1. Sep 3, 2016

### Karol

1. The problem statement, all variables and given/known data

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.

2. Relevant equations
Friction: $f=\mu N$

3. 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:
$$h=L\frac{\mu(M+m)\sin(\alpha)-\frac{1}{2}M\cos(\alpha)}{m\cos(\alpha)}$$

2. Sep 3, 2016

### BvU

hi,
can you explain the $\sin\alpha$ in your $R_x$ ?

3. Sep 3, 2016

### Karol

Thanks BvU, solved