sin 51 0.30cos51 5.0 kg sin 21 0.30cos 21 9.80 m s2 7.0 kg a g 2.2 m s2 5. Giancoli Chapter 5, Problem 30 FT (a) Given that mB is moving down, mA must be moving up the incline, and so the force of kinetic friction on mA will be directed down the incline. Since the blocks are tied together, they will both have the same acceleration, and so a yB a xA a. Write Newton’s FN FT Ffr xA m Bg yB m Bg mAg second law for each mass. FyB mB g FT mBa FT mB g mBa F F xA FT mA g sin Ffr mA a yA FN mA g cos 0 FN mA g cos Take the information from the two y equations and substitute into the x equation to solve for the acceleration. mB g mBa mA g sin k mA g cos mAa a (b) mB g mA g sin mA g k g cos mA mB 1 2 12 g 1 sin k g cos 9.80 m s 1 sin 34 0.15cos 34 1.6 m s 2 2 To have an acceleration of zero, the expression for the acceleration must be zero. a 12 g 1 sin k cos 0 1 sin k cos 0 k 1 sin cos 1 sin 34 cos 34 0.53 6. Giancoli Chapter 5, Problem 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 twoblock system, there is no friction on the system, and so F M m a describes the horizontal motion of the system. Thus the upper block has a vertical acceleration of 0 and a yA FN Ffr horizontal acceleration of F M m . Write Newton’s second law for the mg 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. 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 M m sin cos mg cos sin M m m 7. Giancoli Chapter 5, Problem 44 FT (a) At the bottom of the motion, a free-body diagram of the bucket would be as shown. Since the bucket is moving in a circle, there must be a net force on it towards the center of the circle, and a centripetal acceleration. Write Newton’s second law for the bucket, with up as the positive direction. FR FT mg ma m v 2 r v r FT mg m 1.10 m 25.0 N 2.00 kg 9.80 m s 2 2.00 kg 1.723 1.7 m s (b) A free-body diagram of the bucket at the top of the motion is shown. Since the bucket is moving in a circle, there must be a net force on it towards the center of the circle, and a centripetal acceleration. Write Newton’s second law for the bucket, with down as the positive direction. F R FT mg ma m v 2 r v mg FT r FT mg m If the tension is to be zero, then v r 0 mg rg 1.10 m 9.80 m s 2 3.28 m s m The bucket must move faster than 3.28 m/s in order for the rope not to go slack. 8. Giancoli Chapter 5, Problem 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. mg y x F F N 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 FN mg 9.80 m s 7.6 m 8.6 m s 2 9. Giancoli Chapter 5, Problem 48 To experience a gravity-type force, objects must be on the inside of the outer wall of the tube, so that there can be a centripetal force to move the objects in a circle. See the free-body diagram for an object on the inside of the outer wall, and a portion of the tube. The normal force of contact between the object and the wall must be maintaining the circular motion. Write Newton’s second law for the radial direction. FR FN ma mv2 r FN If this is to have the same effect as Earth gravity, then we must also have that FN mg . Equate the two expressions for normal force and solve for the speed. FN m v 2 r mg v gr 9.80 m s 550 m 73.42 m s 2 86, 400 s 3 1836 rev d 1.8 10 rev d 2 550 m 1 d 73.42 m s 1 rev 10. Giancoli Chapter 5, Problem 54 If the masses are in line and both have the same frequency of rotation, then they will always stay in line. Consider a free-body FNB diagram for both masses, from a side view, at the instant that they FTB are to the left of the post. Note that the same tension that pulls mB inward on mass 2 pulls outward on mass 1, by Newton’s third law. Also notice that since there is no vertical acceleration, the normal m Bg force on each mass is equal to its weight. Write Newton’s second law for the horizontal direction for both masses, noting that they are in uniform circular motion. FRA FTA FTB mAaA mA vA2 rA FRB FTB mBaB mB vB2 rB The speeds can be expressed in terms of the frequency as follows: rev 2 r 2 rf . v f sec 1 rev FTB mB vB2 rB mB 2 rB f rB 4 2mBrB f 2 2 FTA FTB mA vA2 rA 4 mBrB f 2 mA 2 rA f rA 4 2 f 2 mA rA mBrB 2 FNA FTB FTA mA mAg