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Dynamics_Part80

# Dynamics_Part80 -

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Unformatted text preview: Problem Set 10: Problem 2. Problem: Blocks A and B have mass M and m, respectively. They are connected by a cord-and-pulley system and are released from rest. (a) Determine the equilibrium displacement, xeq , of Block A. (b) Explain why the displacement of Block B is twice that of Block A. HINT: Identify what is being conserved. (c) Letting T denote cable tension, verify that M x = 2T - k(x + xeq ) and 2m = mg - T x (d) Derive a single differential equation for the system and determine the natural frequency. Solution: (a) The forces acting on the blocks are as shown in the following figure. Clearly, the cable-tension force acts on Block A through both the top and bottom cables. A ......................................................................... M . . .. . .. . . ... ... . . . . . . . . .......... . ......... . ............................ .................................. .................... .... ............................... ............................... ............................... ............................... ............................... ............................... ............................... ............................... ............................... ............................... ............................... ............................... ............................... ............................... ............................... ............... ............................... ............................... ............................... ................ ............................... ................ ............................... ............................... ............................... . . . . . . . . . . . . . . . . . ... ... .. .. . . .. . T ................................................................................................. ................................................................................................. ................................................................................................. ................................................................................................. ................................................................................................. ................................................. ................................................................................................. ................................................................................................. ................................................................................................. ................................................ ..................................................................................................... ... .... . eq .................................................................................................................. . .. . . .. .. . ................................................................................................. .. .... ...... ......................................................................................................... .. ................................................................................................................. ................................................................................................. .... . ................................................................................................. .............. . .. ................................................................................................................. ................................................................................................. . ................................................................................................. ................................................................................................. . ................................................................................................. ................................................................................................. ................................................................................................. . ................................................................................................. ................................................................................................. ................................................................................................. ................................................................................................. B kx T T m mg In equilibrium, the cable-tension force balances the weight of Block B, wherefore T = mg So, in equilibrium, the spring force on Block A balances twice the cable-tension force, so that its equilibrium displacement, xeq , is given by kxeq = 2T Therefore, we conclude that xeq = 2mg k = kxeq = 2mg (b) Conservation of cable length means that Block B moves twice as far as Block A. ...
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