# Analyzing limiting case in a statics problem

homework-and-exercisesnewtonian-mechanics

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?

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