Question

✨✨❣️hey guys can anybody solve this

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 mm2 and 2.0 mm,
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.

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Answer 1

Given :-

Cross-sectional area of wire A, $\sf a_1$ = $\sf 1.0\: mm^{2} = 1.0 \times 10^{-6} m^{2}$

Cross-sectional area of wire B, $\sf a_2$ = $\sf 2 \: mm^{2} = 2 \times 10^{-6} m^{2}$

To Find :-

The point along the rod should a mass m be suspended in order to produce equal stresses.

Point along the rod should a mass m be suspended in order to  produce equal strains in both steel and aluminium wires.

Solution :-

Given that,

Cross-sectional area of wire A, $\sf a_1$ = $\sf 1.0\: mm^{2} = 1.0 \times 10^{-6} m^{2}$

Cross-sectional area of wire B, $\sf a_2$ = $\sf 2 \: mm^{2} = 2 \times 10^{-6} m^{2}$

We know that,

Young’s modulus for steel, $\sf Y_1$ = $\sf 2 \times 10^{11} Nm^{-2}$

Young’s modulus for aluminium, $\sf Y_2$ = $\sf 7.0 \times 10^{10} Nm^{-2}$

Solving (a),

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 = $\sf \dfrac{Force}{Area}$

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

$\longrightarrow \sf \dfrac{F_1}{F_2} =\dfrac{F_2}{a_2}$

Where,

$\sf F_1$ = Force acting on wire A

And, $\sf F_2$ = Force acting on wire B

$\longrightarrow \sf \dfrac{F_1}{F_2} =\dfrac{a_1}{a_2}=\dfrac{1}{2} \qquad ...(1)$

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

$\sf F_1y=F_2(1.5 -y)$

$\implies \sf \dfrac{F_1}{F_2} =\dfrac{(1.5-y)}{y} \qquad ...(2)$

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

$\sf \longrightarrow \dfrac{(1.5-y)}{y} =\dfrac{1}{2}$

$\implies \sf 2 (1.5 - y) = y$

$\implies \sf y = 1 \: m$

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.

Solving (b),

We know,

Young’s modulus = $\sf \dfrac{Stress}{Strain}$

$\longrightarrow \sf \dfrac{Strain=Stress}{Young's \ Modulus} =\dfrac{y/a}{y}$

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

$\longrightarrow \sf \dfrac{(F_1/a_1)}{Y_1} =\dfrac{(F_2/a_2)}{Y_2}$

$\implies \sf \dfrac{F_1}{F_2} =\dfrac{a_1Y_1}{a_2Y_2}$

$\implies \sf \dfrac{a_1}{a_2}=\dfrac{1}{2}$

$\implies \sf \dfrac{F_1}{F_2} = \bigg( \dfrac{1}{2} \bigg) \bigg(\dfrac{2 \times 10^{11}}{7 \times 10^{10}} \bigg)=\dfrac{10}{7} \qquad ...(3)$

Let the mass m be hung on the stick at a distance $\sf y_1$ 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:

$\longrightarrow \sf F_1y_1 = F_2 (1.5 - y_1)$

$\implies \sf \dfrac{F_1}{F_2} =\dfrac{(1.5-y_1)}{y_1} \qquad ...(4)$

From  equations (3) and (4), we get:

$\implies \sf \dfrac{(1.05-y_1)}{y_1} =\dfrac{10}{7}$

$\implies \sf 7(1.05 -y_1) = 10y_1$

$\implies \sf y_1 = 0.432\: m$

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.