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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.
Asked by Pratheema Last Modified
Answer 
Apoorva Purohit
(a) 0.7 m from the steel-wire end
(b) 0.432 m from the steel-wire end
Cross-sectional area of wire A, a1 = 1.0 mm2 = 1.0 × 10–6 m2
Cross-sectional area of wire B, a2 = 2.0 mm2 = 2.0 × 10–6 m2
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.
If the two wires have equal stresses, then:
Where,
F1 = Force exerted on the steel wire
F2 = Force exerted on the aluminum wire
The situation is shown in the following figure.
Taking torque about the point of suspension, we have:
Using equations (i) and (ii), we can write:
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) 
If the strain in the two wires is equal, then:
Taking torque about the point where mass m, is suspended at a distance y1 from the side where wire A attached, we get:
F1y1 = F2 (1.05 – y1)
… (iii)
Using equations (iii) and (iv), we get:
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.
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