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Center of gravity-torque

  1. Dec 9, 2005 #1
    Please help this is due by Monday at 8:00am Eastern time. I don't know if I am did this problem correctly, please help.

    Here is the question:

    A bicycle is resting against a small step whose height is h=0.120m. The weight and radius of the wheel are W= 25.0 N, and r=0.340m. A horizontal force F is applied to the axle of the wheel. As the magnitude of F increases, there comes a time when the wheel just begins to rise up and loses contact with the ground. What is the magnitude of the force when this happens?

    This is what I did:

    =-F(0.120m)+(25.0 N)*(0.340)=0
    F=(25.0 N)*(0.340m)/(0.120m)=70.83= F=71 N

    Please tell me if this is correct or not before Monday, I am a new member to this site so please bear with me--Thanks!
  2. jcsd
  3. Dec 9, 2005 #2
    No, when you chose the axis of rotation, you got the wrong distances.
  4. Dec 10, 2005 #3
    Resolve the vectors to perpindicular

    From yours:
    This is what I did:

    =-F(0.120m)+(25.0 N)*(0.340)=0
    F=(25.0 N)*(0.340m)/(0.120m)=70.83= F=71 N


    I believe you have to resolve the two forces acting in this equilibreum to the perpindicular and paralell vectors. The perp vectors act to give torque, the parallell ones do nothing at all on the equilibreum.

    the dist from step up to center is .220m

    Using trig: sin(@) =.220/.340
    @=40.32 degrees (below the 0 degree plane) (I'll use 40 degrees)

    Get the perpindicular component of the 25 N force: Lay out the triangle on paper and find all the angles:
    cos 40 = perp force/25 N
    perp force = 19.15 N

    Get the relationship for the perp component for the applied force:
    cos 50 = perp component force/force applied horizontally.

    As you pointed out, the two perpindicular forces resolved from (one from the wheel weight down, one from the applied force horiz) must equal 0 for equilibreum, same ccw and cw torque. so: that means that the perp force component must equal 19.15 N for the applied force, since this is what the other is. SO:

    cos 50 = perp component force/force applied horizontally.

    F applied horiz x cos50 =19.15
    F applied horiz = 29.79 N <ans

    Let me know what you think. Again its been awhile, this seems like a small number to me. I am sure though that you can not just use the vectors given and sum the torques to 0 without considering the angles.
  5. Dec 10, 2005 #4
    yes, that's what i got.
  6. Dec 11, 2005 #5

    Thanks for your help, I'll check it with my teacher on Monday and let you know.
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