A particle of mass 400g is moving under the action of two forces \(F_{1} = (35N, 210°), F_{2} = (35\sqrt{3} N, 300°)\) and a resistance of 40N. Find the magnitude of the
(a) resultant of \(F_{1}\) and \(F_{2}\).
(b) resultant force acting on the particle.
(a) resultant of \(F_{1}\) and \(F_{2}\).
(b) resultant force acting on the particle.
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Correct Answer: Option n
Explanation:
(a) \(F_{1} = \begin{pmatrix} 35 \cos 210° \\ 35 \sin 210° \end{pmatrix}\)
\(F_{2} = \begin{pmatrix} 35\sqrt{3} \cos 300° \\ 35\sqrt{3} \sin 300° \end{pmatrix}\)
\(R = F_{1} + F_{2}\)
= \(\begin{pmatrix} -35 \cos 30 \\ -35 \sin 30 \end{pmatrix} + \begin{pmatrix} 35\sqrt{3} \cos 60 \\ -35\sqrt{3} \sin 60 \end{pmatrix}\)
= \(\begin{pmatrix} -30.31 \\ -17.5 \end{pmatrix} + \begin{pmatrix} 30.31 \\ -52.5 \end{pmatrix}\)
= \(\begin{pmatrix} 0 \\ -70 \end{pmatrix}\)
Magnitude of resultant of \(F_{1}\) and \(F_{2}\) = 70N.
(b) Resultant acting on the particle = resultant of \(F_{1}\) and \(F_{2}\) less resistance.
= 70N - 40N = 30N.
(a) \(F_{1} = \begin{pmatrix} 35 \cos 210° \\ 35 \sin 210° \end{pmatrix}\)
\(F_{2} = \begin{pmatrix} 35\sqrt{3} \cos 300° \\ 35\sqrt{3} \sin 300° \end{pmatrix}\)
\(R = F_{1} + F_{2}\)
= \(\begin{pmatrix} -35 \cos 30 \\ -35 \sin 30 \end{pmatrix} + \begin{pmatrix} 35\sqrt{3} \cos 60 \\ -35\sqrt{3} \sin 60 \end{pmatrix}\)
= \(\begin{pmatrix} -30.31 \\ -17.5 \end{pmatrix} + \begin{pmatrix} 30.31 \\ -52.5 \end{pmatrix}\)
= \(\begin{pmatrix} 0 \\ -70 \end{pmatrix}\)
Magnitude of resultant of \(F_{1}\) and \(F_{2}\) = 70N.
(b) Resultant acting on the particle = resultant of \(F_{1}\) and \(F_{2}\) less resistance.
= 70N - 40N = 30N.