A ladder of length 2L and mass m is placed leaning against a wall, making an angle t with the floor. The coefficient of friction between all surfaces is c. At what angle t does the ladder begin to slip?

Firstly draw a free body diagram of the ladder, showing its weight and the contact forces at either end. We'll call end A the top end and end B the bottom end.
The next thing to do is to eliminate all the unknown forces, finding them in terms of m, g, and c.
As the ladder will be tending to slip down the wall, the friction at A will be acting upwards. As the ladder will be tending to slip away from the wall, the friction at B will be acting inwards.
For the ladder to be on the point of slipping, F = cR (1), where F is the friction force, c the coefficient of friction, and R the reaction force at that point.
Resolving forces horizontally:
R_A = F_B
so R_A = cR_B
and using (1): F_A = c²R_B (2)
Resolving forces vertically:
F_A + R_B = mg
Using (2): c²R_B + R_B = mg
hence R_B = mg/(1 + c²) (3)
Now we look at the moment equilibrium to find the angle. We take the equilibrium at A to eliminate the forces at A because earlier we found all the forces in terms of R_B - this way we have less substituting to do.
Moment equilibrium about A:
mgLcos(t) + cR_B2Lsin(t) = 2Lcos(t)R_B
Substituting (3), cancelling and rearranging gives
t = tan^-1[(1-c²)/2c]