The two blocks${}^{(m=16\mathrm{kg}\mathrm{and}M=88\mathrm{kg})}$in Fig. 6-38 are not attached to each other. The coefficient of static friction between the blocks is ${}^{{\mathbf{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{38}}$,but the surface beneath the larger block is frictionless. What is the minimum magnitude of the horizontal force ${}^{\mathbf{F}}$required to keep the smaller block from slipping down the larger block?

Mass of the two blocks,$m=16\mathrm{kg}$ and$M=88\mathrm{kg},{\mu }_s=0.38$

Use the concept of friction and Newton's first law of motion. According to Newton’s first law, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

Formula:

$F_{Net}=\sum _{}ma$

where, F is the net force, m is mass and a is an acceleration.

The free-body diagrams for the two blocks:

Treating the two blocks together as a single system (sliding across a frictionless floor), apply Newton’s second law (with +x rightward) to find an expression for the acceleration:

$$\begin{gathered} F=m_{total}a \\ a=\frac{F}{m_{total}} \\ =\frac{F}{m+M} \end{gathered}$$

This is equivalent to having analyzed the two blocks individually and then combined their equations. Now, when the small block has been analyzed individually, apply Newton’s second law to the x and y axes, substitute in the above expression for a,

$$\begin{gathered} F-F\text{'}=ma \\ F\text{'}=F-m\left(\frac{F}{m+M}\right) \\ {f}_s-mg=0 \\ {\mu }_{s}F\text{'}-mg=0 \end{gathered}$$

These expressions are combined (to eliminate F') and arrive at,

$$\begin{gathered} F=\frac{mg}{{\mu }_{s\left(1-{\displaystyle \frac{m}{m+M}}\right)}} \\ =4.9\times {10}^{2}N \end{gathered}$$

Hence, the minimum magnitude of the horizontal force required to keep the smaller block from slipping down the larger block is${}^{4.9\times 10\mathrm{N}}$.