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Vector Analysis. Physics – 11.1.1 Vectors. 5a. N. 5b. N. N. 6. 7. If this problem is to be solved mathematically we must establish two formulae. One in the vertical and one in the horizontal;. The key is knowing that the resultant force is 30N at 0 º. Hence; Resolving Horizontally;
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Vector Analysis Physics – 11.1.1 Vectors
5a N
5b N N
7 If this problem is to be solved mathematically we must establish two formulae. One in the vertical and one in the horizontal; The key is knowing that the resultant force is 30N at 0º. Hence; Resolving Horizontally; 30N = FCos + 60NCos(360-120) 30N = FCos + 60N x -0.5 30N = FCos - 30N 60N = FCos Eq 1 Resolving Vertically; 0 = FSin + 60NSin(360-120) 0 = FSin + 60NSin(240) 0 = FSin + 60N x 3/2 0 = FSin -51.96N 51.96N = FSin Eq 2 30N So by dividing Eq 2 by Eq 1 FSin / FCos = 51.95N/ 60N tan = 0.866 = 40.89 F = 60N / cos = 79.4N Or F = 51.96N / sin = 78.96N
7 Alternative method using cosine and sine rule…. Make a triangle of vectors where we are trying to find length c and angle A; 30N b = 30N 120 A a = 60N c = F Sine Rule Use reciprocal version of above SinC / c = SinA / a (60N x Sin 120) / 79.4N) = SinA Sin-1 (60N x Sin 120) / 79.4N) =A A = 40.876 = 41 Cosine Rule c2 = a2 + b2 - 2abCosC c2 = 60N2 + 30N2 – 2x60Nx30N xCos120 c2 = 4500N2 +1800N2 c2 = 6300N c = 79.4N c = F = 79.4N
hyp opp = hyp x sin adj = hyp x cos opp adj Reasoning! When using a resolving triangle to solve this type of problem. We have to turn it on its side to make it fit the problem; This means that the formulae for resolving V & H are reversed for this situation as we are working out and not (90-) So reverse angles.
Wires example – Correct So by turning it on its side we can use the same formulae (which is confusing but correct) right side Left side
Can do by using (90- ) as angle; If we take the outside angle instead of the inner angle we can do this and use the triangle; (90-)