a. Determine the force in member CJ, in kN.

b. Determine th force in member EJ in kN.

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(2) 10 \mathrm{kN}aut section (1) -(1)\begin{array}{l}G=\tan ^{-1}(1 / 1) \\Q=45^{\circ}\end{array}Hence CJ is prependicular to A C\sum M_{A}=0A C=\sqrt{4^{2}+y^{2}}=4 \sqrt{2}\begin{array}{r}F_{00} \times 0+F_{k 5} \times 0+10 \times 4+10 \times 8+F_{\rho 5} \times 4 \sqrt{2}=0 \\F_{C S}=-21.213 \mathrm{kN} \\F_{\mathrm{CJ}}=21.213 \mathrm{kN} \text { compression }\end{array}Now Cut section (2) (2)By EO ES IS\begin{array}{c}\sum M_{G}=0 \quad 10 \times 4+F_{25} \times 0+E_{E D} \times 0+F_{E S} \times 4 \sqrt{2}=0 \\ F_{E S}=-7.07 \mathrm{kH} \\ F_{E S}=7.07 \mathrm{kN} \text { compression }\end{array} ...