91Ó°ÊÓ

It is given that,

$\text{Massofblock}=M\text{ }\text{kg}$

$\text{Massofrope}=m_k$

According to the Newton’s second law of motion:

${\mathit{F}}_{\mathbf{n}\mathbf{e}\mathbf{t}}\mathbf{=}\mathbf{∑}\mathit{M}\mathit{a}$

Here,F is 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:

$F=(M+m)a$

Rewrite the equation for acceleration as:

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

Hence, the acceleration of rope and block is

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

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

i.e.$F_r$

$F_r=Ma$

Hence,

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

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

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

Hence, derive the equation for the tension as:

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

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