In Fig. 6-33, two blocks are connected over a pulley. The mass of block A is$\mathbf{10}\mathbf{}\mathbf{kg}$ , and the coefficient of kinetic friction between A and the incline is $\mathbf{0}\mathbf{.}\mathbf{20}$ . Angle $\mathbf{\theta }$ of the incline is $\mathbf{30}\mathbf{°}$ . Block A slides down the incline at constant speed. What is the mass of block B? Assume the connecting rope has negligible mass. (The pulley’s function is only to redirect the rope)

Mass of block is A, ${\mathrm{m}}_{\mathrm{A}}=10\mathrm{kg}$

Coefficient of kinetic friction between the incline and Body A,${\mathrm{\mu }}_{\mathrm{k}}=0.20$

Angle of incline, $\mathrm{\theta }=30°$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. To find the acceleration of the body A, use Newton’s 2nd law of motion.

Formula:

${\mathrm{F}}_{\mathrm{net}}=\sum _{}\mathrm{ma}$

where, F is the net force, m is mass and a is an acceleration.

Free Body Diagram of the body A and body B:

Block A slides down the incline at constant speed.Thus, the acceleration of block A is 0.By using Newton’s 2nd law of motion along the vertical direction (along y) to the system,(positive x axis along the incline and the positive y along the vertical direction),

$$\begin{gathered} \mathrm{N}={\mathrm{m}}_{\mathrm{A}}\mathrm{gcos}\left(30\right)=0 \\ \mathrm{N}={\mathrm{m}}_{\mathrm{A}}\mathrm{gcos}\left(30\right) \end{gathered}$$

Thus, the frictional force,

$$\begin{gathered} {\mathrm{f}}_{\mathrm{k}}={\mathrm{\mu }}_{\mathrm{k}}\mathrm{N} \\ ={\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_{\mathrm{A}}\mathrm{gcos}\left(30\right) \end{gathered}$$

Now, applying Newton’s 2nd along horizontal direction to the Body A having massm,

$$\begin{gathered} \mathrm{T}+{\mathrm{f}}_{\mathrm{k}}‐{\mathrm{m}}_{\mathrm{A}}\mathrm{g}\sin \left(30\right)=0 \\ \mathrm{T}+{\mathrm{\mu }}_{\mathrm{k}}‐{\mathrm{m}}_{\mathrm{A}}\mathrm{g}\sin \left(30\right)‐{\mathrm{m}}_{\mathrm{A}}\mathrm{g}\sin \left(30\right)=0 \end{gathered}$$ (1)Similarly, apply to the Body B having mass m’,

${\mathrm{m}}_{\mathrm{B}}\mathrm{g}-\mathrm{T}=0$ (ii)

Adding equation (i) and (ii),

$$\begin{gathered} {\mathrm{m}}_{\mathrm{A}}\mathrm{g}+{\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_{\mathrm{A}}\mathrm{g}\sin \left(30\right)‐{\mathrm{m}}_{\mathrm{A}}\mathrm{g}\sin \left(30\right)=0 \\ {\mathrm{m}}_{\mathrm{B}}={\mathrm{m}}_{\mathrm{A}}\left(\sin \left(30\right)‐{\mathrm{\mu }}_{\mathrm{k}}\cos \left(30\right)\right) \\ =3.3\mathrm{kg} \end{gathered}$$

Therefore, the mass of block B is 3.3 kg.