Answer to Question #164680 in Mechanics | Relativity for Physics

Question

A uniform 80.0 N ladder 4.0 m long is placed against a frictionless wall with its base situated 2.0 from the wall. Find the forces exerted by the wall and by the ground on the ladder.

Answer 1

The equations of static equilibrium

become:

$F-N_2=0$ (horizontal)

$N_1-mg=0(vertical)$ and

$N_2Lsin\theta-\frac{1} {2} mgLcos\theta=0$ (torque about the bottom of the ladder)

When $mg=80N, cos \theta=\frac{2.0}{4.0}=0.5$

$N_1$ (the force exerted by the ground) =80N

$N_2$ (the force exerted by the wall) =$\frac{1} {2} mgcos\theta=\frac{1} {2} ×80×0.5$

$N_2=20N$

F(The friction force exerted by the ground) =$N_2=20N.$