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 mmand 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.

Asked by Abhisek | 1 year ago |  190

##### Solution :-

Given,

Cross-sectional area of wire A, a1 = 1.0 mm= 1.0 × 10–6 m2

Cross-sectional area of wire B, a2 = 2 mm= 2 × 10–6 m2

We know, Young’s modulus for steel, Y1 = 2 × 1011 Nm–2

Young’s modulus for aluminium, Y2 = 7.0 ×1010 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,

F1 = Force acting on wire A

and F2 = 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:

F1y = F2 (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 y1 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 :

F1y= F2 (1.5 – y1)

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

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

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

7(1.05 – y1)  =  10y1

y1 = 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.

Answered by Pragya Singh | 1 year ago

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