A uniform beam of mass `m` is inclined at an angle `theta` to the horizontal. Its upper end produces a ninety degree bend in a very rough rope tied to a wall, and its lower end rests on a rough floor (a) If the coefficient of static friction between beam and floor is `mu_(s)` determine an expression for the maximum mass `M` that can be suspended from the top before the beam slips. (b) Determine the magnitude of the reaction force at the floor and the magnitude of the force exerted by the beam on the rope at `P` in terms of `m, M` and `mu_(s)`
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Explanation

We can use `sumF_(x)=sumF_(y)=0` and `sumtau=0` with pivot point at the contact on the floor.

Then `sumF_(x)=T-mu_(s)n=0`
`sumF_(y)=n-Mg-mg=0`
`sumtau=Mg(Lcostheta)+mg(L/2costheta)-T(Lsintheta)=0`
Solving the above equations gives
`M=m/2((2mu_(s)sintheta-costheta)/(costheta-mu_(s)sintheta))`
This answer is the maximum value of `M` if `mu_(s)ltcottheta`. If `mu_(s)ge cot theta`, the mass `M` can increase without limit. it has no maximum value, and part b. cannot be answered an stated either. In the case `mu_(s)ltcottheta,` we proceed.
b. At the floor we have the normal force in the `y`-direction and frictional force in the `x`-direction. The reaction force then is
`R=sqrt(n^(2)+(mu_(s)n)^(2))=(M+m)g sqrt(1+mu_(s)^(2))`
At point `P`, the force of the beam on the rope is
`F=sqrt(T^(2)+(Mg)^(2))=gsqrt(M^(2)+mu_(s)^(2)(M+m)^(2))`