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Question

The figure shows a bicycle wheel resting against a small step whose height is h = 0.140 m. The weight and radius of the wheel are W = 22.0 N and r = 0.390 m. 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?

Answer 1

Consider a FBD of the wheel at the instant the wheel lifts
There is the horizontal force, F
and the reaction force of the wheel against the step
Assuming no acceleration of the wheel,
the reaction force will have a horizontal
component of -F and a vertical component of 22 N

Now, sum the torques at the ground directly below the axle
take clockwise rotation as positive.

First, we need to do some geometry and trig to find the moment arm and angle of the reaction force at the step
The horizontal distance is
L=sqrt(r^2-(r-0.140)^2)
or
L=sqrt(0.39^2-(0.39-0.140)^2)
L=0.299 m
the moment arm from the point on the ground directly below the axle and the step is
D=sqrt(0.299^2+0.140^2)
D=0.330 m

The angle from horizontal to the moment arm is
th=asin(0.140/0.330)
25.1 degrees


Now sum the torques
F*0.39-22*cos(25.1)*0.330-
F*sin(25.1)*0.330

F=26.3 N

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Answer 2

22 N, assuming that the force is applied straight up.

Answer 3

I confirm the results of odu83.