A block of mass Mis pulled along a horizontal frictionless surface by a rope of mass 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,

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

According to the Newton’s second law of motion:

${\mathbf{F}}_{\mathbf{net}}\mathbf{=}\mathbf{\sum }_{\mathbf{}}\mathbf{Ma}$

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