Forces \(F_{1} (18N, 330°), F_{2} (10N, 090°)\) and \(F_{3} (25N, 180°)\) act on a body...
Forces \(F_{1} (18N, 330°), F_{2} (10N, 090°)\) and \(F_{3} (25N, 180°)\) act on a body at rest. Find, correct to one decimal place, the magnitude and direction of the resultant force.
Explanation
\(F_{1} = (18N, 330°) ; F_{2} = (10N, 090°) ; F_{3} = (25N, 180°)\)
\(R = F_{1} + F_{2} + F_{3}\)
= \(\begin{pmatrix} 18 \cos 330 \\ 18 \sin 330 \end{pmatrix} + \begin{pmatrix} 10 \cos 90 \\ 10 \sin 90 \end{pmatrix} + \begin{pmatrix} 25 \cos 180 \\ 25 \sin 180 \end{pmatrix}\)
= \(\begin{pmatrix} 18 \cos 30 \\ -18 \sin 30 \end{pmatrix} + \begin{pmatrix} 10 \times 0 \\ 10 \times 1 \end{pmatrix} + \begin{pmatrix} -25 \\ 0 \end{pmatrix}\)
= \(\begin{pmatrix} 18 \times 0.866 \\ -18 \times 0.5 \end{pmatrix} + \begin{pmatrix} 0 \\ 10 \end{pmatrix} + \begin{pmatrix} -25 \\ 0 \end{pmatrix}\)
= \(\begin{pmatrix} 15.588 - 25 \\ -9 + 10 \end{pmatrix}\)
= \(\begin{pmatrix} -9.41 \\ 1 \end{pmatrix}\)
\(R = \sqrt{(-9.41)^{2} + 1^{2}}\)
= \(\sqrt{88.5481 + 1}\)
= \(\sqrt{89.5481}\)
\(R = 9.463N\)
Direction: Let \(\theta\) be the angle resultant makes with the x - axis.
\(\tan \theta = \frac{1}{-9.41} = -0.1062\)
\(\theta = \tan^{-1} (-0.1063) = -6.07°\)
The resultant has magnitude 9.463N and makes an angle of 6.07° with the negative x- axis.