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Unformatted text preview: 446. (a) See drawing (b) If we select all three blocks as the system, we have F m1 Fx = max: F = (m1 + m2 + m3)a, which gives a = F/(m1 + m2 + m3). x (c) For the three blocks individually, for Fx = max we have F Fnet1 = m1a = m1F/(m1 + m2 + m3); Fnet2 = m2a = m2F/(m1 + m2 + m3); Fnet3 = m3a = m3F/(m1 + m2 + m3). (d) From the force diagram for block 1 we have F21 Fnet1 = F F12 = m1a, which gives F12 = F m1a = F m1F/(m1 + m2 + m3) = F(m2 + m3)/(m1 + m2 + m3). This is also F21 (Newton's third law). From the force diagram for block 2 we have F32 Fnet2 = F21 F23 = m2a, which gives F23 = F21 m2a = F m1a m2a = F (m1 + m2)F/(m1 + m2 + m3) = Fm3/(m1 + m2 + m3). This is also F32 (Newton's third law). (e) When we use the given values, we get a = F/(m1 + m2 + m3) = (96.0 N)/(12.0 kg + 12.0 kg + 12.0 kg) = 2.67 m/s2. Fnet1 = m1a = (12.0 kg)(2.67 m/s2) = 32.0 N. Fnet2 = m2a = (12.0 kg)(2.67 m/s2) = 32.0 N. Fnet3 = m3a = (12.0 kg)(2.67 m/s2) = 32.0 N. Because the blocks have the same mass and the same acceleration, we expect Fnet1 = Fnet2 = Fnet3 = 32.0 N. For the forces between the blocks we have F21 = F12 = F m1a = 96.0 N (12.0 kg)(2.67 m/s2) = 64.0 N. F32 = F23 = F m1a m2a = 96.0 N (12.0 kg)(2.67 m/s2) (12.0 kg)(2.67 m/s2) = m2 FN1 m3 F12 m 1g FN2 m 2g FN3 F23 m 3g 32.0 N. ...
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This note was uploaded on 04/07/2008 for the course PHY 113 taught by Professor Jordan during the Spring '08 term at Rochester.
 Spring '08
 Jordan

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