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

`\mu = (1)/(2tan\theta)`

Step-by-step 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((L)/(2)cos\theta) = N_2(L sin\theta)`

so we will have

`N_2 = (mg)/(2tan\theta)`

now from first equation we have

`\mu (mg) = (mg)/(2tan\theta)`

`\mu = (1)/(2tan\theta)`