We need to calculate the minimum force $F$ required to prevent block $m_1$ (23 kg) from slipping down while it is held against block $m_2$ (77 kg) on a frictionless surface.

Step-by-Step Solution:

1. Forces on Block $m_1$:

• The weight of block $m_1$ causes a downward force: $W_1 = m_1 g = 23 \times 9.8 = 225.4 \, \text{N}$ where $g$ is the acceleration due to gravity (9.8 m/s²).

• The force of static friction between block $m_1$ and block $m_2$ resists this downward motion. The static frictional force $f_s$ is given by: $f_s = \mu_s N$ where $\mu_s = 0.53$ is the coefficient of static friction and $N$ is the normal force exerted by block $m_2$ on block $m_1$.

2. Normal Force $N$:

The normal force $N$ is provided by the horizontal force $F$ pushing block $m_1$ against block $m_2$. Hence: $N = F$

3. Friction Condition:

To prevent block $m_1$ from slipping down, the static friction must balance the weight of block $m_1$: $f_s \geq W_1$ Substituting for $f_s$ and $W_1$: $\mu_s F \geq 225.4$ Solving for $F$: $F \geq \frac{225.4}{0.53} = 425.28 \, \text{N}$

Conclusion:

The minimum force $F$ required to hold block $m_1$ against block $m_2$ without slipping is 425.28 N.

Let me know if you need further details or have any questions.

Here are some additional related questions:

1. What happens if the coefficient of friction is increased?
2. How does the mass of block $m_2$ affect the required force $F$?
3. What would happen if the surface beneath block $m_2$ were not frictionless?
4. How would the situation change if the direction of $F$ was at an angle?
5. What would the required force be if $m_1$ had a different mass?

Tip: When solving friction-related problems, always identify the forces acting on the object and their direction to correctly apply the frictional force formula.