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Homework Help: Very Simple force problem

  1. Dec 31, 2008 #1
    1. The problem statement, all variables and given/known data
    a 4kg box rests on top of a 3kg box. There is a horizontalrope attached to the top box that can pull the boxes. What is the shortest amount of time the boxes can travel 5 meters together if the static friction coefficient between them is 0.6 and the kinetic friction coefficient between the bottom box and the ground is 0.2?

    2. Relevant equations

    3. The attempt at a solution

    Well I am pretty sure I've already solved the problem, it's just that my solution kind of surprises me so I wanna ask a few questions.
    So I solved it by considering the bottom box
    It has a force in the direction of movement due to static friction, so that force would be 4g x 0.6 (4g is the magnitude of the force of the top box on the bottom one due to gravity)
    it also has a force opposin movemet that would be -7g x 0.2 (-7g is the force of both boxes on the ground and therefore the normal force of the groudn on the lower box is 7g)

    So acceleration of b is Fnet / mass of b = (0.6x4g - 0.2x7g )/3 = g/3

    so shortest amount of time for 5 meters is 5 = 0.5 x g/3 x t^2 -----> t = root (5 / (0.5g/3)) = 1.75 seconds

    Now my question is .... i'm kind of surprised that my answer doesn't involve considerin the forces actin on the top box at all...such as the tenion of the rope...When i first looked at the problem i thought i would have to create a system of linear equations (from the x and y components of the forces acting on the 2 boxes) with my unknowns being the tension of the rope and the acceleration and I would have to solve for both. However...I didn't end up considering any forces from the top box and only needed its mass...Did i do something wrong? Cause my answer seems legit to me.
  2. jcsd
  3. Dec 31, 2008 #2
    since the blocks are moving together, the acceleration is the same for each block...so it's only necessary to solve for acceleration in one of the blocks, in this case the 3kg block since we dont know the tension force
  4. Dec 31, 2008 #3


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    Don't you need to accelerate both boxes?

    The total mass being accelerated is 7 kg and not 3 kg then isn't it?
  5. Dec 31, 2008 #4
    That would be if you consider the entire system together... if you are looking at only the x-component of forces acting on the lower box by itself, you would only consider the mass of the lower box...i just checked the posted solutions and the answer is indeed 1.75s ... i guess i just find it suprising that the tension didn't come into play...i thought i would have to find a maximum possible tension before the boxes start slipping, and find the acceleration that corresponds to that max tension
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