Two vertical rods of equal lengths, one of steel and the other of copper, are suspended from the ceiling at a distance `l` apart and are connected rigidly to a rigid horizontal bar at their lower ends. If `A_(S)` and `A_(C)` be their respective cross-sectional areas, and `Y_(S)` and `Y_(C)`,their respective Young's moduli of elasticities, where should a vertical force `F` be applied to the horizontal bar in order that the bar remains horizontal?`

Explanation

Let the force `F` be applied at a distance `x` from the steel bar, measured along the horizontal bar. Let `F_(S)` and `F_(C)` be the loads on steel and copper rods, respectively. Then
`F_(S)+F_(C)=F`……….vii
Since the rigid horizontal bar remains horizontal, the extensions produced in the two rods and hence strains remain the same.
That is
`(F_(s))/(A_(s)Y_(s))=(F_(C))/(A_(C)Y_(C))`..........viii
Solving eqn vii and viii we get
`F_(S)=(FA_(S)Y_(S))/(A_(S)Y_(S)+A_(C)Y_(C))`
and `F_(C)=(FA_(C)Y_(C))/(A_(S)Y_(S)+A_(C)Y_(C))`
now taking moments about steel bar
`F_(C)l=F.x=(F_(C))/Fl=(A_(C)Y_(C)l)/(A_(S)Y_(S)+A_(C)Y_(C))`
or `l/(1+((A_(S))/(A_(C)))((Y_(S))/(Y_(C))))`