Question

ExamA 4.30 kg sign hangs from two wires. Theleft wire exerts a 31.0 N force at 122°.What is the magnitude of the force exertedby the second wire?* F2wmagnitude (N)Enter

Answer 1

We have

Forces on x-axis

`\begin{gathered} F1\cos (122)+F2\cos (\theta)=0 \\ 31\cos (122)+F2\cos (\theta)=0 \end{gathered}`

Forces on y-axis

`\begin{gathered} F1\sin (122)+F2\sin (\theta)-W=0 \\ 31\sin (122)+F2\sin (\theta)=4.30(9.8) \end{gathered}`

then

`F2=(-31\cos(122))/(\cos(\theta))`
`F2=(4.30(9.8)-31\sin (122))/(\sin \theta)`

then we make the next equality

`(-31\cos(122))/(\cos(\theta))=(4.30-31\sin(122))/(\sin\theta)`
`(\sin(\theta))/(\cos(\theta))=(4.30(9.8)-31\sin (122))/(-31\cos (122))`
`\tan (\theta)=((4.30\cdot9.8)-31\sin (122))/(-31\cos (122))`
`\begin{gathered} \tan (\theta)=0.96 \\ \theta=\tan (0.96) \end{gathered}`
`\theta=43.98\text{ \degree}`

then

`F2=(-31\cos (122))/(\cos (43.98))=22.83N`

the magnitude of the second wire is 22.83N