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?

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davidlcastiblanco davidlcastiblanco · Answer 1 · 2019-11-12T03:49:45+01:00

Answer:

[tex]F=27.39N[/tex]

Explanation:

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

Σ[tex]T=T_{N}+T_{f}+T_{W}[/tex]

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

[tex]T_{N}=0[/tex]

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

[tex]T_{f}=F*(r-h)[/tex]

And the T of gravity according to the image

[tex]T_{W}=W*(\sqrt{r^2-(r-h)^2}[/tex]

Σ[tex]T=0[/tex]

[tex]T_{N}+T_{f}+T_{W}=0[/tex]

[tex]F*(r-h)+W*(\sqrt{r^2-(r-h)^2}=0[/tex]

[tex]F=\frac{W*(\sqrt{r^2-(r-h)^2}}{r-h}[/tex]

[tex]F=\frac{24.9 N*(\sqrt{0.336^2-(0.336-0.110)^2}}{(0.336-0.11)}[/tex]

[tex]F=27.39N[/tex]