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If cable \(C B\) is subjected to a tension that is twice that of cable  \(C A,\) determine the angle \(\theta\) for equilibrium of the 10-kg cylinder. Also, what are the tensions in wires \(C A\) and \(C B\) ?

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

\(2Tcos \theta-T cos 30=0\)

\( 2 T cos \theta=T cos 30 \)

\( \cos \theta=0.433 \)

\(∴ \theta=\cos ^{-1} 0.433=64.3^{\circ} \)

\(+\uparrow\sum F_{y}=0 \quad \longrightarrow 2 T sin \theta+T sin 30- W=0 \)

\(2.3T=98.1\longrightarrow T=\frac{98.1}{2.3}=42.6N\)

\(F_{AC}=T=42.6N\)

\( F_{C B}=2 T=2 * 42.6=85.2 N \)

The spring has a stiffness of  \(k=800 \mathrm{N} / \mathrm{m}\) and an unstretched length of  \(200 \mathrm{mm}\) . Determine the foree in cables \(B C\)  and  \(B D\) when the spring is held in the position shown.

\( s=l-l_{0}=0.5-0.2=0.3 \mathrm{m} \)

\( F_{s p}=k s=800 * 0.3=240 \mathrm{N} \)

\(\stackrel{+}{\rightarrow}\sum F_{x}=0\longrightarrow F_{BD}\frac{4}{5}+F_{BC} cos 45=0\)

\(0.7071\ F_{BC}+0.8\ F_{BD}=240\longrightarrow \) 

\( +\uparrow\sum F_{y}=0 \rightarrow F_{BC} sin 45-F_{BD}\frac{3}{5}=0 \)

\( F_{B C}=0.8485 \ F_{BD} \longrightarrow \)

\(∴ \quad F_{B D}=171 \quad N \)

\(∴ \quad F_{B C}=145 \quad N \)

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