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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. |
Answer» Solution :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. |
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