Question

Two weightless cables of equal length and equal cross-sectional area hang from a ceiling. They are connected by a horizontal light bar of length 1.0 m. A downward force $F$ is applied at a point located $x$ meters from Cable–1. The moduli of elasticity of Cable–1 and Cable–2 are 50 GPa and 200 GPa respectively. If the deformation in both cables is equal, find $x$ (in m), rounded to one decimal place.

Hint:

Equal deformation in parallel members means force divides in proportion to their stiffness: \[ \frac{T_1}{T_2} = \frac{E_1}{E_2}. \]

Answer 1

Correct Answer: 0.8

Let the forces in Cable–1 and Cable–2 be: \[ T_1,\quad T_2. \] Because the bar is in horizontal static equilibrium under the downward force $F$: \[ T_1 + T_2 = F. \] Taking moments about Cable–1: \[ T_2(1.0) = F x \] \[ T_2 = Fx. \] Thus: \[ T_1 = F - Fx = F(1 - x). \] --- Equal deformation condition For axial deformation: \[ \delta = \frac{T L}{AE} \] Since both cables have same length and same area: \[ \frac{T_1}{E_1} = \frac{T_2}{E_2} \] Given: \[ E_1 = 50\ \text{GPa},\quad E_2 = 200\ \text{GPa} \] Substitute: \[ \frac{F(1-x)}{50} = \frac{Fx}{200} \] Cancel \(F\): \[ \frac{1-x}{50} = \frac{x}{200} \] Cross–multiply: \[ 200(1 - x) = 50x \] \[ 200 - 200x = 50x \] \[ 200 = 250x \] \[ x = \frac{200}{250} = 0.8\ \text{m} \] Rounded to one decimal: \[ \boxed{0.8\ \text{m}} \]