Two traffic signals are temporarily suspended from a cable as shown. Knowing that the signal at B weighs 400 N, determine the weight of the signal at C.
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Two traffic signals are temporarily suspended from a cable as shown. Knowing that the signal at B weighs 400 N, determine the weight of the signal at C.
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Answer:
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Explanation:
Answer:
The weight of the signal at C is [tex]223.299 \mathrm{~N}[/tex]
Explanation:
Revolving force at B
Step-1
[tex]\sin \theta=\frac{1.5}{\sqrt{(1.5)^{2}+(3.6)^{2}}}=6.712 \times 10^{-3}[/tex]
[tex]\cos \theta=\frac{3.6}{\sqrt{(1.5)^{2}+(-6)^{2}}}=0.923[/tex]
[tex]\begin{aligned}&\sin \phi=\frac{0.4}{\sqrt{(0.4)^{2}+(3.4)^{2}}}=2.0392 \times 10^{-3} \\&\cos p=\frac{3.4}{\sqrt{(0.4)^{2}+(3.4)^{2}}}=0.993\end{aligned}[/tex]
Step-2
[tex]\begin{aligned}&T_{1} \cos \theta=T_{2} \cos \phi&T_{1} \sin \theta+T_{2} \sin \phi=m g\end{aligned}[/tex]
[tex]T_{1}=1.0758 T_{2}[/tex]
[tex]T_{2}=32.4 \times 10^{3}[/tex]
Step-3
[tex]T_{2} \sin \phi+T_{3} \sin \alpha=W[/tex]
[tex]32.4 \times 10^{3} 0.993=T_{3}(0.9998)[/tex]
Substitute the value
[tex]W=223.299 \mathrm{~N}[/tex]