A block of mass$\mathit{M}$is pulled along a horizontal friction less surface by a rope of mass$\mathit{m}$, as shown in Fig. 5-63. A horizontal force$\overrightarrow{\mathbf{F}}$acts on one end of the rope.(a) Show that the rope must sag, even if only by an imperceptible amount. Then, assuming that the sag is negligible, find (b) the acceleration of rope and block, (c) the force on the block from the rope, and (d) the tension in the rope at its midpoint.

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,

$F_{net}=\sum _{}Ma$

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