Masses $M_1$, $M_2$ and $M_3$ are connected by strings of negligible mass which pass over massless and friction less pulleys $P_1$ and $P_2$ as shown in fig The masses move such that the portion of the string between $P_1$ and $P_2$ in parallel to the inclined plane and the portion of the string between $P_2$ and $M_3$ is horizontal. The masses $M_2$ and $M_3$ are $4.0\,\text{kg}$ each and the coefficient of kinetic friction between the masses and the surfaces is $0.25$. The inclined plane makes an angle of $37^\circ$ with the horizontal. If the mass $M_1$ moves downwards with a uniform velocity, find (i) the mass of $M_1$ (ii) The tension in the horizontal portion of the string $(g=9.8\,\text{m/sec}^{2}, \sin 37^\circ \simeq 3/5)$

Step 1: Identify the given parameters and equilibrium condition.

Since mass $M_1$ moves downwards with a uniform velocity, the acceleration of the entire system is zero.

$$M_2 = M_3 = 4.0\text{ kg}$$ $$\mu_k = 0.25$$ $$g = 9.8\text{ m/s}^2$$ $$\sin 37^\circ = \frac{3}{5}$$ $$\cos 37^\circ = \frac{4}{5}$$

Step 2: Analyze the forces acting on mass $M_3$.

Mass $M_3$ is on the horizontal surface and moves towards the pulley $P_2$ with zero acceleration.

$$N_3 = M_3 g$$ $$T_2 = f_3 = \mu_k N_3$$ $$T_2 = \mu_k M_3 g$$

Step 3: Calculate the tension $T_2$ in the horizontal portion of the string.

Substitute the given values into the equation for $T_2$.

$$T_2 = 0.25 \times 4.0 \times 9.8$$ $$T_2 = 9.8\text{ N}$$

Step 4: Analyze the forces acting on mass $M_2$.

Mass $M_2$ moves up the inclined plane with zero acceleration.

$$N_2 = M_2 g \cos 37^\circ$$ $$f_2 = \mu_k N_2 = \mu_k M_2 g \cos 37^\circ$$ $$T_1 - T_2 - M_2 g \sin 37^\circ - f_2 = 0$$

Step 5: Express the tension $T_1$ in terms of the other forces.

Rearrange the equilibrium equation for $M_2$.

$$T_1 = T_2 + M_2 g \sin 37^\circ + \mu_k M_2 g \cos 37^\circ$$

Step 6: Calculate the components of the forces on $M_2$.

Compute the gravitational and frictional force components along the incline.

$$M_2 g \sin 37^\circ = 4.0 \times 9.8 \times \frac{3}{5} = 23.52\text{ N}$$ $$\mu_k M_2 g \cos 37^\circ = 0.25 \times 4.0 \times 9.8 \times \frac{4}{5} = 7.84\text{ N}$$

Step 7: Calculate the tension $T_1$.

Substitute the calculated values into the expression for $T_1$.

$$T_1 = 9.8 + 23.52 + 7.84$$ $$T_1 = 41.16\text{ N}$$

Step 8: Analyze the forces on mass $M_1$ and find its mass.

Since $M_1$ moves downwards with uniform velocity, the upward tension balances its weight.

$$T_1 = M_1 g$$ $$M_1 = \frac{T_1}{g}$$ $$M_1 = \frac{41.16}{9.8}$$ $$M_1 = 4.2\text{ kg}$$

Answer: (i) Mass of $M_1 = 4.2\text{ kg}$, (ii) Tension in the horizontal portion of the string $T_2 = 9.8\text{ N}$