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Balancing forces

  1. Dec 3, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]
    What is the minimum force required to hold the rod at point A?

    2. Relevant equations


    3. The attempt at a solution
    If no one is holding the rod, point B would get the force mg from the rod. So the normal force from the hinge to the rod is mg, pointing up. This normal force would create a torque τ, which will cause the rod to rotate.

    τ = (L/2)(mg cos θ)

    The force that holds the rod in place counters this torque by having the component

    Fτ = mg cos θ

    that is perpendicular to the rod, in the clockwise direction. However, now the three forces acting on the rod--gravity, the normal force from the hinge, and the force from the hand--no longer sum up to zero. What is wrong?

    - Thanks
     
  2. jcsd
  3. Dec 3, 2008 #2

    turin

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    Homework Helper

    Forget about B. Just use torques, and take for granted that B is the pivot point. Note: there is a range of forces in different direction that will hold the rod at A. You need to find the minimum.
     
  4. Dec 3, 2008 #3
    Ok, suppose there is a force Fa at A, and we some the torque about B:

    (L/2)(mg cosθ) - LFa = 0

    So the tangential component of Fa must be 0.5 mg cosθ. Suppose I let Fa to have no radial component and let the hinge do the rest, then:

    Fbx = - Fasinθ

    Fby = mg - Facosθ

    Since the torque was summed at B, the force at B won't change the net torque. This shows that the x and y direction forces are also balanced. So Fa must be 0.5 mg cosθ, and it is tangent to the rod in clockwise direction.
     
    Last edited: Dec 3, 2008
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