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Determine the tension developed in cables \(A B\) , \(A C\), and \(A D\) .
\( +\swarrow \sum F_{x}=0 \)
\( F_{A B}+F_{A D} \cos 120-F_{A C} \cos 60 \sin 30=0\) ①
\( \stackrel{+}\rightarrow \sum F_{ y}=0 \)
\( -F_{A D} \cos 60+F_{A C} \cos 60 \cos 30=0 \) ②
\( +\uparrow \sum F_{z}=0 \)
\( F_{A D} \cos 45+F_{A C} \cos 60=300 \) ③
\( ∴ F_{A B}=139 N \)
\(∴ F_{A D}=176 N \)
\(∴ F_{A C}=203 N \)
The three cables are used to support the \(800-\mathrm{N}\) lamp. Determine the force developed in each cable for equilibrium.
\( \vec{r}_{A D}=\vec{r}_{D} \vec{r}_{A} \)
\( =(4 k)-(2 i+4 j) \)
\( =-2 \hat i-4 \hat j+4 k \)
\( \left|r_{A D}\right|=6 m \)
\(\vec{U}_{AD}=|\frac{\vec{r}_{AD}}{r_{AD}}|=-\frac{1}{3}\hat i-\frac{2}{3}j+\frac{2}{3}k \) \( \vec{F_{D}}=-\frac{1}{3} F_{D} \hat{\imath}- \frac{2}{3} F_{D} \hat j+\frac{2}{3} F_{D} k \)
\( \vec{F_{B}}=0 \hat{\imath}+\underline{F_{B}}\hat j+0 k \)
\( \vec{F_{c}}=F_{c} \hat i+0 j+0 k \)
\( \vec{F_{w}}=0 i+0 j-800 k \)
\( \frac{2}{3} F_{D}-800=0 \rightarrow F_{D}=1200 N \)
\( -\frac{1}{3} F_{D}+F_{c}=0 \rightarrow F_{c}=400 N \)
\(-\frac{2}{3}(F_{D})+F_{B}=0\rightarrow F_{B}=800 N\)
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Determine the tension developed in cables \(A B\) , \(A C\), and \(A D\) .
\( +\swarrow \sum F_{x}=0 \)
\( F_{A B}+F_{A D} \cos 120-F_{A C} \cos 60 \sin 30=0\) ①
\( \stackrel{+}\rightarrow \sum F_{ y}=0 \)
\( -F_{A D} \cos 60+F_{A C} \cos 60 \cos 30=0 \) ②
\( +\uparrow \sum F_{z}=0 \)
\( F_{A D} \cos 45+F_{A C} \cos 60=300 \) ③
\( ∴ F_{A B}=139 N \)
\(∴ F_{A D}=176 N \)
\(∴ F_{A C}=203 N \)
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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