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 crosssectional 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.

Answer 1

(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

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.

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:

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.