In the diagram shown, the blocks $A$, $B$ and $C$ weights, $3\,\text{kg}$, $4\,\text{kg}$ and $5\,\text{kg}$ respectively. The coefficient of sliding friction between any two surface is $0.25$. $A$ is held at rest by a massless rigid rod fixed to the wall while $B$ and $C$ are connected by a light flexible cord passing around a frictionless pulley. Find the force $F$ necessary to drag $C$ along the horizontal surface to the left at constant speed. Assume that the arrangement shown in the diagram, $B$ on $C$ and $A$ on $B$, is maintained all through. $(g = 9.8\,\text{m/s}^{2})$

Step 1: Identify the given parameters.

The masses of the blocks and the coefficient of friction are given as follows:

$$m_A = 3\,\text{kg},\quad m_B = 4\,\text{kg},\quad m_C = 5\,\text{kg}$$ $$\mu = 0.25$$ $$g = 9.8\,\text{m/s}^{2}$$

Step 2: Analyze the motion of each block.

Since block $A$ is fixed to the wall, it remains at rest, while the cord forces block $B$ to move to the right at speed $v$ when block $C$ moves to the left at speed $v$.

$$v_A = 0$$ $$v_B = v \quad \text{(to the right)}$$ $$v_C = v \quad \text{(to the left)}$$

Step 3: Determine the normal forces at each interface.

Calculate the normal forces acting between the contacting surfaces.

$$N_{AB} = m_A g$$ $$N_{BC} = (m_A + m_B)g$$ $$N_{C} = (m_A + m_B + m_C)g$$

Step 4: Calculate the friction forces.

Determine the sliding friction forces acting at each interface.

$$f_1 = \mu m_A g$$ $$f_2 = \mu (m_A + m_B)g$$ $$f_3 = \mu (m_A + m_B + m_C)g$$

Substituting the given values:

$$f_1 = 0.25 \times 3 \times 9.8 = 7.35\,\text{N}$$ $$f_2 = 0.25 \times (3 + 4) \times 9.8 = 17.15\,\text{N}$$ $$f_3 = 0.25 \times (3 + 4 + 5) \times 9.8 = 29.4\,\text{N}$$

Step 5: Apply equilibrium conditions for block $B$.

Since block $B$ moves at a constant speed, the net horizontal force acting on it is zero.

$$T - f_1 - f_2 = 0$$ $$T = f_1 + f_2$$ $$T = 7.35 + 17.15 = 24.5\,\text{N}$$

Step 6: Apply equilibrium conditions for block $C$.

Since block $C$ moves to the left at a constant speed, the net horizontal force acting on it is zero.

$$F - T - f_2 - f_3 = 0$$ $$F = T + f_2 + f_3$$ $$F = 24.5 + 17.15 + 29.4 = 71.05\,\text{N}$$

Alternatively, the force can be expressed in terms of the masses:

$$F = \mu g (4m_A + 3m_B + m_C)$$

Answer:

The force $F$ necessary to drag $C$ to the left at constant speed is $71.05\,\text{N}$.