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AleksAgata [21]
3 years ago
10

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 a

re 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?

Physics
1 answer:
Pavel [41]3 years ago
4 0

Answer:

F=27.39N

Explanation:

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

ΣT=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}

ΣT=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

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