In the figure masses $m_1$, $m_2$ and $M$ are $20\,\text{kg}$, $5\,\text{kg}$ and $50\,\text{kg}$ respectively. The coefficient of friction between $M$ and ground is zero. The coefficient of friction between $m_1$ and $M$ and that between $m_2$ and ground is $0.3$. The pulleys and the strings are massless. The string is perfectly horizontal between $P_1$ and $m_1$ and also between $P_2$ and $m_2$. The string is perfectly vertical between $P_1$ and $P_2$. An external horizontal force $F$ is applied to the mass $M$. Take $g = 10\,\text{m/s}^{2}$. (a) Draw a free body diagram for mass $M$, clearly showing all the forces. (b) Let the magnitude of the force of friction between $m_1$ and $M$ be $f_1$ and that between $m_2$ and ground be $f_2$. For a particular $F$ it is found that $f_1 = 2f_2$. Find $f_1$ and $f_2$. Write equations of motion of all the masses. Find $F$, tension in the string and acceleration of the masses.

Step 1: Identify the given parameters and maximum friction forces.

The given parameters are $m_1 = 20\text{ kg}$, $m_2 = 5\text{ kg}$, $M = 50\text{ kg}$, $\mu = 0.3$, and $g = 10\text{ m/s}^2$.

$$f_{1,\max} = \mu m_1 g$$ $$f_{1,\max} = 0.3 \times 20 \times 10 = 60\text{ N}$$ $$f_{2,\max} = \mu m_2 g$$ $$f_{2,\max} = 0.3 \times 5 \times 10 = 15\text{ N}$$

Step 2: Analyze the forces on mass $M$.

The horizontal forces acting on $M$ are the external force $F$ to the right, the friction force $f_1$ from $m_1$ to the left, and the tension forces from the pulleys $P_1$ and $P_2$ which cancel each other out.

$$(T \to, T \downarrow) + (T \leftarrow, T \uparrow) = 0$$

The vertical forces acting on $M$ are gravity $Mg$ downwards, the normal force $N_1 = m_1 g$ from $m_1$ downwards, and the normal force $N_g$ from the ground upwards.

Step 3: Determine the kinematic constraint.

The total length of the string $\ell$ is expressed in terms of the positions of the masses and the pulleys.

$$\ell = (x_1 - x_{P_1}) + (x_{P_2} - x_2) + \text{constant}$$

Since the pulleys are fixed to $M$, their positions are $x_{P_1} = X_M + c_1$ and $x_{P_2} = X_M + c_2$.

$$\ell = x_1 - x_2 + \text{constant}$$

Differentiating twice with respect to time yields equal accelerations for the masses.

$$a_1 = a_2 = a_M = a$$

Step 4: Write the equations of motion for all masses.

Write the horizontal force equations for $m_1$, $m_2$, and $M$.

$$f_1 - T = m_1 a$$ $$T - f_2 = m_2 a$$ $$F - f_1 = M a$$

Step 5: Solve for the friction forces and acceleration.

Add the equations of motion for $m_1$ and $m_2$.

$$f_1 - f_2 = (m_1 + m_2) a$$

Substitute the given condition $f_1 = 2f_2$.

$$2f_2 - f_2 = (20 + 5) a$$ $$f_2 = 25a$$ $$f_1 = 50a$$

Since $f_2$ reaches its maximum limit of $15\text{ N}$ first, we set $f_2 = 15\text{ N}$.

$$a = \frac{15}{25} = 0.6\text{ m/s}^2$$ $$f_1 = 2(15) = 30\text{ N}$$

Step 6: Calculate the tension $T$ and the external force $F$.

Substitute the values into the equations of motion.

$$T = f_2 + m_2 a$$ $$T = 15 + 5(0.6) = 18\text{ N}$$ $$F = f_1 + M a$$ $$F = 30 + 50(0.6) = 60\text{ N}$$

Answer: (a) The horizontal forces on $M$ are $F$ (right) and $f_1$ (left); the vertical forces are $Mg$ (down), $N_1 = m_1 g$ (down), and $N_g$ (up); the tension forces from the pulleys cancel out. (b) $f_1 = 30\text{ N}$, $f_2 = 15\text{ N}$, $T = 18\text{ N}$, $F = 60\text{ N}$, and $a = 0.6\text{ m/s}^2$.