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Friction: A box at rest on a slope

  1. Dec 18, 2008 #1
    1. The problem statement, all variables and given/known data
    Mass of 22kg on a 45o to the horizontal.
    [tex]\mu[/tex]s = 0.78
    [tex]\mu[/tex]k = 0.65
    Determine the magnitude of the largest force that can be applied upward parallel to the ramp to have the box remain at rest.
    AND
    Determine the magnitude of the smallest force applied to the top of the box, perpendicular to the ramp, to have the box remain at rest.

    2. Relevant equations
    All kinematic-based equations
    Newton's Laws
    FF = [tex]\mu[/tex]FN

    3. The attempt at a solution
    I assumed that the magnitude of the maximum force required would be equal to |FNET| + |FF| I got 2.0 x 102 Newtons of force (correct to two significant digits).

    However, the second part, I'm not quite sure of what to do. I thought that if you are applying a force perpendicular to the ramp, you would be increasing the force of gravity perpendicular to the ramp, and since |FN| = |FgN| the normal force would increase. Then I thought the absolute value of the force of friction would have to equal the absolute value of the force of gravity acting parallel to the ramp. In that case, just solve to find the force needed.

    I was told I was wrong.

    (*side note* I didn't find a use for the kinetic coefficient of friction. I assume that it will be useful, but can't think of how)
     
  2. jcsd
  3. Dec 18, 2008 #2

    PhanthomJay

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    Your equation neglects the component of the gravity force acting down the plane, and, in calculating F_f, you first need to calculate the normal force, which is not the same as the box weight.
    The applied force perpendicular to the box increases the normal force, not the gravity force. You need to draw a free body diagram and examine all forces perpendicular and parallel to the slope, then apply newton's laws.
    Neither can I. Don't let the question trick you. I have to go off line now, so please continue with your attempt, and I would think that one of the night shift owl helpers will assist.
     
    Last edited: Dec 18, 2008
  4. Dec 18, 2008 #3
    I calculated my FNET drawing a FBD and looking at all forces. The only relevant ones were the force of gravity acting parallel to the plane and the force of friction, everything else canceled out. I had 34 Newtons down the plane for Net force.

    You made me realize that I did calculate the Normal force wrong (forgot to include cos 45o), which did skew my answer. So force required would be 1.5 x 102 instead.)

    Yeah, for the second part, I didn't think it was clear enough. Since I am increasing the normal force, would I not be correct in thinking that it would increase friction, and that I am trying to find an applied force that increases normal force enough so that the force of friction counteracts the net force? That is where I was told i was wrong.
     
  5. Dec 19, 2008 #4

    PhanthomJay

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    Well, in order for the box to remain at rest, the net force must be zero, which implies that you must apply a force of 34N up the plane to keep it at rest.
    I'm missing you here. How did you get from 34N to 150N?
    you may have been told you were wrong, because you are trying to find an applied force that increases normal force enough so that the force of friction counteracts the force of gravity down the plane, not the net force. The net force must be zero. Please show your work for this calculation, and i'll give it a look.
     
  6. Dec 19, 2008 #5
    Okay, well my scanner is on the fritz at the moment, but as soon as I fix it, sure thing.
     
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