Question

A light rod of length L is suspended from a support horizontally by means of two vertical wires A and B of equal length as shown in Fig. The cross sectional area of A is half that of B and the Youngs modulus of A is twice that of B. A weight W is hung as shown. The value of x so that W produces equal stress in wires A and B is

• A. $\frac{L}{3}$
• B. $\frac{L}{2}$
• C. $\frac{2L}{3}$
• D. $\frac{3L}{4}$

Answer 1

The correct option is C $\frac{2L}{3}$

for equal stress in wire A and B

$\frac{F_A}{A_A}=\frac{F_B}{A_B}$

$\frac{F_A}{F_B}=\frac{A_A}{A_B}=\frac{1}{2}(\because \text{given in problem}\frac{A_A}{A_B}=\frac{1}{2})$

$2F_A=F_B$

$Forequilibrium:$

$W=F_A+F_B$

$W=\frac{F_B}{2}+F_B$

$W=\frac{3F_B}{2}$

$F_B=\frac{2}{3}W$

Now equating torque about point 0.

$Wx=F_{B}L$

$x=\frac{2}{3}L$