A particle of mass 400g is moving under the action of two forces (F_{1} = (35N, 210°), F_{2} = (35sqrt{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.
Explanation
(a) (F_{1} = begin{pmatrix} 35 cos 210° \ 35 sin 210° end{pmatrix})
(F_{2} = begin{pmatrix} 35sqrt{3} cos 300° \ 35sqrt{3} sin 300° end{pmatrix})
(R = F_{1} + F_{2})
= (begin{pmatrix} -35 cos 30 \ -35 sin 30 end{pmatrix} + begin{pmatrix} 35sqrt{3} cos 60 \ -35sqrt{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.