ps2_soln - Physics 105 Problem Set 2 Solutions Problem 1....

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Physics 105 Problem Set 2 Solutions Problem 1. (3 Points) We are asked to consider the situation where there is a block of mass M 1 on top of a block of mass M 2 resting on a table; the coefficient of friction between the table and block 1 is μ k . a) If block 1 accelerates horizontally, the force acting on it is the friction between the blocks. The free body diagrams are shown below. M 1 g M 2 g +N 1 N 2 N 1 F fr 1 F fr 2 Block 1 Block 2 F Figure 1: Free body diagrams for block 1 and block 2. On block 2 there are two contributions to the friction force pointing to the left. One is the kinetic friction that the table exerts on the block, which is f k = μ k ( M 1 + M 2 ) g , and the other is the static friction between the blocks, which is equal and opposite to the friction force acting on block 1. Although the former acts on the lower edge of block 2, for this problem it is only the direction of the forces which play a role. 1
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b) We can treat the two blocks as one rigid object, since they are moving together. Their total mass is M = M 1 + M 2 = 7 kg, so if want to accelerate them at 3 m/s 2 the required net force is F NET = 3 m/s 2 * 7kg= 21N. The ki- netic friction that M 1 feels from the table is f k = 0 . 2 * 9 . 8 m/s 2 * 7kg= 13 . 7N. Since the friction force always acts in the direction opposite to the motion, the force we must exert to achieve the above net force is the sum of these two, namely F = 21N+13 . 7N 35N c) The net force acting on block 1 to make it accelerate at 3 m/s 2 is 3 m/s 2 * 2kg=6N. This all has to be provided by the static friction between the blocks, μ s M 1 g . M 1 g = 9 . 8 m/s 2 * 2 kg = 19 . 6 N so μ s = 6 / 19 . 6 = 0 . 31 Problem 2. (3 points) In the system containing massless, frictionless pulleys and massless rope, we are interested in finding out the motion of the blocks when the coefficient of friction between the masses and the horizontal surfaces is μ . We will assume that the block with mass M 1 is sliding to the right and that the block with mass M 2 is sliding to the left. a) Force diagrams for each mass: Block 1 N M g 1 T N μ x 1 Block 1 Table 2
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Block 2 N μ N Block 2 T Table x 2 2 M g Block 3 M g 2 T Block 3 3 y 3 The relevant forces in the horizontal direction for block 1 and block 2 are the tension, T and the friction, μN , which is directed oppositely to the tension (note that because the blocks are sliding, the friction takes its maximum value). The forces in the vertical direction balance, and these are the weight force, M 1 g or M 2 g and the normal reaction N . Block 3 only has forces in the vertical direction – there is the weight force M 3 g which is directed down, and there is an upward force of 2 T . (Note that the upward force is 2 T rather than T because the rope on either side of the pulley has a tension T and we add both of the contributions). There is only one relevant co-ordinate for each block. For block 1 and 2,
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ps2_soln - Physics 105 Problem Set 2 Solutions Problem 1....

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