Forces(5N, 030(^o)), (PN, 060(^o)), (QN, 150(^o)), (3N, 180(^o)) and (5N, 270(^o)) act on a body . If the system is in quilibrium, find, correct to one decimal place, the values of P and Q
Explanation
They were expected to resolve the forces vertically and horizontally to have; ((^{5 cos 60^o}_{5 sin 60^o})) + ((^{P cos 30^o}_{P sin 30^o})) + ((^{Q cos 60^o}_{-Q sin 60^o})) + ((^{3 cos 90^o}_{-3 sin 90^o})) + ((^{-5 cos 0^o}_{5 sin 0^o})) = ((^0_0))
which simplifies to;
((^{0.866P + 0.5Q}_{0.5P – 0.866Q})) = ((^{2.5}_{-1.33}))
This will yield two equations to be solved, the equations are 0.866P + 0.5Q = 2.5 and 0.5P – 0.866Q = -1.33
When these equations are solved, they will give P = 1.5 and Q = 2.4019