Question

A ladder of length l and mass m is placed against a smooth vertical wall, but the ground is not smooth. Coefficient of friction between the ground and the ladder is $\mu$. The angle $\theta$ at which the ladder will stay in equilibrium is:

• A. $\theta =ta{n}^{-1}\left(\mu \right)$
• B. $\theta =ta{n}^{-1}\left(2\mu \right)$
• C. $\theta =ta{n}^{-1}\left(\frac{\mu }{2}\right)$
• D. none of these

Answer 1

Answer: (D)

none of these

$mg=N_1$
$\mu N_1=N_2$ or $N_2=\mu mg$
Taking moment about O, we get
$\mu mglsin\theta -mg\frac{l}{2}cos\theta =0$
$tan\theta =\frac{1}{2\mu }$ or $\theta =ta{n}^{-1}\left(\frac{1}{2\mu }\right)$