First slide

Stress and strain

Question

A light rod of length 200 cm is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross-section 0.1cm² and the other of brass of cross section 0.2cm². Along the rod at which distance a weight may be hung to produce equal stresses in both the wires ?

Difficult

A

$\frac{4}{3}$ m from steel wire
B

$\frac{4}{3}$ m from brass wire
C

1 m from steel wire
D

$\frac{1}{4}$ m from brass wire

Solution

As stresses are equal,
$\frac{T_{1}}{A_{2}} = \frac{T_{2}}{A_{1}} or \frac{T_{1}}{T_{2}} = \frac{A_{1}}{A_{2}}$
or $\frac{T_{1}}{T_{2}} = \frac{0 \cdot 1}{0 \cdot 2} = \frac{1}{2} or T_{2} = 2 T_{1}$ ……………(1)
For equilibrium of rod,T₁+T₂ =W……………………(2)
Solving eqs. (i) and (ii), we get
$T_{1} = \frac{W}{3} and T_{2} = \frac{2 W}{3}$
Let x be the distance of weight W from steel wire. Taking moment about weight ,
$T_{1} X = T_{2} (2 - x)$
or $\frac{W}{3} \times x = \frac{2 W}{3} \times (2 - x)$
Solving for x. we get $x = \frac{4}{3} m$

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