Question

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.

Short answer

1. Rope must sag because of mass $m$.Theropeis pulled down by Earth’s gravitational.
2. The acceleration of rope and block is$\frac{F}{M+m}$
3. The force on the block from the ropeis$\frac{MF}{M+m}$
4. The tension in the rope at its midpoint is$\frac{(2M+m)F}{2(M+m)}$

Step-by-step solution

Given information

It is given that,

$Massofblock=Mkg$

$Massofrope=m_k$

Determining the concept

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.

(a) Showing that rope must sag

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.

(b) Determining the acceleration of rope and block

By applying Newton’s law:

$F=(M+m)a$

Hence,

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

Hence, the acceleration of rope and block is

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

(c) Determining the force on the block from the rope

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

(d) Determining the tension in the rope at its midpoint

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