What is the maximum height a man can climb on a ladder before it slips?

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AI Thread Summary
The discussion focuses on determining the maximum height a man can climb on a ladder before it slips, considering the ladder's mass, length, and the angle with the floor. The key equations involve balancing forces and torques, leading to a formula for the maximum height based on the coefficients of friction and the masses involved. The final expression for height incorporates the ladder's angle and the frictional forces at play. A clarification was sought regarding the inclusion of the sine component in the reaction force equation, which was subsequently resolved. Understanding these dynamics is crucial for solving similar physics problems involving ladders and friction.
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Homework Statement


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A man of mass m climbs a ladder of mass M and length L. the ladder leans against a friction less wall and makes an angle α with a rough floor (coefficient μ).
To which max height can he climb before the ladder slips.

Homework Equations


Friction: ##f=\mu N##

The Attempt at a Solution


The reaction in the base: ##R_x=(M+m)\mu g\sin(\alpha)##
The perpendicular component of the lader and man's gravity makes torque round the base which is balanced by R, the reaction in the vertical wall:
$$lmg\cos(\alpha)+\frac{L}{2}Mg\cos(\alpha)=RL\sin(\alpha)~~\rightarrow~~R=\frac{\left( lm+\frac{L}{2}M \right)g\cos(\alpha)}{L\sin(\alpha)}$$
$$R=R_x:~~\frac{\left( lm+\frac{L}{2}M \right)g\cos(\alpha)}{L\sin(\alpha)}=(M+m)\mu g\sin(\alpha)$$
The minimal height h:
$$h=l\sin(\alpha)=\frac{2m\mu L\sin^2(\alpha)-ML\cos(\alpha)}{2m}\tan(\alpha)$$
The result must be:
$$h=L\frac{\mu(M+m)\sin(\alpha)-\frac{1}{2}M\cos(\alpha)}{m\cos(\alpha)}$$
 
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hi,
can you explain the ##\sin\alpha## in your ##R_x## ?
 
Thanks BvU, solved
 
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