We need to determine the coefficient of static friction ($\mu_s$) between the upper block and the lower block that would prevent the upper block from slipping when a force is applied to the lower block. Here's the step-by-step process:

1. Free Body Diagram and Forces:

Upper Block:

• Mass of the upper block ($m_1$) = 11 kg
• Weight of the upper block ($W_1 = m_1 g$) = $11 \times 9.8 = 107.8 \, \text{N}$
• The force of friction acting between the upper block and the lower block (if the upper block is not to slip) is due to static friction, and this force is $f_s = \mu_s N_1$, where $N_1 = W_1$ (the normal force between the two blocks).
• This friction force is what accelerates the upper block.

Lower Block:

• Mass of the lower block ($m_2$) = 30 kg
• Weight of the lower block ($W_2 = m_2 g$) = $30 \times 9.8 = 294 \, \text{N}$
• The total normal force on the lower block is $N_2 = W_1 + W_2 = 107.8 + 294 = 401.8 \, \text{N}$
• The friction between the lower block and the floor is kinetic, with a coefficient $\mu_k = 0.58$. The frictional force is $f_k = \mu_k N_2 = 0.58 \times 401.8 = 233.044 \, \text{N}$.
• The net force accelerating the two blocks is the applied force minus this kinetic friction: $F_{\text{net}} = 367 - 233.044 = 133.956 \, \text{N}$.

2. Acceleration of the System:

Both blocks will accelerate together if the upper block does not slip. The total mass of the system is: $m_{\text{total}} = m_1 + m_2 = 11 + 30 = 41 \, \text{kg}$ Using Newton's second law, the acceleration of the system is: $a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{133.956}{41} \approx 3.27 \, \text{m/s}^2$

3. Friction Force Required to Prevent Slipping:

To prevent slipping, the static friction must provide enough force to accelerate the upper block at the same rate as the system. The force required to accelerate the upper block is: $F_{\text{required}} = m_1 a = 11 \times 3.27 = 35.97 \, \text{N}$

4. Static Friction and Coefficient:

The maximum static friction force available between the upper block and the lower block is: $f_s = \mu_s N_1 = \mu_s W_1 = \mu_s \times 107.8 \, \text{N}$ For the upper block not to slip, the static friction must at least equal the force required to accelerate it: $\mu_s \times 107.8 \geq 35.97$ Solving for $\mu_s$: $\mu_s \geq \frac{35.97}{107.8} \approx 0.3337$

Conclusion:

The coefficient of static friction $\mu_s$ between the upper and lower blocks would need to be at least 0.334 for the upper block not to slip on the lower block.

Would you like any further details or clarification?

Related Questions:

1. How does the acceleration change if the force applied on the lower block is increased?
2. What if the coefficient of kinetic friction between the lower block and the floor was 0.7?
3. How does the weight of the upper block affect the frictional forces between the two blocks?
4. What is the maximum force you can apply to the lower block without causing the upper block to slip, given $\mu_s = 0.334$?
5. How would the system behave if the surface was frictionless?

Tip: Always check if the static friction force is enough to prevent relative motion by comparing it with the required force for the desired acceleration.