Position of 3.5 tonne load on cross piece

Idea: the cross piece stays horizontal only if both rods stretch equally. The rods have the same area and length, so the load carried by each rod must be proportional to its $E$.

Let the steel rod carry $P_s$ and the copper rod carry $P_c$, with $P_s + P_c = 3500$ kgf. Equal extension:

$$\frac{P_sL}{AE_s} = \frac{P_cL}{AE_c} \Rightarrow \frac{P_s}{P_c} = \frac{E_s}{E_c} = \frac{2\times10^6}{1\times10^6} = 2$$ $$P_s = \frac{2}{3}(3500) = 2333.3\ \text{kgf}, \qquad P_c = \frac{1}{3}(3500) = 1166.7\ \text{kgf}$$

(With $A = \dfrac{\pi}{4}(2.5)^2 = 4.909$ cm², the extension of each rod is $\dfrac{2333.3\times300}{4.909\times2\times10^6} = 0.0713$ cm, the same for both rods.)

Position of the load

Let the load act at distance $x$ from the steel rod. Taking moments about the steel rod (rods 100 cm apart):

$$3500\,x = P_c(100) = 1166.7\times100 \Rightarrow x = 33.33\ \text{cm}$$

   steel            copper
    |<----- 100 cm ---->|
    |                    |
  ==|====================|==  cross piece
    |  ^                 |
    |<-x->  3.5 t load   |

Answer: the load should be placed 33.33 cm from the steel rod (66.67 cm from the copper rod).