Find the direction of the resultant of the forces in the diagram.

Explanation

(R_{1} = begin{pmatrix} -20 cos 60° \ -20 sin 60° end{pmatrix} ; R_{2} = begin{pmatrix} -10 cos 60° \ 10 sin 60° end{pmatrix} )

(R_{3} = begin{pmatrix} 8sqrt{2} cos 45° \ 8sqrt{2} sin 45° end{pmatrix})

(R = R_{1} + R_{2} + R_{3})

(R = begin{pmatrix} -20 cos 60° \ -20 sin 60° end{pmatrix} + begin{pmatrix} -10 cos 60° \ 10 sin 60° end{pmatrix} + begin{pmatrix} 8sqrt{2} cos 45° \ 8sqrt{2} sin 45° end{pmatrix})

(R = begin{pmatrix} -20 times frac{1}{2} \ -20 times frac{sqrt{3}}{2} end{pmatrix} + begin{pmatrix} -10 times frac{1}{2} \ 10 times frac{sqrt{3}}{2} end{pmatrix} + begin{pmatrix} 8sqrt{2} times frac{1}{sqrt{2}} \ 8sqrt{2} times frac{1}{sqrt{2}} end{pmatrix})

= (begin{pmatrix} – 10 – 5 + 8 \ -10sqrt{3} + 5sqrt{3} + 8 end{pmatrix})

= (begin{pmatrix} -7 \ 8 – 5sqrt{3} end{pmatrix} = begin{pmatrix} -7 \ -0.66 end{pmatrix})

Let (theta) be the direction from the horizontal.

(tan theta = frac{-0.66}{-7})

(tan theta = 0.0943)

(theta = tan^{-1} (0.0943) = 5.39°)