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Q18 (a)  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 mm2 and 2.0 mm2, respectively. At what point along the rod should a mass m be suspended in order to produce  equal stresses 

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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 x distance From the wire A

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

Since the Stress in the wires is equal

\\\frac{F_{A}}{A_{A}}=\frac{F_{B}}{A_{B}}\\ \frac{F_{A}}{1}=\frac{F_{B}}{2}\\ 2F_{A}=F_{B}

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

\\xF_{A}=(1.05-x)F_{B}\\ xF_{A}=(1.05-x)\times 2F_{A}\\ x=2.1-2x\\ x=0.7m

The Load should be suspended at a point 70 cm from a wire A such that there are equal stresses in the two wires.

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