consider the following problem from mechanics by Davd Morin:

"A ladder of length L and mass M has its bottom end attached to the

ground by a pivot. It makes an angle θ with the horizontal and is held

up by a massless stick of length l that is also attached to the ground by a

pivot (see Fig. 2.22). The ladder and the stick are perpendicular to each

other. Find the force that the stick exerts on the ladder."

The solution comes out to be $$F=\frac{MgL}{2l}\sin{\theta}$$

But as $\theta$ approaches $0$ should $F$ not approach $\frac{MgL}{2d}$ where d is the distance between the pivots of the stick and the ladder because then the stick would behave like a small pivot, and will have to balance the torque due to the weight of the ladder?

## Best Answer

Take the sum of the torques abbaut point A , you obtain

$$ M\,g\,\cos(\theta)\frac L2-F\,x=0$$

thus

$$ F=\frac {M\,g\,\cos(\theta)\frac L2}{x}\tag 1$$

with $~x=\frac {l}{\tan(\theta)}~$ you obtain

$$F=\frac{M\,g\,L}{2\,l}\sin(\theta)$$

but if $~\theta\mapsto 0~$ then $~x\mapsto\infty~$ and from equation (1) $~F\mapsto 0~$ as it schuld be .

with

$$ d=\sqrt{x^2+l^2}$$

thus if $~x\mapsto\infty~\Rightarrow~d\mapsto\infty$ and the force $~F$ for $~\theta=0~$ will be zero