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Incline Plane with Friction

  1. Oct 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the force necessary to start the crate moving, given that the mass of the crate is 32 and the coefficient of static friction between the crate and the floor is 0.58. The angle is 21 degrees below horizontal.

    2. Relevant equations



    3. The attempt at a solution

    I don't really understand where to go from here, I've found out the answer but really don't understand why we use the equations we do. Would appreciate any help regarding the logic behind the methods!

    My attempt:

    Weight = 32kg x 9.81 = 313.9 Newtons
    Normal Force = 313.9 cos(21) = 293 Newtons
    The force parallel to the plane = 313.9(sin21) = 112.5N

    The Force of Friction = 293 x .57 = 167N

    Now this is pretty much the farthest I can get without understanding what I'm doing. I think I did the above steps right I'm not completely sure. Would appreciate any help.
     
  2. jcsd
  3. Oct 10, 2012 #2
    so the crate is sitting on some slope, but is not moving because the frictional force acting on it is greater than the force due to gravity pushing it down the slope (which is the weight parallel to the surface)

    so right now Ffric > Fweight

    let's say you apply some force F in the same direction as Fweight

    how big would F need to be in order to make Fweight + F > Ffric?
     
  4. Oct 10, 2012 #3
    I guess any number that you can add to make the force greater than the frictional force? So in our case the force parallel is is 112.5 and the Frictional force is 167. So we need an extra 54.5 Newtons to make the crate move.
     
  5. Oct 10, 2012 #4
    pretty much
     
  6. Oct 10, 2012 #5
    So how would that help us in getting the answer of 250 Newtons?

    In the solution they put,

    (.57) (313) / cos21 - .57sin21 = 250 Newtons

    I don't really get how we end up there from what we have so far.
     
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