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

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SUMMARY

The maximum height a man can climb on a ladder before it slips is determined by the equation: $$h=L\frac{\mu(M+m)\sin(\alpha)-\frac{1}{2}M\cos(\alpha)}{m\cos(\alpha)}$$. This equation incorporates the mass of the man (m), the mass of the ladder (M), the length of the ladder (L), the angle (α), and the coefficient of friction (μ) between the ladder and the floor. The analysis involves balancing forces and torques acting on the ladder, particularly focusing on the reaction forces at the base and the wall.

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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:
$$h=L\frac{\mu(M+m)\sin(\alpha)-\frac{1}{2}M\cos(\alpha)}{m\cos(\alpha)}$$
 
Physics news on Phys.org
hi,
can you explain the ##\sin\alpha## in your ##R_x## ?
 
Thanks BvU, solved
 

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