The problem asks for the magnitude of the horizontal force $F$ required to ensure there is no relative motion between the objects $m_1$ and $m_2$. The condition of "no relative motion" implies that the entire system (cart $M$, mass $m_1$, and mass $m_2$) moves together with a common horizontal acceleration, $a$. Furthermore, it implies that mass $m_2$ has zero vertical acceleration.

We will apply Newton's second law to the components of the system in an inertial frame of reference (the ground).

1. Analyze mass $m_2$: The forces acting on $m_2$ are the gravitational force $m_2g$ downwards and the tension $T$ from the string upwards. Since there is no relative vertical motion, its vertical acceleration is zero. Applying Newton's second law in the vertical direction for $m_2$:

   $$T - m_2g = 0$$ $$T = m_2g$$

2. Analyze mass $m_1$: The only horizontal force acting on $m_1$ is the tension $T$ from the string (friction is ignored). Since $m_1$ moves with the cart, its horizontal acceleration is $a$. Applying Newton's second law in the horizontal direction for $m_1$:

   $$T = m_1 a$$

3. Determine the common acceleration $a$: By equating the two expressions for tension $T$ from steps 1 and 2, we can find the required acceleration of the system.

   $$m_1 a = m_2g$$ $$a = \frac{m_2}{m_1}g$$

4. Determine the applied force $F$: Now, consider the entire system, composed of masses $M$, $m_1$, and $m_2$, as a single object of total mass $M_{total} = M + m_1 + m_2$. The only external horizontal force acting on this combined system is $F$. Applying Newton's second law to the entire system:

   $$F = (M + m_1 + m_2)a$$

   Substituting the expression for $a$ from step 3:

   $$F = (M + m_1 + m_2)\left(\frac{m_2}{m_1}g\right)$$

5. Substitute numerical values: Given $M = 51$ kg, $m_1 = 5$ kg, $m_2 = 4$ kg, and using the standard gravitational acceleration $g = 9.8 \text{ m/s}^2$:

   $$F = (51 \text{ kg} + 5 \text{ kg} + 4 \text{ kg})\left(\frac{4 \text{ kg}}{5 \text{ kg}}(9.8 \text{ m/s}^2)\right)$$ $$F = (60 \text{ kg})(0.8 \times 9.8 \text{ m/s}^2)$$ $$F = (60 \text{ kg})(7.84 \text{ m/s}^2)$$ $$F = 470.4 \text{ N}$$