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