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Newton's 3rd law, tension, friction

  1. Oct 31, 2009 #1
    1. The problem statement, all variables and given/known data
    hw.gif

    2. Relevant equations
    F=m*a
    Fs <= N*us
    Fk = N * uk

    3. The attempt at a solution
    b) T-fs = 120
    T-.2 * 9.8 * (15 + 20 + 25) = 120 => T = 237.6 Newtons

    c) (Fs)2 on 1 = (Fs)1 on 2 = 15 * 9.8* .6 = 88.2 Newtons

    d) (15 + 20) * 9.8 * .6 = 205.8 Newtons

    e) This is where I'm falling apart. Since we are applying a larger tension force than the static friction of both top boxes, then the boxes must slide off. Right?

    f) My professor said they slide off at the same time, but I'm not sure how to show it, or why. I thought the top one would fly off first, since the static friction force is smallest.

    Any help is appreciated!
     
  2. jcsd
  3. Oct 31, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Ask yourself: In order to accelerate m1 (or m1 & m2) what force is required? Is static friction sufficient to provide that force?
    Figure out the maximum acceleration that each box can have without sliding. (Consider the available static friction.)
     
  4. Oct 31, 2009 #3
    Thank you! That helped.

    The accelerations where the boxes start to slide are:

    m1: 88.2 = 15 * a => a = 5.88
    m2: 205.8 = (15+20) * a => a = 5.88

    So they both slide at the same time. And in part e, since the acceleration is less than 5.88, they don't slide off.
     
  5. Oct 31, 2009 #4

    Doc Al

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    Staff: Mentor

    Good!
     
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