91Ó°ÊÓ

It is given that,

$Massofblock=Mkg$

$Massofrope=m_k$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it.Acceleration and tension can be calculated by using Newton’s second law.

Formula:

According to the Newton’s second law of motion,

$F_{net}=\underset{}{∑}Ma$

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

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:

$F=(M+m)a$

Hence,

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

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,

role="math" localid="1655526221452" $F_r=\frac{MF}{(M+m)}$

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,

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

$T=\frac{(2M+m)F}{2(M+m)}$

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