A block A, of weight W, slides down an inclined plane S of slope ${37}^{\circ }$ at a constant velocity, while the plank B, also of weight W, rests on top of A. The plank B is attached by a cord to the top of the plane. The coefficient of kinetic friction $\mathrm{\mu }$ is the same between the surfaces A and B and between S and A. Determine the value of $1/\mathrm{\mu }$.

Since A tends to slip down, frictional forces act on it from both sides up the plane.

Let N be the reaction of the plank on A and N' be the mutual normal action-reaction between A and B. 

From the free-body diagram of A,

${\mathrm{N}}^{\mathrm{\prime }}+\mathrm{mgcos}\mathrm{\alpha }=\mathrm{N}\text{ and }\mathrm{mgsin}\mathrm{\alpha }=\mathrm{\mu }\left(\mathrm{N}+{\mathrm{N}}^{\mathrm{\prime }}\right)$

From the free-body diagram of B

${\mathrm{N}}^{\mathrm{\prime }}=\mathrm{mgcos}\mathrm{\alpha }$

$\mathrm{mgsin}\mathrm{\alpha }+{\mathrm{\mu N}}^{\mathrm{\prime }}=\mathrm{T}$

$\therefore 2\mathrm{mgcos}\mathrm{\alpha }=\mathrm{N}$

and $\mathrm{mgsin}\mathrm{\alpha }=\mathrm{\mu }(2\mathrm{mgcos}\mathrm{\alpha }+\mathrm{mgcos}\mathrm{\alpha })$

or $\mathrm{\mu }=\frac{1}{3}\tan \mathrm{\alpha }=\frac{1}{3}\times \frac{3}{4}=0.25\text{ or }\frac{1}{\mathrm{\mu }}=4$