From the global equilibrium:

Problem $7-52$

$7\mathrm{.}52*$ Determine the forces in members $CD,CE,$ and $EF$ of the truss shown in Fig. P$7-52.$

${\displaystyle \underset{?}{\overset{?}{}}}{F}_x=A_x+2=0$

:$.A_x=-2,$:$.A_x=2[kN]+$

Then,

$M_y=-A_y(18)-2(9)+2(15)+2(12)+2(9)+2(6)+2(3)=0$

:$.A_y=4[kN]$

$M_A=B_y(18)-2(9)-2(15)-2(12)-2(9)-2(6)-2(3)=0$

:$.B_y=6[kN]$

Then,

or you can also use:

${\displaystyle \underset{?}{\overset{?}{}}}{F}_y=A_y+B_y-2-2-2-2-2=0$

:$.B_y=-A_y+10=6[kN]$

Now, let's think about the right section only:

$M_C=-F_{EF}(3)-2(3)-2(6)-2(9)+B_y(12)=0$

:$.F_{EF}=\frac{-2(3)-2(6)-2(9)+B_y(12)}{3}=\frac{-2(3)-2(6)-2(9)+6(12)}{3}=12$

:$.F_{EP}=12[kN](T)$

$M_E=F_{CD}(3)-2(3)-2(6)+B_y(9)=0$

:$.F_{CD}=\frac{2(3)+2(6)-B_y(9)}{3}=-12,$:$.F_{CD}=12[kN](C)$

${\displaystyle \underset{F_x=0}{\overset{?}{}}}$

$-F_{EF}+F_{CD}-F_{CE}cos45=0$

:$.F_{CE}=\frac{-F_{EP}+F_{CD}}{cos45}=\frac{-12+12}{cos45}=0,$:$.F_{CE}=0[kN]$

The results are:

$F_{BF}=12[kN](T)$

$F_{CD}=12[kN](C)$

$F_{CE}=0[kN]$

Bonus Credit $(10$ extra points$)$ HomeworkIII

Homework #$16$

Solve the following problem by the method of sections.

$7-58$ Determine the forces in members $EF,EJ,$ and $JK$ of the truss shown in Fig. P$7-58.$

Please solve number $16$ homework and try to use simalar to exact format as proffesor and if possible please do on paper or neatly on technology and if thier are any signs of chat gbt i will have to report it

Thank! Try your best!!