A uniform ladder with mass $ m_2 $ and length $ L $ rests against a smooth wall. A do-it-yourself enthusiast of mass $ m_1 $ stands on the ladder a distance $ d $ from the bottom (measured along the ladder). The ladder makes an angle $ \theta $ with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude $ f $ between the floor and the ladder. $ N_1 $ is the magnitude of the normal force exerted by the wall on the ladder, and $ N_2 $ is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive.

Question

A uniform ladder with mass $ m_2 $ and length $ L $ rests against a smooth wall. A do-it-yourself enthusiast of mass $ m_1 $ stands on the ladder a distance $ d $ from the bottom (measured along the ladder). The ladder makes an angle $ \theta $ with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude $ f $ between the floor and the ladder. $ N_1 $ is the magnitude of the normal force exerted by the wall on the ladder, and $ N_2 $ is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive.(a) What is the minimum coefficient of static friction $ \mu_{\text{min}} $ required between the ladder and the ground so that the ladder does not slip?
Express $ \mu_{\text{min}} $ in terms of $ m_1 $, $ m_2 $, $ d $, $ L $, and $ \theta $.
, (b) Suppose that the actual coefficient of friction is one and a half times as large as the value of $ \mu_{\text{min}} $. That is, $ \mu_s = \dfrac{3}{2} \mu_{\text{min}} $. Under these circumstances, what is the magnitude of the force of friction $ f $ that the floor applies to the ladder?
Express your answer in terms of $ m_1 $, $ m_2 $, $ d $, $ L $, $ g $, and $ \theta $.
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Jason · Accepted Answer

$\mu_{\min} = \frac{\left(\tfrac{m_2 L}{2} + m_1 d\right)\cos\theta}{(m_1+m_2)L\sin\theta}$
$f = \frac{g\cos\theta\left(\tfrac{m_2 L}{2} + m_1 d\right)}{L\sin\theta}$

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