Rotational motion question

A uniform ladder of length 3.25m and weighing 250 N is placed against a smooth vertical wall with its lower end 1.25m from the wall.If coefficient of friction b/w ladder and floor is 0.3, then, find the frictional force acting on the ladder at the point of contact between ladder and floor.

I ll solve it using vars - u can substitute the values.

Let the ladder of length l make an angle t with horizontal (mass = m).
Find t using trigonometry.

2 Normal rxns exist -
N1 (at base contact pt) - vertical
N2 (at wall contact pt) - horizontal

Other forces :-
mg at the center of mass (middle of ladder) - downwards
f (static friction) - horizontal, at base pt - in a direction to balance force - N2

Now, net Force is 0 =>
N1 = mg
N2 = f

Due to rotational equilibrium, conserving net torque about the center of mass we get,
f(l/2)sin(t) - N(l/2)cos(t) + f(l/2)tan(t) = 0

Using above eqns,
=> 2*f*tan(t) = mg

Thus, f = mg / 2 tan(t).

Note, we dont require co-efficient of friction as its static friction and a self adjusting force - which can be obtained only indirectly.