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.
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 \(F_{R}=600 \mathrm{N}\).
\( F_{1}=\left(\underbrace{F_{1} \cos \theta}_{x} i+\underbrace{F_{1} \sin \theta}_{y} j\right) N \)
\( F_{2}=350 i N \)\( (350 i+0 j) N \)
\( F_{3}=\left(\begin{array}{lll}{0 i} & {-} & {100 j}\end{array}\right) N \)
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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