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-)