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Coefficient of static friction for tongs.

  1. May 29, 2013 #1
    1. The problem statement, all variables and given/known data

    I´ll show the problem with a picture:

    2s0ldfp.jpg

    2. Relevant equations

    M=rxf

    Sum Forces = 0

    Sum Moments= 0

    3. The attempt at a solution

    I recognize that I am completely lost with this problem, so I´ll try to analyze every force acting on every body to show where are my doubts.

    When I analyze forces and moments acting at the part T1, I see a red force F2 applied to it, it is vertical, T1 acts on the tube through F3, I think that F3 is much closer to the axis of rotation at point A, so F3 should be bigger than F2, here arises my first question.

    The equation of moments for T1 tells us that F3 is bigger than F2, but What happens with the forces?

    I assume that there must be an equal and opposite reaction from the tube against T1, but it is difficult to imagine, that force must come from the tong T2 acting through F4, which I assume that is equal to F7, so on the y axis F3 and F7 should compensate, and on the x axis the friction should be equal to F7x+F3x. IF that is true, there is a net vertical force acting to T1 which is F2 and that can´t be true in static conditions.

    But I don´t know how to calculate the minimum value since Friction=μ N, I don´t know μ but I don´t know N too, I know that F5 and F6 should compensate forces throught x axis but I don´t know how to calculate this.

    Any suggestions, comments ... are welcome.
     
  2. jcsd
  3. May 29, 2013 #2
    Draw a free body diagram of the tube. The tong surfaces are tangent to the tube. Use geometry to identify the two contact points of the tongs with the tube. These are the only locations where forces are being exerted on the tube. These forces can be resolved into components normal and tangent to the tube.

    Chet
     
  4. May 29, 2013 #3
    Well I draw that but since N are orthogonal to the friction force F I cannot understand how to solve the statics equations.

    I´m sorry if I am not able to see the solution.
     

    Attached Files:

  5. May 29, 2013 #4
    No problem. I can help you some more. Draw lines from the center of the tube to the tangent points with the tube. What angle do these lines make with a vertical line (y-direction) through the center of the tube? Let the normal component of each contact force of the tongs with the tube be N. When the the tube is just about to slip out of the tongs, the tangential components of the contact forces will be μN. Draw these 4 components of the contact forces on your free body diagram. Note the directions that each of the 4 components is pointing. Resolve each of the 4 components into components in the x- and y-directions. By symmetry, the components in the y-direction should cancel and sum to zero. Sum the components in the x-direction, and set the sum to zero (equilibrium in the x-direction just prior to slippage).

    Chet
     
  6. May 30, 2013 #5
    Thank you very much!!!

    I got the right result, I´ll show the equation:

    -N sin(10) +μN cos(10) +μN cos(10) -N sin(10)=0

    Finally:

    μ= sin(10)/ cos(10) = 0.1763

    This problem has helped me a lot to understand how friction forces act, thanks!!!
     
  7. May 30, 2013 #6
    Wow!! You're welcome. It's very gratifying to receive a "thank you" like this.

    Chet
     
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