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Two Boxes + Friction

  1. Oct 7, 2008 #1
    I understand how to get the question right, but my friend posed an alternate method that is wrong and I was wondering why his method doesn't work.

    1. The problem statement, all variables and given/known data
    The coefficient of static friction is 0.75 between two blocks stacked on one another. The coefficient of kinetic friction between the lower block and the floor is 0.20. Force F vector acting on the upper block to the right causes both blocks to cross a distance of 4.0 m, starting from rest. What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?

    The upper box is 4kg and the lower box is 3kg.

    2. Relevant equations
    Upper box: Fxtotal = F - fs = m1*a (my guess is I made a mistake here since it wasn't pertinent to my solution)
    Lower box: Fxtotal = fs - fk = m2*a

    where a1 = a2 = a because of motion together.

    Upper box: Fytotal = N = m1*g
    Lower box: Fytotal = N = m1*g + m2*g

    3. The attempt at a solution
    I simply solved the lower equation for a and plugged it into the proper kinematic equation. However, my friend wrote from intuition F - fk = (m1 + m2)*a, which happens to be a combination of my two Fxtotal equations, and said that to maximize the distance F would have to equal fs, thus getting fs - fk = (m1 + m2)*a, which clearly would yield a response different from mine. I was wondering if anyone could clear up for us why it's wrong. I figure it has to either do with his assumption that F = fs or with my first equation and his overall equation (or both).

    Thanks!
     
  2. jcsd
  3. Oct 7, 2008 #2

    LowlyPion

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    Homework Helper

    Your equation on the lower box overlooks that whatever acceleration there is must include the mass of both boxes.

    The clearer way to view it for me is to look at what the maximum force may be applied. And that is simply .75*m1*g

    From that you subtract the frictional force which is the combined weight times .2.

    So the net force available then is .75*m1*g - .2*g*(m1 + m2)
    That in turn equals the force available to accelerate the two boxes as a system, namely

    .75*m1*g - .2*g*(m1 + m2) = a * (m1 + m2)

    Then figure your time the usual way with kinematics.
     
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