### Q.18:- 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 mm^{2} and 2.0 mm^{2}, 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**

Cross-sectional area of wire **A**, *a*_{1} = 1.0 mm^{2 }= 1.0 × 10^{–6} m^{2}

Cross-sectional area of wire **B**, *a*_{2} = 2.0 mm^{2 }= 2.0 × 10^{–6} m^{2}

Young’s modulus for steel, *Y*_{1} = 2 × 10^{11 }Nm^{–2}

Young’s modulus for aluminium, *Y*_{2} = 7.0 ×10^{10 }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.

Stress in the wire = Force / Area = *F* / *a*

If the two wires have equal stresses, then:

*F*_{1} / *a*_{1} = *F*_{2} / *a*_{2}

Where,

*F*_{1} = Force exerted on the steel wire

*F*_{2} = Force exerted on the aluminum wire

*F*_{1} / *F*_{2} = *a*_{1} / *a*_{2} = 1 / 2 ….**(i)**

The situation is shown in the following figure:

Taking torque about the point of suspension, we have:

*F*_{1}*y* = *F*_{2} (1.05 – *y*)

*F*_{1} / *F*_{2} = (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:

(*F*_{1}/*a*_{1}) / *Y*_{1} = (*F*_{2}/*a*_{2}) / *Y*_{2}

*F*_{1} / *F*_{2} = *a*_{1}*Y*_{1} / *a*_{2}*Y*_{2}

*a*_{1} / *a*_{2} = 1/2

*F*_{1} / *F*_{2} = (1 / 2) (2 × 10^{11} / 7 × 10^{10}) = 10 / 7 …….**(iii)**

Taking torque about the point where mass *m*, is suspended at a distance *y*_{1} from the side where wire **A** attached, we get:

*F*_{1}*y*_{1 }= *F*_{2} (1.05 – *y*_{1})

*F*_{1} / *F*_{2} = (1.05 – *y*_{1}) / y_{1} ….**(iii)**

Using equations **(iii)** and **(iv)**, we get:

(1.05 – *y*_{1}) / y_{1} = 10 / 7

7(1.05 – y_{1}) = 10*y*_{1}

*y*_{1} = 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.