Question

A load of $10\text{kN}$ is supported from a pulley which in turn is supported by a rope of cross-sectional area ${10}^3{\text{mm}}^2$ and modulus of elasticity ${10}^3{\text{N/mm}}^2$ as shown in the figure. Neglecting friction at the pulley, downward deflection of the load $\left(\text{in mm}\right)$ is

• A. $3.75$
• B. $4.25$
• C. $2.75$
• D. $4.00$

Answer 1

The correct option is A $3.75$

Given:

$A={10}^3{\text{mm}}^2;Y={10}^3{\text{N/mm}}^2$

FBD of the given arrangement


From equilibrium of forces,

$T_0={10}^4\text{N}$ (from FBD)

Also,

$T+T=T_0$

$\Rightarrow T=\frac{T_0}{2}=0.5\times {10}^4\text{N}.......\left(1\right)$

We know that, from stress-strain relation,

$Y=\frac{FL}{A\Delta L}$

$\Rightarrow \Delta L=\frac{FL}{AY}$

So, extension in left rope,

$\Delta L=\frac{0.5\times {10}^4\times 600}{{10}^3\times {10}^3}$

$\Rightarrow \Delta L_1=3\text{mm}$

Further, extension in right rope,

$\Delta L=\frac{0.5\times 10\times 900}{{10}^3\times {10}^3}$

$\Rightarrow \Delta L_2=4.5\text{mm}$

So, total extension in the rope $=\Delta L$

$\Delta L=\Delta L_1+\Delta L_2=3+4.5=7.5\text{mm}$

Hence, downward deflection $\left(\delta \right)$ of the load is $\frac{\Delta L}{2}$

$\therefore \delta =\frac{\Delta L}{2}=\frac{7.5}{2}=3.75\text{mm}$

$\begin{array}{c}\text{Why this question ?}\\ \text{Concept - If load goes down by x then rope on left}\\ \text{and right side of the pulley will also so down by x, thus}\\ \text{in total, extension in rope will be}2x\text{(say y)}\\ \therefore x=\frac{y}{2}\end{array}$