Chapter 4 - Force System Resultants - Section 4.8 - Further Simplification of a Force and Couple System - Problems - Page 188: 129

$$ \begin{aligned} & F_C=600 \mathrm{~N} \\ & F_D=500 \mathrm{~N} \end{aligned} $$

$$ \begin{gathered} \Sigma M_x=0 ; \quad F_D(0.4)+600(0.4)-F_C(0.4)-500(0.4)=0 \\ F_C-F_D=100 &(1)\\ \Sigma M_y=0 ; \quad 500(0.2)+600(0.2)-F_C(0.2)-F_D(0.2)=0 \\ F_C+F_D=1100 &(2) \end{gathered} $$ Solving Eqs. (1) and (2) yields: $$ F_C=600 \mathrm{~N} \quad F_D=500 \mathrm{~N} $$