Answer:
[tex]\mu = \frac{1}{2tan\theta}[/tex]
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
[tex]\mu N_1 = N_2[/tex]
by vertical force balance we have
[tex]N_1 = mg[/tex]
now by torque balance about contact point on ground we will have
[tex]mg(\frac{L}{2}cos\theta) = N_2(L sin\theta)[/tex]
so we will have
[tex]N_2 = \frac{mg}{2tan\theta}[/tex]
now from first equation we have
[tex]\mu (mg) = \frac{mg}{2tan\theta}[/tex]
[tex]\mu = \frac{1}{2tan\theta}[/tex]
