Position of 2t force for zero net elongation

Let the bar have uniform area $A$ and modulus $E$. The bar is fixed at the left end, and the loads are (positions from the fixed end): 1.5 t (→) at 1 m, 1.5 t (→) at 3 m, 1.5 t (←) at the free end (5 m). The 2 t force is applied at distance $x$ from the fixed end; to cancel a net shortening it must act away from the wall (→).

Force in each portion (tension +ve) without the 2 t force

Take the sum of the forces to the right of the section:

$$\delta_{without} = \frac{1}{AE}\left[1.5(1) + 0(2) + (-1.5)(2)\right] = \frac{-1.5}{AE}\ \text{(t·m)}$$

The bar is net shortened, so a tensile 2 t force must lengthen it by $1.5/AE$.

Effect of the 2 t force at distance $x$

It adds $+2$ t of tension in every portion between the wall and $x$, so it adds $2x/AE$ to the length:

$$-1.5 + 2x = 0 \Rightarrow x = 0.75\ \text{m}$$

Since $x = 0.75 < 1$ m, the force lies in the first portion, as assumed.

Check: portions become 0 – 0.75 m: $+3.5$ t; 0.75 – 1 m: $+1.5$ t; 1 – 3 m: 0; 3 – 5 m: $-1.5$ t. Sum $= 3.5(0.75) + 1.5(0.25) + 0 - 1.5(2) = 2.625 + 0.375 - 3 = 0$.

Answer: $x = 0.75$ m from the fixed end, with the 2 t force acting away from the fixed end.