91Ó°ÊÓ

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,

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

Here ${\mathit{μ}}_{\mathbf{k}}$ 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 $3w\sin \mathrm{θ}$ and $3w\cos \mathrm{θ}$. 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 ${\mathrm{μ}}_k$.

The frictional force between S and A is given by,

$$\begin{gathered} f_A={\mathrm{μ}}_k\left(F_n\right) \\ ={\mathrm{μ}}_k\left(4w\cos \mathrm{θ}\right) \end{gathered}$$

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

$F_{n_1}=w\cos \mathrm{θ}$

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

$F_{n_2}=4w\cos \mathrm{θ}$

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

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

The frictional force between A and B is given by,

$f_B={\mathrm{μ}}_k\left(w\cos \mathrm{θ}\right)$

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

$\underset{}{∑}{F}_x=ma$

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

Substitute $3w\sin \mathrm{θ}-5w{\mathrm{μ}}_k\cos \mathrm{θ}$ for $\underset{}{∑}{F}_x$, and $36.9°$for $\mathrm{θ}$in the above equation.

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

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