Question

A large box of mass $M$is pulled across a horizontal, frictionless surface by a horizontal rope with tension $T$. A small box of mass m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are ${\mu }_s$and ${\mu }_k$, respectively. Find an expression for the maximum tension $T_{max}$for which the small box rides on top of the large box without slipping.

Short answer

The expression is$T_{max}={\mu }_{s}g(M+m)$

Step-by-step solution

Step 1. The small box acceleration

First we isolate the F.B.D of each mass,

we start with the smaller one who is on top of the bigger box

The small box need to accelerate toto induce a force to make it slip

Step 2. Acceleration needed for the small box to start slipping 

Now let take the F.B.D of the large box, the large box is moving in a frictionless surface

$$\begin{gathered} \sum _{}{F}_{xM}=0 \\ {T}_{max}-ma=0 \\ {T}_{max}=m_{total}{a}_{max} \end{gathered}$$

We will use the acceleration needed for the small box to start slipping, because the reference frame of the small box is the large box

$$\begin{gathered} T_{max}=(M+m)\left({\mu }_{s}g\right) \\ {T}_{max}={\mu }_{s}g(M+m) \end{gathered}$$