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The crate has a weight of  \(2.75 \mathrm{kN}\) . Determine the  force in each supporting cable.

\(\stackrel{+}{\rightarrow} \sum F_{x}=0 \)

\( F_{c}\left(\frac{4}{5}\right)-F_{B} \cos 30=0 \leftarrow \)

\( +\uparrow \sum F_{y}=0 \)

\( F_{B} \sin 30+F_{c}\left(\frac{3}{5}\right)-2.75=0 \)

\(0.5 \mathrm F_{B}+\left(1.08 \mathrm{F}_{B}\right)(0.6)=2.75\longleftarrow(2) \)

\( ∴ F_{B}=2.395 \mathrm{kN} \)

\(∴ F_{c}=2.59 \mathrm{kN} \)

If the mass of the girder is  \(3 \mathrm{Mg}\) and its center of mass is located at point  \(G,\) determine the tension developed incables  \(A B,\) \(B C,\) and  \(B D\)  for equilibrium. 

\( \uparrow \mathrm{F}_{\mathrm{AB}} \)

\( \downarrow T_{A B} \)

\( +\uparrow \sum F_{y}=0 \rightarrow F_{A B}-T_{A B}=0 \)

\(∴ T_{A B}=F_{A B}=W \)

\( w=\left(3 * 10^{3}\right)*9 \cdot 81=29.43 \mathrm{kN} \)

\( T_{A B}=29.43 \mathrm{kN} \)

\(\stackrel{+}{\rightarrow} \sum F_{x}=0 \)

\(T_{BD} sin 30 - T_{BC} sin 45 = 0\)

\( T_{BD} =1.414 \ T_{BC} \)

\(+\uparrow  \sum F_{y}=0\longrightarrow T_{AB}-T_{BC} cos 45-T_{BD} cos 30=0\)

\(29.43-0.707 \ T_{BC}-(1.414 \ T_{BC}) 0.866=0\)

\(∴ T_{BC}=15.24KN \longrightarrow T_{BD}=1.414 (15.24)=21.54KN\)

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