Two blocks, a smaller one of mass m and a larger one of mass M, are in contact. A horizontal force F is applied to the larger block. The coefficient of static friction between the blocks is 00b5s. Wha

For the two blocks to move together without slipping, the static friction force ($f_s$) exerted by the larger block on the smaller block must provide the acceleration for the smaller block.
In the vertical direction, the normal force ($N_m$) acting on the small block is equal to its weight:

$$N_m = mg$$

In the horizontal direction, the static friction force is the only force causing the small block to accelerate. According to Newton's second law:

$$f_s = ma$$

The maximum static friction force ($f_{s,max}$) that can act on the small block is given by:

$$f_{s,max} = \mu_s N_m = \mu_s mg$$

For the blocks to move together without slipping, the acceleration ($a$) of the small block must be such that $f_s \le f_{s,max}$. The maximum acceleration ($a_{max}$) occurs when $f_s = f_{s,max}$:

$$ma_{max} = \mu_s mg$$ $$a_{max} = \mu_s g$$

Given $\mu_s = 0.6$ and $g = 9.81 \text{ m/s}^2$:

$$a_{max} = 0.6 \times 9.81 \text{ m/s}^2$$ $$a_{max} = 5.886 \text{ m/s}^2$$

When the blocks move together without slipping, they can be treated as a single system with a total mass of $(m+M)$.
The external force $F$ is applied to this combined system. Since the horizontal surface is frictionless, $F$ is the net external force on the combined system.
According to Newton's second law for the combined system:

$$F = (m+M)a$$

We are looking for the maximum force ($F_{max}$) that can be applied such that they move together. This occurs when the system accelerates at $a_{max}$.

$$F_{max} = (m+M)a_{max}$$

Given $M = 3m$, the total mass is $m+M = m+3m = 4m$.

$$F_{max} = (4m)a_{max}$$

Substitute the value of $a_{max}$:

$$F_{max} = (4m)(5.886 \text{ m/s}^2)$$ $$F_{max} = 23.544m \text{ N}$$

The maximum force $F$ that can be applied to the lower block such that the two blocks move together without slipping is $\boxed{23.544m \text{ N}}$.