Question

The drawing shows a bicycle wheel resting against a small step whose height is h = 0.110 m. The weight and radius of the wheel are W = 24.9 N and r = 0.336 m, respectively. A horizontal force vector F is applied to the axle of the wheel. As the magnitude of vector 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

`F=27.39N`

Step-by-step explanation:

Take sum of torques at the point the step touches the wheel, that eliminates two torques

Σ
`T=T_(N)+T_(f)+T_(W)`

Since we are looking for when the wheel just starts to rise up N-> 0 so no torque due to normal force

`T_(N)=0`

The perpendicular lever arm for the F force is R-h

`T_(f)=F*(r-h)`

And the T of gravity according to the image

`T_(W)=W*(√(r^2-(r-h)^2)`

Σ
`T=0`

`T_(N)+T_(f)+T_(W)=0`

`F*(r-h)+W*(√(r^2-(r-h)^2)=0`

`F=(W*(√(r^2-(r-h)^2))/(r-h)`

`F=(24.9 N*(√(0.336^2-(0.336-0.110)^2))/((0.336-0.11))`

`F=27.39N`