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Torque around a curb - bicyle forces problem

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data
    A bicycle tire with radius R is attempting to go over a curb of height h. If a force F is applied to the center of the wheel, at an angle θ above the horizontal, what angle results in a force with the smallest magnitude to raise the wheel over the curb?

    2. Relevant equations
    ƩT=0
    ƩF=0 (?)


    3. The attempt at a solution
    Having done this problem for a solely horizontal force, I tried to copy the same idea, but separated F into x and y components (giving Fcosθ and Fsinθ). The distance to Fg and Fy was √(2Rh-h^2), and the distance to Fx is R-h. After doing this, Fg is + and the other two are negative. The net torque equation I get it
    0=mg*√(2Rh-h^2)-Fy*√(2Rh-h^2)-Fx*(R-h)
    I don't know where to go from here (or if this is correct).
     
  2. jcsd
  3. Apr 3, 2013 #2
    Try thinking about the problem this way.

    First, Assume the tire pivots about the point where the tire makes contact with the curb. The torque due to gravity is your first term mg*√(2Rh-h^2). Now the angle of the minimum force will be where the force is perpendicular to the line from the center of the wheel to the pivot point.

    Try setting FR = mg*√(2Rh-h^2) to find the force.
     
  4. Apr 3, 2013 #3
    How do I know that the angle perpendicular is the smallest? Also by FR do you mean the applied force? Sorry! I've been on this problem so long I don't even know what I'm thinking anymore
     
  5. Apr 3, 2013 #4
    FR is force X tire radius. You will get maximum torque when the force is perpendicular to the line from the tire center to the point of pivot, i.e. where the tire meets the curb. You can probably use geometry to fine the perpendicular angle.
     
  6. Apr 3, 2013 #5
    Oh! I understand. So to figure out the angle, it's probably easier to use geometrical arguments than net torque?
     
  7. Apr 3, 2013 #6
    I would think so. The angle would be a function of h and R. I think it would involve a arctan function probably. The force will be in terms of h, R, m, and g.
     
  8. Apr 3, 2013 #7
    The answer provided gives an arcsin function with just r and h, but I'm going to try it for a bit and see what I can get.
     
  9. Apr 3, 2013 #8
    I think you are correct. As I look at it, I see something like arcsin (R-h)/R -90 or something similar.
     
  10. Apr 3, 2013 #9
    Yes! That worked out. Thank you so much!
     
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