Answer to Question #107263 in Mechanics | Relativity for Aaron

Question

A uniform ladder 10m long and weighing 59N rets against a smooth vertical wall. If the ladder is just on the verge of slipping when it makes a 50 degree angle with the ground, find the coefficient of static friction between the ladder and the ground.

Answer 1

As per the given question,

Length of the uniform ladder(L)=10m

Weight of the ladder =59N

Angle between the ladder and the ground =$50^\circ$

Let the reaction on the bottom of the ladder $N_1$ and reaction on the wall $N_2$

Now, taking the torque about the center of the ladder

$N_1\times\dfrac{L}{2} \sin 50-\mu N_1\dfrac{L}{2}\cos 50=N_2\times \dfrac{L}{2}\cos 50$

$N_2=N_1\tan 50-\mu N_1--------(i)$

Now, balancing horizontal force

$N_2=\mu N_1-------(ii)$

From equation (i) and (ii)

$2\mu N_1=N_1\tan 50$

$\mu=\dfrac{\tan 50}{2}=0.59$

$\mu =0.59$