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Unformatted text preview: h ic e8 e k# 5 i b iC a aha 5 Problem 24 ( a ): Let up be the positive direction. A student with weight w = 550 N has mass m = w g = 56 kg. The forces on the student are the weight w down and a normal force N = 450 N up. (A bathroom scale doesnt actually measure your weight: it measures the force with which it is squeezed between your feet and the floor, which is the normal force between you and the floor. This is usually your weight.) Thus Newtons second law says that N w = ma a = N w m = (450 N) (550 N) 56 kg = 1 . 8 m / s 2 . The elevator is accelerating down. ( b ): Now we have a = N w m = (670 N) (550 N) 56 kg = 2 . 1 m / s 2 . The elevator is accelerating up. ( c ): Now we have a = N w m = (0 N) (550 N) 56 kg = 9 . 8 m / s 2 . The elevator is in free fall, and the student should worry. ( d ): For the elevator and student combined, there are two forces: the tension T of the cable pulling up and the weight W = Mg = (850 kg)(9 . 8 m / s 2 ) = 8300 N down. Newtons second law says that T Mg = Ma T = M ( g + a ) . For part ( a ) we have a = 1 . 8 m / s 2 , so T = (850 kg)(8 . 0 m / s 2 ) = 6800 N . For part ( c ) we have a = 9 . 8 m / s 2 , so T = 0. 5 Problem 33 ( a ): Let us use coordinates parallel and perpendicular to the slope. Let up be the positive directions. The angle of the slope is = Tan 1 parenleftbigg 2 . 50 m 4 . 75 m parenrightbigg = 27 . 8 . Let m 1 be the larger, bottom mass and m 2 be the smaller, upper mass. The perpendicular forces on m 2 are N 2 m 2 g cos = 0 N 2 = m 2 g cos , where N 2 is the normal force between m 1 and m 2 . Here we have used the fact that the perpendicular acceleration is 0. The parallel acceleration is also 0, because the problem states that the boxes are moving with constant velocity. The parallel forces are F f, 2 m 2 g sin = 0 F f = m 2 g sin . Here F f, 2 is the force of friction exerted on m 2 by m 1 , and we guess that it points uphill. (If we had guessed wrong, we would just have gotten a negative number.) Now we examine the perpendicular forces on m 1 : N 1 N 2 m 1 g cos = 0 N 1 = N 2 + m 1 g cos = ( m 1 + m 2 ) g cos , 1 where N 1 is the normal force between m 1 and the ramp. Note that because m 1 is pushing m 2 up with force N 2 , m 2 is pushing m 1 down with the same strength. The parallel forces are F f, 1 F f, 2 + F m 1 g sin = 0 , where F is the force with which we are pulling, and F f, 1 is the friction force. Note that because the boxesis the friction force....
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This note was uploaded on 07/12/2008 for the course PHY 121 taught by Professor Chamberlin during the Summer '08 term at ASU.
 Summer '08
 Chamberlin
 Force, Mass, Work

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