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

Answer:

$F=27.39N$

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

$T_{W}=W*(\sqrt{r^2-(r-h)^2}$

Σ$T=0$

$T_{N}+T_{f}+T_{W}=0$

$F*(r-h)+W*(\sqrt{r^2-(r-h)^2}=0$

$F=\frac{W*(\sqrt{r^2-(r-h)^2}}{r-h}$

$F=\frac{24.9 N*(\sqrt{0.336^2-(0.336-0.110)^2}}{(0.336-0.11)}$

$F=27.39N$