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 the figure. 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.

Asked by Abhisek | 1 year ago | 190

Given,

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 mm^{2 }= 2 × 10^{–6} m^{2}

We know, 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}

**(i) **Let a mass m be hung on the stick at a distance y from the end where wire A is attached.

Stress in the wire

= \( \dfrac{Force}{Area }=\dfrac{ F}{a}\)

Now it is given that the two wires have equal stresses ;

\( \dfrac{F_1}{F_2}=\dfrac{F_1}{F_2}\)

Where,

F_{1} = Force acting on wire A

and F_{2} = Force acting on wire B

\( \dfrac{F_1}{F_2}=\dfrac{F_1}{F_2}=\dfrac{1}{2}\) . . . . . . . . ( 1 )

The above situation can be represented as :

Moment of forces about the point of suspension, we have:

F_{1}y = F_{2} (1.5 – y)

\( \dfrac{F_1}{F_2}\) = \( \dfrac{(1.5 – y)}{y}\) . . . . . . . ( 2 )

Using equation ( 1 ) and equation ( 2 ), we can write:

\( \dfrac{(1.5 – y)}{y}=\dfrac{1}{2}\)

2 (1.5 – y) = y

y = 1 m

Therefore, the mass needs to be hung at a distance of 1m from the end where wire A is attached in order to produce equal stress in the two wires.

**(ii)** We know,

Young’s modulus =\( \dfrac{Stress}{Strain}\)

= Strain

= \(\dfrac{Stress}{Young’s\; modulus }\)

= ( \( \dfrac{F}{a}.\dfrac{1}{Y}\))

It is given that the strain in the two wires is equal :

\( \dfrac{F_1}{a_1}.\dfrac{1}{Y_1}= \dfrac{F_1}{a_2}.\dfrac{1}{Y_2}\)

\( \dfrac{F_1}{F_2}=\dfrac{a_1Y_1}{a_2Y_2}\)

\( \)\( \dfrac{a_1}{a_2}=\dfrac{1}{2}\)

\( \dfrac{F_1}{F_2}\) = (\( \dfrac{1}{2}\)) (\( \dfrac{(2 × 10^{11}}{7 × 10^{10}}\))

= \( \dfrac{10}{7}\). . . . . . . . ( 3 )

Let the mass m be hung on the stick at a distance y_{1} from the end where the steel wire is attached in order to produce equal strain

Taking the moment of force about the point where mass m is suspended :

F_{1}y_{1 }= F_{2} (1.5 – y_{1})

\( \dfrac{F_1}{F_2}\) = \( \dfrac{(1.5 – y_1) }{Y_1}\) . . . . . . . . . . . ( 4 )

From equations ( 3 ) and ( 4 ), we get:

\( \dfrac{(1.05 – y_1) }{Y_1}\)= \( \dfrac{10}{7}\)

7(1.05 – y_{1}) = 10y_{1}

y_{1} = 0.432 m

Therefore, the mass needs to be hung at a distance of 0.432 m from the end where wire A is attached in order to produce equal strain in the two wires.

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