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Determine the tension developed in cables  \(A B\) , \(A C\), and  \(A D\) .

\( +\swarrow \sum F_{x}=0 \)

\( F_{A B}+F_{A D} \cos 120-F_{A C} \cos 60 \sin 30=0\)  

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

\( -F_{A D} \cos 60+F_{A C} \cos 60 \cos 30=0 \)        

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

\( F_{A D} \cos 45+F_{A C} \cos 60=300 \)               

\( ∴ F_{A B}=139 N \)

\(∴ F_{A D}=176 N \)

\(∴ F_{A C}=203 N \)

The three cables are used to support the  \(800-\mathrm{N}\) lamp. Determine the force developed in each cable for equilibrium.

\( \vec{r}_{A D}=\vec{r}_{D} \vec{r}_{A} \)

\( =(4 k)-(2 i+4 j) \)

\( =-2 \hat i-4 \hat  j+4 k \)

\( \left|r_{A D}\right|=6 m \)

\(\vec{U}_{AD}=|\frac{\vec{r}_{AD}}{r_{AD}}|=-\frac{1}{3}\hat i-\frac{2}{3}j+\frac{2}{3}k \)
\( \vec{F_{D}}=-\frac{1}{3} F_{D} \hat{\imath}- \frac{2}{3} F_{D} \hat j+\frac{2}{3} F_{D} k \)

\( \vec{F_{B}}=0 \hat{\imath}+\underline{F_{B}}\hat j+0 k \)

\( \vec{F_{c}}=F_{c} \hat i+0 j+0 k \)

\( \vec{F_{w}}=0 i+0 j-800 k \)

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

\( \frac{2}{3} F_{D}-800=0 \rightarrow F_{D}=1200 N \)

\( +\swarrow \sum F_{x}=0 \)

\( -\frac{1}{3} F_{D}+F_{c}=0 \rightarrow F_{c}=400 N \)

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

\(-\frac{2}{3}(F_{D})+F_{B}=0\rightarrow F_{B}=800 N\)

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