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If \(F_{1}=600 \mathrm{N}\) and \(\phi=30^{\circ}\), determine themagnitude of the resultant force acting on the eyebolt and its direction measured clockwise from the positive \(x\) axis. 

x- Direction 9-

\( F_{1} \cos \phi+F_{2} \cos 60-F_{3}\left(\frac{3}{5}\right) \)

\( = 600 \cos 30+500 \cos 60-450\left(\frac{3}{5}\right)=(499.6 \mathrm{N}) \)

 

y- Direction 9-

\( F_{1} \sin \phi-F_{2} \sin 60-F_{3}\left(\frac{4}{5}\right)= \)

\( =600 \sin 30-500 \sin 60-450\left(\frac{4}{5}\right)=-493 N=(493 N) \)

\( ∴ F_{R}=\sqrt{(499.6)^{2}+(493)^{2}}=702 N \)

\( ∴ \tan \theta=\frac{493}{499 \cdot 6} \rightarrow \theta=44 \cdot 6^{\circ} \)

Express each of the three forces acting on the bracket in Cartesian vector form with respect to the  \(x\) and \(y\) axes. Determine the magnitude and direction \(\theta\) of \(\mathbf{F}_{1}\) so thatthe resultant force is directed along the positive \(x^{\prime}\) axis and has a magnitude of \(F_{R}=600 \mathrm{N}\).

 

\( F_{1}=\left(\underbrace{F_{1} \cos \theta}_{x} i+\underbrace{F_{1} \sin \theta}_{y} j\right) N \)

\( F_{2}=350 i N \)   \( (350 i+0 j) N \)

\( F_{3}=\left(\begin{array}{lll}{0 i} & {-} & {100 j}\end{array}\right) N \)

\( +\rightarrow \) \( \sum F_{x}=F_{R} \cos 30 \longrightarrow \)\( F_{1} \cos \theta \pm 350=600 \cos 30 \)

                                            \( \longrightarrow F_{1} \cos \theta=169.6 \longrightarrow (1) \)

\( +\uparrow   \sum F_{y}=F_{R} \sin 30 \longrightarrow   F_{1} \sin \theta=100=600 \sin 30 \)

                                           \( \rightarrow F_{1} \sin \theta=400 \longrightarrow (2) \)

\( ∴ \tan \theta = \frac{400}{169 \cdot 6}=\theta=67^{\circ} \rightarrow \theta=670 \)

\( ∴ F_{1}=434 N \)

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