In Fig. 12-51, a uniform plank, with a lengthLof $\mathbf{6}\mathbf{.10}\mathit{m}$ and a weight of$\mathbf{445}\mathit{N}$ , rests on the ground and against a frictionless roller at the top of a wall of height $\mathit{h}\mathbf{=}\mathbf{3}\mathbf{.05}\mathit{m}$. The plank remains in equilibrium for any value of $\mathit{\theta }\mathbf{\ge }\mathbf{70}\mathbf{°}$ but slips if $\mathit{\theta }\mathbf{<}\mathbf{70}\mathbf{°}$. Find the coefficient of static friction between the plank and the ground.

The weight of plank is,$W=445N$.

The length of the plank is,$L=6.10m$.

The plank is at height,$h=3.05m$.

The angle of inclination is,$\theta ={70}^{\circ }$.

To find the coefficient acceleration, use the condition of equilibrium.Using this condition, we can write the equation for the net force and net torque. Solving these equations, we can find the coefficient of friction.

Here$F_f$ is the frictional force, $F$ is the external/applied force, and $F_N$ is the normal force.

The net force in the horizontal direction can be written as,

$\begin{array}{c}F\sin \theta -F_f=0\\ F_f=F\sin \theta \end{array}$ (1)

The net force in the verticle direction can be written as,

$\begin{array}{c}F\cos \theta -W+F_N=0\\ F\cos \theta =W-F_N\\ F=\frac{W-F_N}{\cos \theta }\end{array}$

(2)

Substitute the value from equation 2 into equation 1, and we get,

$F_f=\tan \theta (W-F_N)$ (3)

The equation for the net torque can be written as,

$F_{N}d-F_{f}h-W(d-\frac{L}{2}\cos \theta )=0$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{F}_{N}d-\tan \theta (W-F_N)h-W(d-\frac{L}{2}\cos \theta )=0\\ F_N=\frac{d-\left(\frac{L}{2}\right)\cos \theta +h\tan \theta }{d+h\tan \theta }\\ F_N=W(1-\frac{L}{2h}{\cos }^2\theta \sin \theta )\end{array}$ (4)

Substitute the values in equation 3, and we get,

$F_f=\frac{WL{\sin }^2\theta \cos \theta }{2h}$(5)

Coefficient of static friction between plank and ground can be written as,

$\mu =F_f/F_N$

From equations 4 and 5, substitute the values in the above expression, and we get,

$\mu =\frac{L{\sin }^2\theta \cos \theta }{(2h-L\sin \theta {\cos }^2\theta )}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}\mu =\frac{6.10\times {\sin }^2{70}^{\circ }\cos {70}^{\circ }}{(2\times 3.05-6.10\sin {70}^{\circ }{\cos }^2{70}^{\circ })}\\ =0.34\end{array}$

Thus, the coefficient of static friction between plank and ground is,.$\mu =0.34$