Imagine two boxes stacked on top of each other.
The bottom box is larger than the top box and it rests on a surface.
So:
Between the boxes is a friction force constant $\mu_s$ and between the bottom box and the surface is the friction force constant $0,24$.
What is the maximum force I can pull the bottom box (box 1) with in order for box 2 not to slide.
I have NO idea how to start this problem.
All I got is some gravity forces acting on the boxes and the traction force on the bottom box represented as:
$9,8 \cdot (b_1 + b_2)$ and traction: $0,24 \cdot 9,8 \cdot (b_1 + b_2)$
Hints:
For the bottom mass, you have the following force equation in x-axis (why?)
$$T - \mu_s m_{top}g - 0.24 (m_{top}+m_{bottom})g = (m_{top}+m_{bottom})a \tag{1}$$ where $a$ is the common acceleration if the top body does not slip.
Writing equations for the top mass you have:
$$\mu_s m_{top}g = m_{top}a \implies a \le \mu_sg \quad \text{for not slipping} \tag{2}$$
Use the bound from $(2)$ in $(1)$ to get the maximum force for not slipping.