# Q18 (b)  A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas of wires A and B are $1.0 mm^2 \: \:and \: \: \: 2.0 mm^2$ respectively. At what point along the rod should a mass m be suspended in order to produce equal strains in both steel and aluminium wires.

S Sayak

Cross-Sectional Area of wire A is AA = 1 mm2

Cross-Sectional Area of wire B is AB = 2 mm2

Let the Mass m be suspended at y distance From the wire A

Let the Tension in the two wires A and B be FA and FB respectively

Since the Strain in the wires is equal

$\\\frac{F_{A}}{A_{A}Y_{S}}=\frac{F_{B}}{A_{B}Y_{Al}}\\ \frac{F_{A}}{F_{B}}=\frac{A_{A}Y_{S}}{A_{B}Y_{Al}}\\ \frac{F_{A}}{F_{B}}=\frac{1\times 2\times 10^{11}}{2\times 7\times 10^{10}}\\ \frac{F_{A}}{F_{B}}=\frac{10}{7}$

Equating moments of the Tension in the wires about the point where mass m is suspended we have

$\\yF_{A}=(1.05-y)F_{B}\\ yF_{A}=(1.05-y)\times \frac{7}{10}F_{A}\\ 10y=7.35-7y\\ y=0.432m$

The Load should be suspended at a point 43.2 cm from the wire A such that there is an equal strain in the two wires.

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