Thread: Mechanics, linear motion help
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Old January 10th, 2009, 06:47 AM
HallsofIvy HallsofIvy is offline
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I answered (a) then saw that you had already done that! Very good.

For (c), a "range" of values for \mu must have both an upper and lower bound. The upper bound is easy: \mu cannot be so large as to have no motion at all. The total force on the system is mg(3\mu- 1) (you must have calculated that when you did (a)) and the masses will not move if that is negative mg(3\mu-1)< 0 leads to \mu< 1/3. Now you are told that the mass A, which, as you calculated for (b) has speed \sqrt{\frac{gh(1-3u)}{2}} when B stops falling and so the only force on mass A is the friction force 3mg\mu. The acceleration due to that is 3g\mu and that must stop A before it reaches the pulley.

A is initially distance 2h from the pulley. After B has fallen a distance h, A will have moved distance h also and so will be distance h from the pulley. Using decceleration 3g\mu to determine the time required for A to stop. Calculate the distance A will have traveled. Find the value of \mu so that distance would be exactly h. That is the lower bound for \mu.
Last edited by HallsofIvy; January 10th, 2009 at 07:02 AM.
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