Block A, with weight 3w, slides down an inclined plane S of slope angle 36.9° at a constant speed while plank B, with weight w, rests on top of A. The plank is attached by a cord to the wall (Fig. P5.97).

(a) Draw a diagram of all the forces acting on block A.

(b) If the coefficient of kinetic friction is the same between Aand B and between S and A, determine its value.

According to Newton’s second law, the linear force is given by,

$\mathit{F}\mathbf{=}\mathit{m}\mathit{a}$

Here, F is linear force, m is the mass of the object, and a is the acceleration of the object.

The kinetic friction force is given by,

${\mathit{f}}_{\mathbf{k}}\mathbf{=}{\mathit{\mu }}_{\mathbf{k}}\mathit{N}$

Here ${\mathit{\mu }}_{\mathbf{k}}$ is the coefficient of kinetic friction and is the normal force.

Given data:

The free-body diagram of the block A is shown below.

From the above figure, it is obtained that the normal force $F_n$acting along the upward direction, the force due to gravity 3w acting along the downward direction. The components of force due to gravity are $3w\sin \theta$ and $3w\cos \theta$. The force $f_A$ is the force on top of A .

Let the coefficient of kinetic friction between A and B , and between S and A be ${\mu }_k$.

The frictional force between S and A is given by,

$$\begin{gathered} f_A={\mu }_k\left(F_n\right) \\ ={\mu }_k\left(4w\cos \theta \right) \end{gathered}$$

The normal force between the block and the plank is given by,

$F_{n_1}=w\cos \theta$

The normal force between the block and the inclined plane is given by,

$F_{n_2}=4w\cos \theta$

The net normal force between the block and the plank is given by,

$$\begin{gathered} F_n=F_{n_1}+F_{n_2} \\ =w\cos \theta +4w\cos \theta \\ =5w\cos \theta \end{gathered}$$

The frictional force between A and B is given by,

$f_B={\mu }_k\left(w\cos \theta \right)$

Since A is sliding down at constant speed, so calculate the net force on the block A by using Newton’s second law.

$\sum _{}{F}_x=ma$

Here $m_A$is the mass and a is the acceleration.

Substitute $3w\sin \theta -5w{\mu }_k\cos \theta$ for $\sum _{}{F}_x$, and $36.9°$for $\theta$in the above equation.

$$\begin{gathered} 3w\sin \theta -5w{\mu }_k\cos \theta =0 \\ {\mu }_k=\frac{3}{5}\tan \theta \\ =\frac{3}{5}\tan \left(36.9°\right) \\ {\mu }_k=0.451 \end{gathered}$$

Therefore, the value of the coefficient of kinetic friction is 0.451.