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Simple Coefficient of Friction Question

  1. May 21, 2010 #1
    1. The problem statement, all variables and given/known data

    A ring on a rod of weight 8 N. A 7 N force at an angle of 45 degrees to the rod is acting away from the ring. The system is in limiting equilibrium. Find the coefficent of friction between the rod and the ring.

    2. Relevant equations

    Quite obvious.

    3. The attempt at a solution

    Could someone explain to me as to why the correct solution is 1.62 and not 0.62? If the former is the case doesn't the COF change everytime a different force is applied?
     
  2. jcsd
  3. May 21, 2010 #2

    PhanthomJay

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    May we have a picture please? Or a better description of the problem? 1.6 is an unusually high friction coefficient, which is independent of the loading. But I don't understand the problem.
     
  4. May 21, 2010 #3
    A ring of weight 8 Newtons is threaded on a rod. An external force of 7 Newtons is acting at an angle of 45 degrees from the horizontal on the ring and is trying to pull it to the right. The system is in limiting equilibrium. The first part asks us to show that the normal contact force exerted by the rod on the ring is 3.05 N which can easily be found by subtracting the vertical component (7sin45) of the external force from the weight (8 N). Then in the next part it asks us to find the COF.

    My solution was to first find the maximum value of Friction by equating the Horizontal component of the external force (7cos45) with friction since the system is in limiting equilibrium. The problem comes when using F = mew x R. I used the R when an external force is not being applied (8 N) and divided 7cos45 by it to find 0.62. But most other people who took the exam divided it by the R found in the previous part (3.05) to find 1.62 as the COF. I wanna know which method is correct and why.
     
  5. May 21, 2010 #4

    PhanthomJay

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    At limiting equilibrium, the ring is just on the verge of moving, so the static friction force must equal the horizontal component of the applied force, and be at its maximum value, F_f = uN, where N is the normal contact force between the rod and ring. So if F_f = 7 cos 45, and N = 3.05, then u= 1.6 . You can't use N = 8 if N is actually 3.05.
     
  6. May 21, 2010 #5
    Ohhhh I get it now. So when calculating Mu, we always take the value of N when the system is in limiting equilibrium. Is that correct?
     
  7. May 21, 2010 #6

    PhanthomJay

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    Yes. For objects in equilibrium, the static friction force, F_f, is always less than or equal to u_s(N). At limiting equilibrium, it is exactlty equal to u_s(N). u_s is the coefficient of static friction which is a function of the surfaces of the materials in contact only. N is always the normal force.
     
  8. May 22, 2010 #7
    Ok thanks. That clears up a lot.
     
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