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Force and Friction on an Inclined Plane

  1. Apr 15, 2004 #1
    This question is driving me insane please help.

    A 25 kg block of ice slides down an incline 2.5 m long and 1.5 m high. A man pushes up the ice parallel to the incline so it slides down at a constant speed. The coefficient of friction between the ice and incline is 0.10. Find the force exerted by the man.
     
  2. jcsd
  3. Apr 15, 2004 #2

    Doc Al

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    Staff: Mentor

    Sliding at constant speed implies equilibrium: the forces in any direction add to zero.

    Start by identifying all the forces on the block of ice. I count four.
     
  4. May 12, 2005 #3
    ok I have the same problem.

    The four forces I have found are:

    Fg = 147.1N (directly down)
    FN = 195.9N (perpendicular to the slope)
    Ff = 19.6N (up parallel the slope)
    F = 147.1N (down parallel the slope)

    just want to make sure here..

    now I add the 4 vectors together and find out what Force I need to add (up parallel to the slope) in order to make the net force = 0

    ?


    thanks
     
  5. May 12, 2005 #4
    First, you have to draw your "free body diagram". (Did you do that bullroar?)
    There are trig functions that should be applied here because the block is at an incline of [tex]\theta[/tex].
    Once you draw it, you should have something like::

    [tex] \Sigma F_{x} = F_{g}\sin\theta + F_{push} + f_{k} = ma[/tex]
    [tex] \Sigma F_{y} = n - F_{g}\cos\theta = 0[/tex]
     
  6. May 12, 2005 #5

    Doc Al

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    Staff: Mentor

    Not sure where you are getting these numbers. The four forces are:
    (1) weight = mg, acting down. This will have components perpendicular and parallel to the incline. If you call the angle that the incline makes with the horizontal [itex]\theta[/itex], then [itex]W_{parallel} = mg \sin \theta[/itex] down the incline and [itex]W_{perpendicular} = mg \cos \theta[/itex] into the incline.

    (2) normal force, N, which is the reaction to the forces pulling the block against the incline. [itex]N = mg \cos \theta[/itex], acting out of the plane. (This comes from setting the net force perpendicular to the plane equal to zero.)

    (3) friction force = [itex]\mu N = \mu mg \cos \theta[/itex] acting up the plane.

    (4) the applied force F that acts up the incline.​

    To find the applied force, do as PhysicsinCalifornia advised: set the net force parallel to the plane equal to zero: [itex]mg \sin \theta - \mu mg \cos \theta - F = 0[/itex].
     
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