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Homework Help: 2 Forces with Kinetic Friction

  1. Feb 4, 2008 #1
    The drawing shows a 28.8-kg crate that is initially at rest. Note that the view is one looking down on the top of the crate. Two forces, F1 and F2, are applied to the crate, and it begins to move. The coefficient of kinetic friction between the crate and the floor is k = 0.374. Determine the (a) magnitude and (b) direction (relative to the x axis) of the acceleration of the crate.

    1. The problem statement, all variables and given/known data
    k(kinetic friction)=.374

    2. Relevant equations
    fk(magnitude of kinetic friction)= (k)(Fn)
    ax= (-fk)/m

    3. The attempt at a solution

    I have tried a few things, but i am not sure really what equation to go after, since i feel i should be using something with the angles in it.

    any help is appreciated.

    Attached Files:

  2. jcsd
  3. Feb 5, 2008 #2

    Doc Al

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    Staff: Mentor

    Start by finding the net applied force. Add up the F1 and F2 force vectors. Don't forget that they are vectors, not just numbers.
  4. Feb 5, 2008 #3
    so for getting the magnitude to the right, along the x axis shouldnt i do
    cos(55) (88) =50.47
    then add 54


    If this is right, then where do i go from here.
  5. Feb 5, 2008 #4
    That's the sum of the vectors on the x-axis. To complete the vector addition you need to consider y-axis too.
  6. Feb 5, 2008 #5
    would that be (sin(55))(88) for Y vector?
  7. Feb 5, 2008 #6

    Doc Al

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    Staff: Mentor

    That's correct. That's the y-component of the applied force.

    Use the components that you've calculated for the total applied force to find the magnitude and direction of the total applied force. The direction of the applied force will also be the direction of the acceleration.

    To find the magnitude of the acceleration you will have to include friction, which opposes the motion and acts opposite to the applied force. What's the friction force? What's the total force (not just the applied force) acting on the box?
  8. Feb 5, 2008 #7
    Ok i got it,
    thanks everyone i just worked it out and it was correct.
  9. Feb 5, 2008 #8
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