__Question 9.18__: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__:**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:

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 X 10

^{11}/ 7 X 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}) = 10y

_{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.

## No comments:

## Post a Comment