Question

 a ladder is set to rest against a very slippery wall at an angle of 55 °. How large must the coefficient of friction between the ladder and the floor be in order for the ladder to remain upright?

Answer 1

Let’s consider the forces acting on the ladder: the weight of the ladder $mg$ directed downward, the normal force $F_W$ directed from the wall, the normal force $F_N$ directed from the floor and the force of friction $F_{fr}$ directed toward the wall. In order to ladder remains upright and not slipping, the sum of these forces must be equal to zero. Applying Newton’s laws we get:

$$F_{fr}=F_W,$$ $$F_N=mg.$$

The sum of moments of forces around the pivot point at the bottom of the ladder equals:

$$\sum \tau=0,$$ $$\tau_{ladder}-\tau_{wall}=0,$$ $$F_WLsin\theta-\dfrac{L}{2}mgcos\theta=0,$$ $$\mu_smgsin\theta-\dfrac{1}{2}mgcos\theta=0,$$ $$\mu_s=\dfrac{1}{2}cot\theta=\dfrac{1}{2}cot55^{\circ}=0.35$$

Answer:

$\mu_s=0.35$