A bicycle wheel of radius R and mass M is at rest against a step of height r = 0

Question

A bicycle wheel of radius R and mass M is at rest against a step of height r = 0.35R, as illustrated in the figure below. Find the minimum horizontal force F that must be applied to the axle to make the wheel start to rise up over the step. (Use g, R, and M as necessary.)

F= ______________

Explanation / Answer

If wheel starts to rise up over the step, the point of contact O between wheel and step becomes axis of rotation. Let C = center of wheel , P = point directly below C but at same level as top of the step. Then, the torque about O due to weight of wheel is (1) ..... Tw = M * g * OP where ( OP )^2 = ( OC )^2 - ( PC )^2 = R^2 - ( PC )^2 (2) ..... PC = R - 3 * R / 4 = R / 4 .......... ( OP )^2 = R^2 - ( R / 4 )^2 = R^2 - R^2 / 16 = 15 * ( R^2 ) / 16 .......... OP = R * SQRT(15) / 4 It follows that (1) can be rewritten as (3) ..... Tw = M * g * R * SQRT(15) / 4 The torque due to the hiorizontal force F is (4) ..... Thf = F * ( PC ) = F * R / 4 [ see (2) above ] To make the wheel rise up the step, Thf >= Tw, which yields .......... F * R / 4 >= M * g * R * SQRT(15) / 4 (5) ..... F >= M * g * SQRT(15) The minimum horizontal force F that must be applied to the axle to make the wheel rise up over the step is F = M * g * SQRT(15).=37.993*M

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