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.

Answer 1

Answer: (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.