Question

A load of $10kN$ is supported from a pulley which in turn is supported by a rope of sectional area, $1\times {10}^{3}m{m}^2$ and modulus of elasticity ${10}^{3}Nm{m}^{-2}$, as shown in Fig. 5.18. Neglecting the friction at the pulley, determine the deflection of the load is $x+0.75mm$. Find $x$

Answer 1

Let T be the tension in the rope.
Then 2T = 10kN $\Rightarrow$ = T = 5 kN
Hence longitudinal stress in the rope is
$\sigma =\frac{T}{A}=\frac{5kN}{{10}^{3}m{m}^2}=5N/m{m}^2$
Extension in the rope $=\frac{Stress}{Y}\times L$
$=\frac{5N/m{m}^2}{{10}^{3}N/m{m}^2}\times 1500mm=7.5mm$
Therefore, deflection of the load $\delta =\frac{7.5}{2}=3.75mm.$