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;
30NThe key is knowing that the resultant force is 30N at 0º.
Hence;
Resolving Horizontally;
30N = FCos + 60NCos(360-120)30N = FCos + 60N x -0.530N = FCos - 30N60N = FCos Eq 1
Resolving Vertically;
0 = FSin + 60NSin(360-120) 0 = FSin + 60NSin(240) 0 = FSin + 60N x 3/2 0 = FSin -51.96N51.96N = FSin Eq 2
So by dividing Eq 2 by Eq 1
FSin / FCos = 51.95N / 60Ntan = 0.866 = 40.89
F = 60N / cos = 79.4N
Or
F = 51.96N / sin = 78.96N
7 Alternative method using cosine and sine rule….
30NMake a triangle of vectors where we are trying to find length c and angle A;
120
c = F
b = 30N
a = 60N
A
Cosine Rule
c2 = a2 + b2 - 2abCosCc2 = 60N2 + 30N2 – 2x60Nx30N xCos120c2 = 4500N2 +1800N2
c2 = 6300Nc = 79.4Nc = F = 79.4N
Sine RuleUse 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
SinC
c
SinA
a
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.
opp
adj
hyp opp = hyp x sin
adj = hyp x cos
Wires example – Correct
Vertically1.20N = T1cos30+ T2cos601.20N = 0.866T1+ 0.5T2 Eq 1
HorizontallyT1sin30 = T2sin600.5T1= 0.866T2
T1= 1.732T2 Eq 2
Sub 2 into 1;1.20N = 0.866T1+ 0.5T2
1.20N = 0.866 x 1.732T2 +0.5T2
1.20N = 1.99T2
T2 = 0.60N
& T1= 1.732T2
T1= 1.732 x 0.6N
T1= 1.04N
So by turning it on its side we can use the same formulae (which is confusing but correct)
Left side right side
WRONG ANSWER – does not take this into account!!!
Vertically1.20N = T1sin30+ T2sin601.20N = 0.5T1+ 0.866T2 Eq 1
HorizontallyT1cos30 = T2cos600.866T1= 0.5T2
T1= 0.577T2 Eq 2
Sub 2 into 1;1.20N = 0.866T1+ 0.5T2
1.20N = 0.866 x 0.577T2 +0.5T2
1.20N = 1T2
T2 = 1.2N
& T1= 0.577T2
T1= 0.577x1.2N
T1= 0.69N
Can do by using (90- ) as angle;
Vertically1.20N = T1sin60+ T2sin301.20N = 0.866T1+ 0.5T2 Eq 1
HorizontallyT1cos60 = T2cos300.5T1= 0.866T2
T1= 1.732T2 Eq 2
Sub 2 into 1;1.20N = 0.866T1+ 0.5T2
1.20N = 0.866 x 1.732T2 +0.5T2
1.20N = 1.99T2
T2 = 0.60N
& T1= 1.732T2
T1= 1.732 x 0.6N
T1= 1.04N
If we take the outside angle instead of the inner angle we can do this and use the triangle;
(90-)
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