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Coefficient of Friction, and frictional force

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data
    1st problem: If and object has a weight of 100 N and a frictional force of 50 N exerted on it, what is the coefficient of friction?

    2nd problem: A Block with a mass of 50 kg, is pulled with a force of 10 N at an angle of 10 degrees to the horizon, if the coefficient of friction is 0.1, what is the frictional force?


    2. Relevant equations
    F = uN
    with F = Frictional force
    u = mew, coefficient of friction
    N = normal force
    Sin and Cos functions


    3. The attempt at a solution

    1) 50N = u*100N
    50N/100N = u = 0.5
    The solutions i found for this online says the answer is 0.2, which seems incorrect to me. Is my answer of 0.5 true?

    2) F = (0.1) * [(50kg*9.8)-10sin(10)]
    F = 48.8

    the solution i found online to problem two uses the cos instead of sin function. This does not make any sense to me as the sin is what will give you the upwards force of the rope pulling the block right? Cos would give you the horizontal force which does not affect the normal, is my approach correct?
     
  2. jcsd
  3. Oct 15, 2011 #2
    I am not sure for the first one as it seems there are other things that must be assumed that I am unsure of.

    For the second, if the coefficient of friction is static friction (doesn't really matter if it is kinetic as static is usually larger) and the object is initially at rest (which should imply that it is static friction):

    ---
    F_x = 10N * cos 10
    F_y = 10N * sin 10

    n = m*g - 10N * sin 10

    F_s,max = 0.1 * (m*g - 10N * sin 10) = 0.1 * (50kg*9.8m/s^2 - 10N * sin 10) = 48.83N

    10N * cos 10 = 9.848 N
    ---

    What you must note is that mew of 0.1 is its max value. That is, if the force of 10N in the horizontal direction is less than mew,max * n, then mew * n = applied horizontal force.

    So the frictional force should be 10N * cos 10 = ~9.85 N. This implies that we are taking some mew < 0.1 to make the net force zero.

    If you used mew max, then your net force would be 9.85N - 48.84N = -38.99N, which implies that the object begins to accelerate in the opposite direction than the force which does not make sense. That is, if the object was initially at rest.
     
  4. Oct 15, 2011 #3
    I dont really follow what your saying on the second part. The general equation the solution that i found is using i s

    F = 0.1 * {(m*g)-10cos(10)}
    Which comes out to an answer of 48N
     
  5. Oct 15, 2011 #4

    PhanthomJay

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    As sefrez noted, how can you get a frictional force in the negative x direction of 48 N if the pulling force in the x direction is only 10cos10 in the positive direction? The static friction force friction force is always less than uN unless the object is just on the verge of moving relative to the surface it is in contact with (in which case it is equal to uN, but that is not the case here). Use Newton'1st Law to find the friction force.
     
  6. Oct 15, 2011 #5
    How does using cosine in the Force of friction equation even impact this in any way? that still doesnt make any sense because the cos tells you the horizontal force acting on the block, which as i said in my first question does not effect the normal force, which wouldnt affect the friction
     
  7. Oct 15, 2011 #6
    also the problem statement doesn't give any mention of it being static of kinetic friction
     
  8. Oct 15, 2011 #7

    PhanthomJay

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    Since the horizontal force (10 sin 10) is less than the maximum limiting value of the friction force, the block cannot move, and therfore, static friction applies, and static friction is less than or equal to uN. To find the static friction force, use Newton 1 in the x direction. It will not depend on N.
     
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