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Coefficient of static friction calculation

  1. Oct 13, 2013 #1
    1. The problem statement, all variables and given/known data
    I have a box on a horizontal plane.
    Given data: mass = 0.25kg
    height of object = 14 cm
    width of object = 9cm
    Calculate the coefficient of static friction.

    2. Relevant equations
    μs = Ff/N (N is the normalforce)
    F = Ff = μ*N
    mg = 2.45 Newton

    3. The attempt at a solution
    I apply an unknown force on the object from the left side at the height of x to see where it starts to tip over instead of slide. Looks like it's at 13 cm. x = 13

    The vertical forces:
    N cancels out mg since the object is not accelerating up or down, so N=mg.
    μk is irrelevant in this case so I'll just write μ when I mean μs.

    The horizontal forces:
    Ff is acting in the opposite direction of the applied Force
    F = Ff = μ*N
    L is the length to the middle 4.5cm
    mg*L - F*x = 0
    I substitute in μ*mg instead where F*x is
    mg*L - μ*mg = 0

    This is where I think I might be going wrong:
    μ = mg*L/mg
    μ = L
    Which would mean the coefficient of static friction is 4.5 or 0.45 which is too high.
    When I angle (at 12 degrees) the same object on the same surface and calculate the μs it became 0.21

    If I go this way it's slightly better but probably wrong since one m disappears:
    μ = mg*L/g = 0.24
  2. jcsd
  3. Oct 13, 2013 #2


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    Remember that the forces that generally generate torques about the tipping point are:
    The unknown horizontal force, gravity, and the normal force.
    The frictional force act along the ground and generates no torque at all about the tipping point.

    The box begins to tip when a) The normal force acts through the tipping point (generating no torque), and there is an equal, but opposite torque from the force F and gravity.

    From this relation, you may determine F, and thus the coefficient for maximal static friction.

    This equation "mg*L - μ*mg = 0" is dead wrong, your first term is a torque, the other a force

    It should read mgL=(mu)*mg*x, that is: (mu)=L/x, roughly 0.35 by my reckoning.
  4. Oct 13, 2013 #3
    Thanks a lot, that helps me.
    The equation "mg*L - μ*mg = 0" was what my teacher wrote down for me when he was trying to help. First we have
    mg*L - F*x = 0
    and as mentioned earlier F = Ff = μ*N and N=mg
    that's how I got the "dead wrong equation". He was using it to separate μ.
  5. Oct 13, 2013 #4


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    It is unfortunate that your teacher made that silly mistake. It is easy to do this sort of thing, and should not be held against him (unless he has a general tendency to mix up forces and torques)

    F*x=(mu)mg*x, agreed?
  6. Oct 13, 2013 #5
    I did some testing and I would get the same μs on the angled surface test as in the horizontal test if I applied the force at 14 cm. That is μs = 0.32
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