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=\frac{-31\cos(122)}{\cos(\theta)}$$F2=\frac{4.30(9.8)-31\sin (122)}{\sin \theta}$

then we make the next equality

$\frac{-31\cos(122)}{\cos(\theta)}=\frac{4.30-31\sin(122)}{\sin\theta}$$\frac{\sin(\theta)}{\cos(\theta)}=\frac{4.30(9.8)-31\sin (122)}{-31\cos (122)}$$\tan (\theta)=\frac{(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

$F2=\frac{-31\cos (122)}{\cos (43.98)}=22.83N$

the magnitude of the second wire is 22.83N