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Two boxes on one another

  1. Oct 4, 2009 #1
    The problem statement, all variables and given/known data

    The coefficient of static friction is 0.636 between the two blocks shown. The coefficient of kinetic friction between the lower block and the floor is 0.138. Force F causes both blocks to cross a distance of 7.00 m, starting from rest. What is the least amount of time in which the motion can be completed without the top block sliding on the lower block, if the mass of the lower block is 1.14 kg and the mass of the upper block is 2.80 kg?

    http://capa.physics.mcmaster.ca/figures/kn/Graph08/kn-pic0832.png [Broken]


    The attempt at a solution

    Fmax= UsN
    =(.636)(2.80*9.81)
    =17.4696 N

    F(bottom) = UkN
    = (.138)((2.80 + 1.14) * 9.81)
    =5.3339

    Fnet = 17.4696 - 5.3339
    =12.1357

    F=ma
    a=F/m
    =12.1357/(1.14+2.8)
    =3.0801

    d=v1*t + .5at^2
    t=square root of (d/(.5a))
    = square root of (7/(.5(3.0801)))
    =2.13 s


    where am i going wrong?
    thank you
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 4, 2009 #2
    this is the maximum net force that can accelerate the bottom block.
    you should consider only the acceleration of the bottom block
    in your next calculation.
     
  4. Oct 4, 2009 #3
    Thank you.

    But what do you mean consider on the bottom block. Do you mean use only the weight of the bottom block?
     
  5. Oct 4, 2009 #4
    yes. Since you found the maximum net force on the bottom block, then you need to use the mass and the acceleration of the bottom block as well in F = ma
     
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