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 figure.

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.

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 ×10¹⁰Nm^–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
F₁ / F₂ = a₁ / a₂  =  1 / 2    ….(i)
The situation is shown in the following figure:

Taking torque about the point of suspension, we have:
F₁y = F₂ (1.05 – y)
F₁ / F₂ = (1.05 – y) / y    ……(ii)
Using equations (i) and (ii), we can write:
(1.05 – y) / y  = 1 / 2
2(1.05 – y)  =  y
y = 0.7 m
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) Young’s modulus = Stress / Strain
Strain = Stress / Young’s modulus  =  (F/a) / Y
If the strain in the two wires is equal, then:
(F₁/a₁) / Y₁  =  (F₂/a₂) / Y₂
F₁ / F₂ = a₁Y₁ / a₂Y₂
a₁ / a₂ = 1/2
F₁ / F₂ = (1 / 2) (2 × 10¹¹ / 7 × 10¹⁰)  =  10 / 7      …….(iii)
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)
F₁ / F₂  =  (1.05 – y₁) / y₁    ….(iii)
Using equations (iii) and (iv), we get:
(1.05 – y₁) / y₁  =  10 / 7
7(1.05 – y₁)  =  10y₁
y₁ = 0.432 m
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