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Homework Help: Coefficient of Static Friction

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

    A crate of weight Fg is pushed by a force P on a horizontal floor. (a) If the coefficient of static friction is μ s and P is directed at angle θ below the horizontal, show that the minimum value of P that will move the
    crate is given by

    P = usFgSecθ / (1 - usTanθ)

    (b) Find the minimum value of P that can produce motion when μ s = 0.400,
    If the angle were 68.2° or more, the expression for P would go to infinity and motion would become impossible.

    3. The attempt at a solution

    I was able to figure out how to get to P, but I cannot figure out how to find the minimum value of P. I am assuming that if they want the minimum value of P, theta would be equal to 0, since all of the force would be put along the horizontal. I am not sure where exactly to go from there though.
  2. jcsd
  3. Oct 1, 2012 #2
    Mind showing us what you got as P?
  4. Oct 1, 2012 #3
    well P is the same thing as in my original post.

    P = usFgSecθ / (1 - usTanθ)

    Now I dont know how to find the minimum value of P
  5. Oct 1, 2012 #4
    Do they provide the weight of the object?
  6. Oct 1, 2012 #5
    No they do not
  7. Oct 1, 2012 #6


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    What is the derivative of P with respect to θ ?
  8. Oct 1, 2012 #7
    Woops, sorry, i must have missed it. (I was feeling sleepy when i posted my reply, sorry)

    Finding minimum value requires the knowledge of Calculus, do you know how to find the derivative?

    You can do it in an another way too. Rearrange the equation, write sec and tan in terms of sin and cos, you get:
    For P to be minimum, the denominator should be maximum, so simply differentiate [itex]\cos(\theta)-μ_s\sin(\theta)[/itex] with respect to θ.
    Last edited: Oct 1, 2012
  9. Oct 1, 2012 #8
    Find equation of applied force equal to frictional force.
    You will get an equation that is equal to a constant.
    For minimum value of P, the other factor must be maximum.
    Last edited: Oct 1, 2012
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