A light bar of length 2 m is suspended horizontally by means of two wires of equal lengths connected to its two ends. One wire is of steel having a cross sectional area of `0.1 cm^2`, the other is of brass with a cross sectional area of `0.2 cm^2`,Find the point on the bar from where a weight must be suspended so that both the wires experience (i) the same stress,(ii) the same strain, Given the young's modulus for steel `=2 times 10^11 N.m^-2` and that for brass `=10 times 10^10 N.m^-2`.

Explanation

Let the horizontal bar be AB (fig.1.14). A load W is hung from the point C. Let AC be x. In this condition, let the tension in the steel wire be `T_1` and that in the brass wire be `T_2`.

Stress on the steel wire `=T_1/alpha_1` and stress on the brass wire=`T_2/alpha_2`.
(i) IF the stresses in both the wires are the same, then
`T_1/(0.1times10^-4)=T_2/(0.2times10^-4)or,T_1/T_2=1/2`.....(1)
Since the system is in equilibrium, taking moments about the point C, we get
`T_1x=T_2(2-x)or,T_1/T_2=(2-x)/x`
`therefore1/2=2/x-1or,2/x=3/2or,x=4/3=1.33m`
So the weight should be suspended at a distance of 1.33m from the steel wire.
(ii) Longitudinal strain `=(longitudi na l stress)/Y`
`=(T/a)/Y=T/(aY)`
For the same strain in the two wires,
`T_1/(0.1times10^-4Y_1)=T_2/(0.2times10^-2 Y_2)`
or,`T_1/T_2=(0.1times10^-4times2times10^11)/(0.2times10^-4times10times10^10)=1thereforeT_1=T_2`
Since the system is in equilibrium, taking moments about the point C, we get,
`T_1x=T_2(2-x)or,x=2-xor,x=1m`
So, the weight should be suspended from the mid-point of the horizontal bar.