How to show equlibrium forces in two ladders leaning against each other, other ends resting on rough floor?

Question

Two equal ladders of weight W each are placed so as to lean at A against each other with their ends resting on a rough floor, given the coefficient of friction is u. The ladders at A make an angle 2a with each other. Find what weight on the top would cause them to slip. I am getting only one equation 2N=(2W+w) where N=normal reaction I am getting this equation both using force and using moment equilibrium. Plz suggest

Answer 1

Let $R$ be the normal reaction at the foot of either ladder, $S$ the (symmetrically) horizontal reaction between the ladders at the top, $w$ the weight of each ladder and $W$ be the required weight. Let $2a$ the lenght of each ladder, $2\alpha$ the angle that they make with each other and $\mu$ the friction coefficient.

$W$ will produce a vertical pressure downwards on each ladder equal to $\frac{W}{2}$; hence, for either ladder, resolving horizontally and vertically, we have $$\mu R=S\quad\text{and}\quad R=w+\frac{W}{2}$$

Taking moments about the foot of either ladder, we have $$w\cdot a\sin\alpha+\frac{W}{2}\cdot 2a\sin\alpha=S\cdot 2a\cos\alpha$$ where $2a$ is the length of a ladder.

Hence $$\frac{w+W}{2}\tan\alpha=S=\mu R=\mu \frac{2w+W}{2}$$ whence $$W=w\frac{2\mu-\tan\alpha}{\tan\alpha-\mu}$$

If $\tan\alpha > 2\mu$, the weight is negative, i.e. the ladder would have to be held up in order that there should be equilibrium ; if $\tan \alpha < \mu$, the weight is again negative, and we should then only get limiting equilibrium if the ladder were held up, and in this case the feet would be on the point of moving towards one another.