Q32P - Fundamentals Of Physics

Question

A block of mass M is pulled along a horizontal frictionless surfaceby 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 imperceptibleamount. Then, assuming that the sag is negligible, find 

(b) the accelerationof rope and block,

 (c) the force on the block from the rope, and 

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

Step-by-Step Solution

Answer

1. Rope must sag because of mass mThe rope is 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 rope is $\frac{MF}{M+m}$
4. The tension in the rope at its midpoint is$\frac{(2M+m)F}{2(M+m)}$

Step 1: Given information

It is given that,

$\text{Mass of block}=M \text{kg}$

$\text{Mass of rope}=m_k$

Step 2: Determining the concept

According to the Newton’s second law of motion:

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

Here,F is the net force, M is mass and a is an acceleration.

 

Step 3: (a) Show that rope must sag

Consider the diagram for the force on the block as:

Rope must sag because of mass m.The rope is pulled down by Earth’s gravitational force.

For equilibrium, there would be a component of Tension in upward direction, which indicates rope must sag.

 

Step 4: (b) Determine the acceleration of rope and block

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

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

Step 5: (c) Determine 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, 

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

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

Step 6: (d) Determine thetension in the rope at its midpoint

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 $T=\frac{(2M+m)F}{2(M+m)}$