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Math Help - Mechanics, linear motion help

  1. #1
    Super Member craig's Avatar
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    Mechanics, linear motion help

    Hi, this is probably something basic I am missing but I can't get my head around it.

    Mechanics, linear motion help-tension-question.bmp

    In the question you are told that A and B are particles of mass 3m and m respectively, each attached to a light inextensionable string of length 3h. The string passes over a smooth pulley attached to a rough table. The coefficient of friction between the table and A is u.

    The system is released from rest and begins to move...

    1. Show that until B hits the floor, the acceleration of A is \frac{1}{4}(1-3u)g, this I was able to do.

    2. Find the speed of A immediatly before B hits the floor, managed to do this as well, got it to be \sqrt{\frac{gh(1-3u)}{2}}

    3. After B hits the floor, A continues to move, coming to rest before reaching the pulley. Deduce the range of values of u for the motion above to be possible.

    It is the last bit I can't get my head round, any ideas?

    Thanks in advance

    Craig
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  2. #2
    MHF Contributor

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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 05:02 AM.
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  3. #3
    Super Member craig's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    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.
    Thank you , I don't know why but with all the unknown quantities I just could not get my head around the maths

    Thanks again
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