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
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Correct Answer: Option
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
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