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.

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.

Asked by Abhisek | 1 year ago |  190

1 Answer

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 :

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.

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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