The frame shown supports part of the roof of a small building. Knowing that the tension in the cable is 150 kN, determine the reaction at the fixed end E.
Free-Body Diagram. A free-body diagram of the frame and of the cable BDF is drawn. The reaction at the fixed end E is represented by the force components {E}_{x}\operatorname{and}\ {E}_{y} and the couple {M}_{E}. The other forces acting on the free body are the four 20-kN loads and the 150-kN force exerted at end F of the cable.
Equilibrium Equations. Noting that DF~=~2\overline{(4.5~m)^2~ + ~(6~m)^2} ~=~7.5~m, we write
\stackrel{+}{\to}\Sigma F_{x}\;=\;0\mathrm{:}\qquad\qquad\qquad E_{x}+{\frac{4.5}{7.5}}(150\,\mathrm{kN})\,=\,0 \\ \\ \qquad\qquad\qquad\qquad\qquad\qquad E_{x}=-90.0\,\mathrm{kN}\qquad\qquad\mathrm{E}_{x}=90.0\mathrm{kN}\,Z+\uparrow\Sigma F_{y}\,=\,0\mathrm{:}\qquad\qquad\qquad E_{y}-4(20\,\mathrm{kN})\,-\,\frac6{7.5}(150\,\mathrm{kN})\,=\,0 \\ \\ \qquad\qquad\qquad\qquad\qquad\qquad E_{y}=\ +200\mathrm{~kN}\qquad\qquad\ \ \ E_{y}=\ 200\mathrm{~kN}~X
+{\mathrm l{{\Sigma}}M_{E}=0}\mathrm{:}\qquad\qquad\qquad (20\mathrm{~kN})(7.2\mathrm{~m})\,+\,(20\mathrm{~kN})(5.4\mathrm{~m})\,+\,(20\mathrm{~kN})(3.6\mathrm{~m}) \\ \\ \qquad\qquad\qquad +~(20~\mathrm{kN})(1.8~\mathrm{m})-\frac{6}{7.5}(150~\mathrm{kN})(4.5~\mathrm{m})\,+\,M_{E}\,=\,0 \\ \\ \qquad\qquad\qquad\qquad\qquad\qquad M_{E}=+180.0\,\,\mathrm{kN}\cdot\,\mathrm{m}\qquad\mathrm{M}_{E}=\,180.0\,\,\mathrm{kN}\cdot\,\mathrm{m}\,\,1