A uniform ladder of length `l` rests against a smooth, vertical wall (figure). If the mass of the ladder is `m` and the the coefficient of static friction between the ladder and the ground is `mu_(s) = 0.375` and the minimum angle `theta_(min)` at which the ladder does not slip.

A uniform ladder at rest, leading against a smooth wall. Th ground is rough.

Explanation

Applying the first condition for equilinrium to the ladder, we have
(1) `Sigma F_(x) = f_(s) - P = 0`
(2) `Sigma F_(y) = n - mg = 0`
The first equation tells us that `P = f_(0)`. From the second equation we see that n` = mg`. Furthermore, when the ladder is on the verge of slipping, the force of friction must be a mximum, which is given by `f_(c, max) = mu_(s) n = mu_(s) mg`.
The find `theta_(min)`, we must use the second condition for equilibrium. When we take the torque about an axis through the origin `O` at the bottom of the ladder, we have
`Sigma tau_(0) = P l sin theta - (mg) (l)/(2) cos theta = 0`
This expression gives
`tan theta_(min) = (mg)/(2 P) = (mg)/(2 mu_(s) mg) = (1)/(2 mu_(s)) = 1.25`
`theta_(min) = 53^@`.
Any body experiencing zero net force & zero torque about `C.M` is raid to be in equilibrium.

`F (1)/(2) + f(1)/(2) - Nx = 0`
`N = mg , F = f`
`x = (Fb)/(N) = (Fb)/(mg)`
but `x lt (a)/(2)`
for not toppling `(a)/(2) gt (Fb)/(mg)` or `(a)/(2 b) gt (F)/(mg)`
assuming `F lt mu mg`.