Question
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 (a) equal stresses and b) equal strains in both steel and aluminium wires.
Solution
Given,
Cross-sectional area of wire A = a1 = 1.0 mm2 = 1.0 × 10–6m2
Cross-sectional area of wire B, a2 = 2.0 mm2 = 2.0 × 10–6m2
Young’s modulus for steel, Y1 = 2 × 1011 Nm–2
Young’s modulus for aluminium, Y2 = 7.0 ×1010 Nm–2
a) Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.
Stress in the wire = Force / Area = F / a
If the two wires have equal stresses, then
F1 / a1 = F2 / a2
where,
F1 = Force exerted on the steel wire
F2 = Force exerted on the aluminum wire
In order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m from the end where wire A is attached.
b)
In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire A is attached.
F1 / a1 = F2 / a2
where,
F1 = Force exerted on the steel wire
F2 = Force exerted on the aluminum wire
In order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m from the end where wire A is attached.
b)
In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire A is attached.
