Question

What is the minimum coeffecient of static friction μmin required between the ladder and the ground so that the ladder does not slip?

Answer 1

Answer:

$\mu = \frac{1}{2tan\theta}$

Explanation:

let the ladder is of mass "m" and standing at an angle with the ground

So here by horizontal force balance we will have

$\mu N_1 = N_2$

by vertical force balance we have

$N_1 = mg$

now by torque balance about contact point on ground we will have

$mg(\frac{L}{2}cos\theta) = N_2(L sin\theta)$

so we will have

$N_2 = \frac{mg}{2tan\theta}$

now from first equation we have

$\mu (mg) = \frac{mg}{2tan\theta}$

$\mu = \frac{1}{2tan\theta}$