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Homework Help: Coefficient of static friction problem

  1. Oct 2, 2014 #1
    1. The problem statement, all variables and given/known data
    I got this problem:
    A wedge of mass m = 36.1 kg is held in place on a fixed plane that is inclined by an angle θ = 21.3° with respect to the horizontal. A force F = 302.3 N in the horizontal direction pushes on the wedge, as shown in Figure 4.25a. The coefficient of kinetic friction between the wedge and the plane is μk = 0.159. Assume that the coefficient of static friction is low enough that the net force will move the wedge.

    basically what's there except friction would be the other way around (since its accelerating up the slide.

    2. Relevant equations
    Sum of all forces = ma

    3. The attempt at a solution
    I did a free body diagram and got:
    Summation of forces in y:
    Normal force = mg sin theta + Appliedforce sin theta

    Then for the forces in x:
    mgcos + friction - Aplliedforce sin = 36.1a

    I did solve for everything but strangely the acceleration gives me acceleration = 3.2

    Can any1 help me if im doing something wrong? Like getting cos and sin mixed?
  2. jcsd
  3. Oct 2, 2014 #2


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    Check "Normal force = mg sin theta + Appliedforce sin theta"
    mg is in a different direction than F, so how can it have the same coëfficiënt ?
    also check the angle. Why do you state 21.3 and the text 22 degrees ?

    What is so trange about acceleration 3.2 ?
    You sure the picture and your exercise match ? Why do you want to calculate the acceleration if the exercise says the task is to draw an FBD ?

    Oh, and: welcome to PF :)
  4. Oct 2, 2014 #3
    Oh! Ok, I think fixing the cos and sin around helped me out
    it was mg cost + Forceappliedsin = Normal force (sum of forces in y)
    and mg sin + friction - Forceappliedcos = ma

    I just have another question. The acceleration here would be negative right? My teacher had told me that itw as positive but I wasn't sure why but I'm also pretty sure it's actually negative.

    Thank you for the welcoming too :D

    Just a fast MC question I had a doubt with:
    The Tornado is a carnival ride that consists of a hollow vertical cylinder that rotates rapidly about its vertical axis. As the Tornado rotates, the riders are pressed against the inside wall of the cylinder by the rotation, and the floor of the cylinder drops away. The force that points upward, preventing the riders from falling downward, is
    tension, friction, gravity or normal force.

    I'm pretty sure its friction due to the fact that normal would act perpendicular (to the center of the cylinder) and gravity would point downward and well tension would not exist.
    Am i right?
  5. Oct 2, 2014 #4


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    You are right about your last comment. Note that the normal force is acting as the centripetal force. And the friction force is acting upward.
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