Answer to Question #306968 in Mechanics | Relativity for Aria

Question

A uniform 10 kg ladder 2.5 m long is placed against a frictionless wall with its base on the ground 80 cm from the wall.

Find the magnitude of the force exerted on the wall and on the ground.

Answer 1

Explanations & Calculations

• Since the wall is frictionless, the only force acting on the ladder by the wall is the normal reaction force say $\small R$.
• And the floor should have friction to keep the ladder sliding on the floor so there are 2 forces acting on the ladder from the floor: the normal force and the friction force say $\small S$ and $\small f$ respectively.
• With the horizontal equilibrium, it is obvious that $\small f = R$ and from the vertical equilibrium $\small S= mg$ therefore, only $\small R$ or $\small f$ needs to be found.
• Take moment around the point where the ladder touches the floor.

$\qquad\qquad \begin{aligned} \small R\times\sqrt{(2.5m)^2-(0.8m)^2}-10kg\times9.8\,ms^{-2}\times0.4\,m&=\small 0\\ \small R&=\small 16.6\,N \end{aligned}$

• Then, the force at the wall is $\small R=16.6\,N$
• Force on the ground point is

$\qquad\qquad \begin{aligned} \small F&=\small \sqrt{S^2+f^2}=\sqrt{(10\times9.8)^2+16.6^2}\\ &=\small \end{aligned}$