A steel rod of length `l_(1) = 30 cm` and two identical brass rod of length `l_(2)= 20 cm` each support a light horizontal platform as shown in Fig. Cross-sectional area of each of the three rods is `A = 1 cm^(2)`. A vertically downward force `F= 5000 N` is applied on the platform. Young's modulus of elasticity for steel `Y_(s)=2xx10^(11)Nm^(-2)` and brass `Y_(b) = 1 xx 10^(11) Nm^(-2)`. Find stress (in `MPa`) developed in a. Steel rod b. Brass rod

Answer

`15mPa`

Explanation

Let tension in steel rod be `T_(1)` and in brass rod be `T_(2)` then
`T_(1)+2T_(2)=5000`………i
Equation in both steel and brass rods will be same
`(/_\l_(s)=/_\l_(b))`
`(T_(1)l_(1))/(AY_(1))=(T_(2)l_(2))/(AY_(2))impliesT_(1)=(4T_(2))/3`………ii
From eqn i and ii `T_(1)=2000N`
`T_(2)=1500N`
Stress in steel wire:
`S_(1)=(T_(1))/A=2000/10^(-4)=2xx10^(7)N//m^(2)=20MPa`
Stress in brass wire:
`S_(2)=(T_(2))/A=1500/10^(-4)=1.5xx10^(7)N//m^(2)=15MPa`