Question

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 mm², respectively. At what point along the rod should a mass m be suspended in order to produce equal stresses.

Answer 1

HEY DEAR ... ✌️

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Here's , Your Answer :-

=) Given , Cross-sectional area of wire A, a1 = 1.0 mm^2 = 1.0 × 10^-6m2
Cross-sectional area of wire B, a2 = 2.0 mm^2 = 2.0 × 10^-6 m^2
Young’s modulus for steel, Y1 = 2 × 10^11 Nm^-2
Young’s modulus for aluminium, Y2 = 7.0 ×10^10 Nm^-2

Let a small mass m be suspended to the rod .

So , Stress in the wire = Force / Area  =  F / a

If the two wires have equal stresses, then ,
it would be , F1 / a1  =  F2 / a2

Where , we know ,
F1 = Force exerted on the steel wire
F2 = Force exerted on the aluminum wire
F1 / F2 = a1 / a2  =  1 / 2   -----------(i)

The situation is clearly shown in the figure attached by you .

Now , Taking torque about the point of suspension , So , we have ,
F1y = F2 (1.05 – y)

F1 / F2 = (1.05 – y) / y    ----------(ii)

We got two equations

By using the equations (i) and (ii) , we can write ,

(1.05 – y) / y  = 1 / 2

2(1.05 – y)  =  y

y = 0.7 m. (ans.)

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