(a) 0.7 m from the steel-wire end 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
Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.
F1 = Force exerted on the steel wire
F2 = Force exerted on the aluminum wire
F1/F2 = a1/a2 = 1/2.............(i)
The situation is shown in the following figure.
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.
Taking torque about the point where mass m, is suspended at a distance y1 from the side where wire A attached, 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.