Question

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 1

Answer:

Hi dear

I don't know the answer

Explanation:

Answer 2

Answer:

The weight of the signal at C is $223.299 \mathrm{~N}$

Explanation:

Revolving force  at B

Step-1

$\sin \theta=\frac{1.5}{\sqrt{(1.5)^{2}+(3.6)^{2}}}=6.712 \times 10^{-3}$

$\cos \theta=\frac{3.6}{\sqrt{(1.5)^{2}+(-6)^{2}}}=0.923$

$\begin{aligned}&\sin \phi=\frac{0.4}{\sqrt{(0.4)^{2}+(3.4)^{2}}}=2.0392 \times 10^{-3} \\&\cos p=\frac{3.4}{\sqrt{(0.4)^{2}+(3.4)^{2}}}=0.993\end{aligned}$

Step-2

$\begin{aligned}&T_{1} \cos \theta=T_{2} \cos \phi&T_{1} \sin \theta+T_{2} \sin \phi=m g\end{aligned}$

$T_{1}=1.0758 T_{2}$

$T_{2}=32.4 \times 10^{3}$

Step-3

$T_{2} \sin \phi+T_{3} \sin \alpha=W$

$32.4 \times 10^{3} 0.993=T_{3}(0.9998)$

Substitute the value

$W=223.299 \mathrm{~N}$