A block A, of weight W, slides down an inclined plane S of slope 37∘ 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 μ is the same between the surfaces A and B and between S and A. Determine the value of 1/μ.

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,N′+mgcos⁡α=N  and  mgsin⁡α=μN+N′From the free-body diagram of BN′=mgcos⁡αmgsin⁡α+μN′=T∴ 2mgcos⁡α=Nand mgsin⁡α=μ(2mgcos⁡α+mgcos⁡α)or μ=13tan⁡α=13×34=0.25 or 1μ=4