Forces \(F_{1} = (10 N, 090°), F_{2} = (20 N, 210°)\) and \(F_{3} = (4 N, 330°)\) act on a body at rest on a smooth table. Find, correct to one decimal place, the magnitude of the resultant force.
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Correct Answer: Option n
Explanation:
\(F_{R} = F_{1} + F_{2} + F_{3}\)
\(F_{1} = (10 N, 090°) = \begin{pmatrix} 10 \cos 90 \\ 10 \sin 90 \end{pmatrix} = \begin{pmatrix} 0 \\ 10 \end{pmatrix}\)
\(F_{2} = (20 N, 210°) = \begin{pmatrix} 20 \cos 210 \\ 20 \sin 210 \end{pmatrix} = \begin{pmatrix} -10\sqrt{3} \\ - 10 \end{pmatrix}\)
\(F_{3} = (4 N, 330°) = \begin{pmatrix} 4 \cos 330 \\ 4 \sin 330 \end{pmatrix} = \begin{pmatrix} 2\sqrt{3} \\ -2 \end{pmatrix}\)
\(F_{R} = \begin{pmatrix} 0 \\ 10 \end{pmatrix} + \begin{pmatrix} -10\sqrt{3} \\ -10 \end{pmatrix} + \begin{pmatrix} 2\sqrt{3} \\ -2 \end{pmatrix}\)
= \(\begin{pmatrix} -8\sqrt{3} \\ -2 \end{pmatrix}\)
\(|F_{R}| = \sqrt{(8\sqrt{3})^{2} + (-2)^{2}}\)
= \(\sqrt{192 + 4} \)
= \(\sqrt{196} = 14 N\)
\(F_{R} = F_{1} + F_{2} + F_{3}\)
\(F_{1} = (10 N, 090°) = \begin{pmatrix} 10 \cos 90 \\ 10 \sin 90 \end{pmatrix} = \begin{pmatrix} 0 \\ 10 \end{pmatrix}\)
\(F_{2} = (20 N, 210°) = \begin{pmatrix} 20 \cos 210 \\ 20 \sin 210 \end{pmatrix} = \begin{pmatrix} -10\sqrt{3} \\ - 10 \end{pmatrix}\)
\(F_{3} = (4 N, 330°) = \begin{pmatrix} 4 \cos 330 \\ 4 \sin 330 \end{pmatrix} = \begin{pmatrix} 2\sqrt{3} \\ -2 \end{pmatrix}\)
\(F_{R} = \begin{pmatrix} 0 \\ 10 \end{pmatrix} + \begin{pmatrix} -10\sqrt{3} \\ -10 \end{pmatrix} + \begin{pmatrix} 2\sqrt{3} \\ -2 \end{pmatrix}\)
= \(\begin{pmatrix} -8\sqrt{3} \\ -2 \end{pmatrix}\)
\(|F_{R}| = \sqrt{(8\sqrt{3})^{2} + (-2)^{2}}\)
= \(\sqrt{192 + 4} \)
= \(\sqrt{196} = 14 N\)