Based on this stress, the allowable force magnitude for member $\left(1\right)$ is:

$$\begin{gathered} {\mathrm{F}}_{1,\mathrm{allow}}={\mathrm{\sigma }}_{\mathrm{allow}}{\mathrm{A}}_1 \\ =(25 \mathrm{MPa})(3,080 {\mathrm{mm}}^2) \\ =77,000 \mathrm{N}=77.0 \mathrm{kN} \end{gathered}$$

In the preceding scene, it was shown that the force in member $\left(1\right)$ could be expressed in terms of $\mathrm{w}$ as:

${\mathrm{F}}_1=(-2.2595 \mathrm{m})(\mathrm{w} \mathrm{kN/}\mathrm{m})$

Equate these two expressions and solve for $\mathrm{w}$.
Since only the magnitude of $\mathrm{w}$ is desired, consider the absolute value of ${\mathrm{F}}_1$.

$$\begin{gathered} \left|(-2.2595 \mathrm{m})(\mathrm{w} \mathrm{kN/}\mathrm{m})\right|=77.0 \mathrm{kN} \\ \therefore \mathrm{w}=34.08 \mathrm{kN/}\mathrm{m} \end{gathered}$$