Question

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?

Answer 1

Let the force F be applied at a distance x from the steel bar, measured along the orizontal bar. Let FS and FC 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 $\frac{F_s}{A_s{Y}_s}=\frac{F_c}{A_c{Y}_c}$ (viii)
Solving Eqs. (vii) and (viii), we get
$F_S=\frac{F{A}_S{Y}_S}{A_S{Y}_S+A_C{Y}_C}$
and $F_C=\frac{F{A}_C{Y}_C}{A_S{Y}_S+A_C{Y}_C}$
Now, taking moments about the steel bar,
$F_{C}l=F\cdot x=\frac{F_C}{F}l=\frac{A_C{Y}_{C}l}{A_S{Y}_S+A_C{Y}_C}$
or $x=\frac{l}{1+\left(A_S/A_C\right)\left(Y_S/Y_C\right)}$