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Minimum Force such that box m does not slide down

  1. Feb 23, 2015 #1
    1. The problem statement, all variables and given/known data

    slide.png
    2. Relevant equations

    F = ma; f = µN
    3. The attempt at a solution
    Let F' be the action-reaction force between box m and box M
    f be the friction on small box m
    For small box m:
    F' = ma
    f - mg = 0
    µF' - mg = 0 since f = µF'
    µma - mg = 0 since F' = ma
    a = g / µ = 9.81 m/s2 / 0.2 = 49.05 m/s2
    which is not equal to the given acceleration of 2.0 m/s2
    Please help. Thanks



     
  2. jcsd
  3. Feb 23, 2015 #2

    BvU

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    Science Advisor
    Homework Helper
    Gold Member

    Hello p, welcome to PF :smile:

    This is a most peculiar exercise. What you calculate is correct: an acceleration of no less than 5g is required to make m stick to M with so much friction that it doesn't slide down. I find the exercise wording/picture combination very unsatisfactory: It isn't clear at all where the given 2 m/s2 comes from, and -- as you so justly put it -- 2 m/s2 is not 5g at all.

    If an answer is absolutely required to get to the next level or something like that, then perhaps it's allowed to make some assumptions: for example that the 2 m/s2 is already there (some jet engine inside M that we can't see) and that the horizontal component of F is supposed to provide the remaining a' = 47.05 m/s2 to the ensemble of (m + M) . You apply Fx = (m+M) a' and some trigonometry to go from Fx to |F|.

    Good luck, and maybe you can let us know what came out ?
     
  4. Feb 23, 2015 #3
    Alright. Thanks! :)
     
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