A3 9th Edition Solution

A3 9th Edition Solution - PROBLEM 13.21 The system shown...

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PROBLEM 13.21 The system shown, consisting of a 20-kg collar A and a 10-kg counterweight B , is at rest when a constant 500-N force is applied to collar A . (a) Determine the velocity of A just before it hits the support at C . (b) Solve part a assuming that the counterweight B is replaced by a 98.1-N downward force. Ignore friction and the mass of the pulleys. SOLUTION Kinematics Displacement of A , x A = 0.6 m Displacement of B , x B = 2 × 0.6 = 1.2 m A B x x 2 = A B v v 2 = (a) Potential energy , ( 29 ( 29 1 2 1 2 1 2 B B B A A A x x g m x x g m V V - + - = - ( 29 ( 29 ( 29 ( 29 2 1 81 9 10 6 0 81 9 20 . . . . + - = 0 = Position 1 Position 2
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Kinetic energy , 2 2 1 2 2 1 2 1 B B A A v m v m T T + = - ( 29 ( 29( 29 2 2 2 10 2 1 20 2 1 A A v v + = 2 30 A v = F n is the force that does the work on the system which cause the change of position. The work done, 2 1 dx F n is positive if it is in the same direction as the motion. Energy equation , ( 29 ( 29 1 1 2 2 2 1 T V T V dx F n + - + = ( 29 2 30 0 6 0 500 A v . + = + m/s 162 3 . v A = (b) Potential energy , ( 29 1 2 1 2 A A A x x g m V V - = - ( 29 ( 29 6 0 81 9 20 . . - = Joule 72 117 . - = Kinetic energy , 2 1 2 2 1 A A v m T T = - ( 29 2 20 2 1 A v = 2 10 A v = Energy equation , ( 29 ( 29 1 1 2 2 2 1 T V T V dx F n + - + = ( 29 ( 29 ( 29 2 10 72 117 2 1 1 98 6 0 500 A v . . . . + - = - + + m/s 477 5 . v A = 98.1 N Position 1 Position 2 98.1 N
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PROBLEM 13.19 The two blocks shown are released from rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and
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This note was uploaded on 01/22/2012 for the course CIVE 281 taught by Professor Prof.v.h.chu during the Fall '10 term at McGill.

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A3 9th Edition Solution - PROBLEM 13.21 The system shown...

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