Question

A light rod of length 2m is suspended from the ceiling horizontally by means of two vertical wires of equal lengths. A weight w is hung from the light rod as shown. The rod is hung by means of a steel wire of cross-sectional area a1 = 0.1cm2 and brass wire of cross-sectional area a2 = 0.2 cm2. To

Answer 1

The weight is hung at position x - 4/3 m to produce equal stress.

Explanation:

Given data:

Length of rod = 2 m

Area of cross section of steel wire = 10^-3 m^2

Area of cross section of brass wire = 2 × 10-3 m^2

Young's modulus of Brass = 1 × 10^11 N.m^2

Young's modulus of steel = 2 × 10^11 N.m^2

T1 / A1 = T2 / A2

=> T2 = 2 T1

Στ = 0 about point C.

T1x = T2 ( 2 - x )

x = 4 /3 m

The weight is hung at position x = 4/3 m to produce equal stress.

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A force of 5N acting on a body at a angle of 30 degree with an horizontal direction displaces it horizontally through a distance of 6m .Calculate the work done . ?

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Answer 2

To where i can understand from ur such silly incomplete question..

you r asking us to find resultant youngs modulous???

If it is the case then

$y = \frac{y1 + y2}{2}$

Apne question mein se values vagarah put kro ab isme ....