Download Preview - ZigZag Education

VECTORS
IN MECHANICS
Magnitude and Direction
If a vector, a, is given in component form e.g. a = 4i + 3j, we
can use Pythagoras’ theorem to find the magnitude, and basic
trigonometry to find the direction.
Note, when printed, a
vector is written in bold
print. When handwritten,
vectors are underlined: a
~
a
3
θ
4
The magnitude of a is written as either a or just: a
a or a
It would be handwritten as: ~
Here we have: a =
42  32 = 5
The direction to the horizontal is θ, where tan θ =
3 
θ = 36.9º
4
Example 1: Given the vectors a = 3i + j, and b = 6i – 7j, find the
magnitude and direction of i) 2a + b ii) a – b. Give
the directions as three figure bearings.
p 
Note, the vector pi + qj can be written as a column vector 
q 






 
3
 3
 6
 6   12 



2a + b = 2   +   =  
i) a =  b = 
1
1
 
 7 
 7   5 






2a + b = 122  52 = 13
tan θ = 5
12
12
θ
 θ = 22.6º
The bearing is 90 + 22.6 = 113º ( to the nearest degree).
5
ii)






3
a= 
1
 6 
b = 
 7 
 3 
 a–b= 
8
8
j
θ
3
a–b
=
tan θ = 8
3
i
32  82 = 8.54 (3 s.f.)
 θ = 69.4º
The bearing is 270 + 69.4 = 339º (to the nearest degree).
Finding the components of a vector:
25
Consider a vector of magnitude 25
which makes an angle of 40º to the
horizontal.
j
40º
i
adjacent
The horizontal component can be found: cos 40 =
25
 adjacent = 25 cos 40
opposite
sin 40 =
The vertical component can be found:
25
 opposite = 25 sin 40


19.2 
25 cos 40


Hence the vector is:
 (or 19.2i + 16.1j).
 = 
16.1
25 sin 40 








Example 2: A vector of magnitude 17 is shown. Find the vector in
component form.
j
i
23º
23º
17
17
adjacent
cos 23 =
17
 adjacent = 17 cos 23
opposite
sin 23 =
17
 opposite = 17 sin 23
Note the negative
for both of the
components.
 17 cos 23   15.6 
Hence the vector is:
=


17
sin
23

6.64

 

Example 3: A vector of magnitude 35 is shown. Find the vector in
component form.
35
j
51º
i
51º
35
opposite
sin 51 =
35
 opposite = 35 sin 51
adjacent
cos 51 =
35
 adjacent = 35 cos 51
 35 sin 51  27.2 
Hence the vector is:
= 

35
cos
51

  22.0 
Summary of key points:
If a vector is given in component form, Pythagoras’ theorem can
be used to find the magnitude, and basic trigonometry to find the
direction.
The component of a vector opposite
a given angle is found using sin.
i.e. h = R sin θ
R
h
The component of a vector adjacent
to a given angle is found using cos.
i.e. k = R cos θ
θ
k
This PowerPoint produced by R.Collins; ©ZigZag Education 2010