A particle is under the action of forces (P = (4N, 030°)) and (R = (10N, 300°)). Find the force that will keep the particle in equilibrium.
Explanation
(P = (4N, 030°) ; R = (10N, 300°))
Let (F = (a N, theta)) be the force that will keep the particle in equilibrium.
Then P + R + F = 0.
(begin{pmatrix} 4 cos 30 \ 4 sin 30 end{pmatrix} + begin{pmatrix} 10 cos 300 \ 10 sin 300 end{pmatrix} + begin{a cos theta \ a sin theta end{pmatrix} = begin{pmatrix} 0 \ 0 end{pmatrix}).
(begin{pmatrix} 4(frac{sqrt{3}}{2}) \ 4(frac{1}{2}) end{pmatrix} + begin{pmatrix} 10(frac{1}{2}) \ 10(-frac{sqrt{3}}{2}) end{pmatrix} + begin{pmatrix} a cos theta \ a sin theta end{pmatrix} = begin{pmatrix} 0 \ 0 end{pmatrix})
(begin{pmatrix} 2sqrt{3} + 5 \ 2 – 5sqrt{3} end{pmatrix} + begin{pmatrix} a cos theta \ a sin theta end{pmatrix} = begin{pmatrix} 0 \ 0 begin{pmatrix})
(implies 2sqrt{3} + 5 + a cos theta = 0 )
(2sqrt{3} + 5 = -a cos theta)
(2 – 5sqrt{3} = -a sin theta)
(tan theta = frac{2 – 5sqrt{3}}{2sqrt{3} + 5} = frac{-6.66}{8.486})
(tan theta = -0.7848 implies theta = -38.13°)
= (141.87°)
(-a cos theta = 8.486)
(-a cos (141.87) = -a (-0.7866) = 8.486)
(0.7866a = 8.486 implies a = frac{8.486}{0.7866} = 10.79N)
(therefore F = (10.79N, 141.87°))