91Ó°ÊÓ

The length of the beam is.$2.50\text{ N}$

Weight of beam is$500\text{ N}$.

Given the distance from the floor is $1.50\text{″¾}$

Using the condition for the static equilibrium, you can write the equation for torque. Solving this, you would get force. Using this value of force, you can find the coefficient of friction. The equations are given below

$\begin{array}{l}∑F_x=0\\ ∑F_y=0\\ ∑\mathrm{τ}=0\end{array}$

For the angle between floor and beam

As the beam is diagonal of a triangle and given height is its one side.

$\begin{array}{c}\mathrm{θ}=si{n}^{−1}\left(\frac{1.5}{2.5}\right)\\ \mathrm{θ}=37°\end{array}$

As the system is in equilibrium, we can write the moment of force as

$\begin{array}{l}F×L−\frac{L}{2}×Mgcos\mathrm{θ}=0\\ F=\frac{Mgcos\mathrm{θ}}{2}\\ F=\frac{500\text{ }N×\cos 37°}{2}\\ F=199.6\text{ N}\end{array}$

If we see the diagram, we can write,

$\begin{array}{c}{F}_N=Mg−Fcos\mathrm{θ}\\ =500\text{ }N−\left(200\text{ }N\right)\left(0.87\right)\\ =340\text{ N}\end{array}$

$\begin{array}{c}{F}_{fric}=\mathrm{μ}{F}_N\\ =Fsin37°\\ =120\text{ N}\end{array}$

So,

$\begin{array}{l}{F}_{fric}=\mathrm{μ}{F}_N\\ \mathrm{μ}=\frac{F_{fric}}{F_N}\\ \mathrm{μ}=\frac{120\text{ }N}{340\text{ }N}\\ \mathrm{μ}=0.35\end{array}$