91Ó°ÊÓ

It is given that,

$$\begin{gathered} \mathrm{Mass}\mathrm{of}\mathrm{block}=\mathrm{M}\mathrm{kg} \\ \mathrm{Mass}\mathrm{of}\mathrm{rope}={\mathrm{m}}_{\mathrm{k}} \end{gathered}$$

According to the Newton’s second law of motion:

${\mathbf{F}}_{\mathbf{net}}\mathbf{=}\underset{\mathbf{}}{\mathbf{∑}}\mathbf{Ma}$

Here, Fis the net force, Mis mass and a is an acceleration.

Consider the diagram for the force on the block as:

Rope must sag because of mass m.The ropeis pulled down by Earth’s gravitational force.

For equilibrium, there would be a component of Tension in upward direction, which indicates rope must sag.

By applying Newton’s law:

$F=\left(M+m\right)a$

Rewrite the equation for acceleration as:

$a=\frac{F}{\left(M+m\right)}$

Hence, the acceleration of rope and block is

$a=\frac{F}{\left(M+m\right)}$

If we consider FBD of block only, there is only one force acting on block.

i.e.$F_r$

$F_r=Ma$

Hence,

$F_r=\frac{MF}{\left(M+m\right)}$

Hence, the force on the block from the rope is$F_r=\frac{MF}{\left(M+m\right)}$

At the midpoint of rope, total mass$=M+\left(\frac{m}{2}\right)$

Hence, derive the equation for the tension as:

$$\begin{gathered} T=\left(M+\frac{m}{2}\right)a \\ T=\frac{\left(2M+m\right)F}{2\left(M+m\right)} \end{gathered}$$

Hence, the tension in the rope at its midpoint is$T=\frac{\left(2M+m\right)F}{2\left(M+m\right)}$