Chapter 5, Problem 88A

To determine

To Find: The expression for $F$ in terms of $M,m,{\mu }_s,g$

Answer to Problem 88A

  $F=\frac{g\left(M+m\right)}{{\mu }_s}$

Explanation of Solution

Given:

Mass of bigger block $=M$

Mass of smaller block $=m$

Applied force $=F$

Coefficient of static friction $={\mu }_s$

Formula used:

Newton’s law of motion

  $F=ma$

Where,

  $m$ is the mass

  $a$ is the acceleration

  $F$ is the force

Calculation:

Let the acceleration of the blocks = a

Condition for no fall is:

Frictional force between $m$ and $M$ must be greater than the weight of small box $mg$

  ${\mu }_{s}N={\mu }_{s}ma$

Where,

  $N$ is the normal force

Now applying condition

  $\begin{array}{l}{\mu }_{s}ma\ge mg\\ a=\frac{g}{{\mu }_s}\end{array}$

This is the acceleration of the blocks.

Force applied is:

  $\begin{array}{l}F=\left(m+M\right)a\\ F=\left(m+M\right)\frac{g}{{\mu }_s}\end{array}$

Conclusion:

Required expression for force is $F=\frac{g\left(M+m\right)}{{\mu }_s}$