Answer
$$
=(+797 \mathrm{~N}) \hat{\mathrm{i}}+(265 \mathrm{~N}) \hat{\mathrm{j}}
$$
Work Step by Step
(b) Equilibrium of horizontal and vertical forces on beam $A$ readily yields
$$
G_x=-F_x=797 \mathrm{~N}, \quad G_y=m_A g-F_y=265 \mathrm{~N} .
$$
In unit-vector notation, we have
$$
\vec{G}=G_x \hat{\mathrm{i}}+G_y \hat{\mathrm{j}}=(+797 \mathrm{~N}) \hat{\mathrm{i}}+(265 \mathrm{~N}) \hat{\mathrm{j}} .
$$
