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Bicycle and Pulley, part 1

  1. Dec 18, 2006 #1
    1. The problem statement, all variables and given/known data
    What will the scale read in these 2 examples?
    The incline is 12 degrees. The bicycle weight is 14kg. The cylinder weight is 14kg.

    2. Relevant equations
    Force of 2nd pulley 'arm' is 1/2 weight
    Force on incline is force X sin(angle)

    3. The attempt at a solution
    A scale is attached with wires to the circumference of the bicycle wheel and the bicycle frame. The line of the wire is at a tangent to the wheel. The measurement on the scale is: (14kg) (sin12) = 2.9kg

    Another scale is attached similarly to the circumference of the cylinder, except the scale is attached to a frame mounted to the ground. The measurement of the scale is: .5 (14kg) (sin12) = 1.45kg
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Dec 18, 2006 #2


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    Staff: Mentor

    Could you explain the drawing a bit? I see the tension scale on the cylinder's rope, but don't understand the bicycle part and how it attaches/relates.
  4. Dec 18, 2006 #3
    There is a tension scale on the bike part also. It is connected to the rear tire on one side and seat post on the other side. The scale is supposed to come off as a tangent to the rear tire.
  5. Dec 19, 2006 #4
    I'll go a little further in my interpretation to see if it sounds right.
    In the bicycle there is a force, F1 = m*g*sin12, that is pulling the bicycle down the incline. The same force is pulling down the cylinder.
    On the pulley there are 2 opposing forces. F2, the friction force of the ground on the cylinder and F3, the force of the ground support that is pulling the rope. Therefore F2 + F3 = F1. If this is right, I only need to find the relationship between F2 and F3 to find their values.
    On the bicycle there is only 1 opposing foce, F4. Where the rope is attached to the bike frame, the rope is pulling the same direction as F1, so it does not oppose F1. Therefore F4 = F1
    Last edited: Dec 19, 2006
  6. Dec 19, 2006 #5
    T = rF
    F2 and F3 are both acting on the cylinder, so F2 = F3

    F1 = F4
    F3 = 1/2 F4
  7. Dec 20, 2006 #6


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    Homework Helper

    It's a bit difficult to follow which forces you are talking about, but if I am following them correctly your results look OK to me. For the cylinder, I am assuming the lower strap is attached to the floor, so friction is not a factor, but even if it is not attached and friction keeps it from slipping the reading on the scale would be the same. The upper and lower straps equally share the force needed to keep the bike frorm moving in order for the net torque to be zero, so F2 = F3 looks right.

    For the bicycle the only force keeping the bike from moving is friction, so the magnitude of the torque at the bottom is rF1 and the magnitude at the top must be rF4 = rF1, so I agree with you that F4 = F1 and F4 is half of F2 or F3.
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