A uniform girder AB of length $10m$ and weight $20000N$ rests with its end $A$ on the rough horizontal floor of a factory and with its other end $B$ supported by a cable. The girder makes an angle $α$ with the floor, and the cable is inclined at an angle $β$ to the horizontal. The girder is in limiting equilibrium, and the coefficient of friction between the girder and the floor is $μ$.
We have $tanβ$ = $\frac{20000 - R}{μR}$
where $R$ is the magnitude of the normal force acting on the girder at $A$.
Given $tanα = \frac{3}{4}$ and $tanβ = 5$
Show that:
a) $μ = \frac{2}{7}$
b) find the tension in the cable. (should be $12000N$)
Now:
Resolving horizontally:
$F = Tcosβ$ , where $F$ is the frictional force at $A$ and $T$ is the tension on the cable. Also $F = μR$ (limiting equilibrium).
Resolving vertically:
$R + Tsinβ = W$
What is the relationship of $tanα = \frac{3}{4}$ and $tanβ = 5$ to $μ$?
HINT...you can take moments at, for example, the point of contact of the girder and the floor and obtain $$20000\times5\cos \alpha=T\times10\sin(\beta-\alpha)$$
I hope this helps.