View source for Chapter 10 Problem 93

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==Problem==
[[File:Chapter10Problem93q.png|300px|center|diagram]]
A wheel of mass <math>M</math> has radius <math>R</math>. It is standing vertically on the floor, and we want to exert a horizontal force <math>F</math> at its axle so that it will climb a step against which it rests. The step has height <math>h</math>, where <math>h<R</math>. What minimum force <math>F</math> is needed?
==Solution==
[[File:Chapter10Problem93s.png|300px|right|diagram]]
The wheel is rolling about the point of contact with the step, and so all torques are to be taken about that point. As soon as the wheel is off the floor, there will be only two forces that can exert torques on the wheel – the pulling force and the force of gravity. There will not be a normal force of contact between the wheel and the floor once the wheel is off the floor, and any force on the wheel from the point of the step cannot exert a torque about that very point. Calculate the net torque on the wheel, with clockwise torques positive. The minimum force occurs when the net torque is 0.

<math>\sum\tau=F(R-h)-mg\sqrt{R^2-(R-h)^2}=0</math>

<math>F=\frac{Mg\sqrt{R^2-(R-h)^2}}{R-h}=F=\frac{Mg\sqrt{2Rh-h^2}}{R-h}</math>
<math></math>