Sabah Dr. AL-Naimi
Qatar University
Physics Program
FALL 2009
PHYS191
1.1 Coordinate Systems
A coordinate system used to specify
locations in space consists of:
A fixed reference point O , called the origin.
A set of specified axes with appropriate scales
and labels on the axes.
Instructions that tell us how to label a point in
space relative to the origin and axes.
A common coordinate system is called the
Cartesian Coordinate system, Sometimes
called the Rectangular Coordinate system,
This system labeled with the coordinate
(X,Y).
The Cartesian Coordinate can be expressed
through equations relating the coordinates
(X,Y) to the Plane polar coordinates system
(r, ). As follows:
r
x2 y2
1
1.2 Some properties of vectors
A scalar quantity has only magnitude and no
direction.
A vector quantity has both magnitude and
direction.
Addition & Subtraction:
Dr. Sabah AL-Naimi
Qatar University
Physics Program
FALL 2009
PHYS191
A B B A
A ( B C ) ( A B) C
A B A ( B)
A B B A
Multiplication of a vector by a scalar:
mA nB nB mA
n( A B) nA nB
k ( A B) kA kB
Where n , m & k are scalar quantities.
1.3 Examples
Example (1):
Find the sum of the two vectors A and B lying in x - y plane and given
by:
A 3i 5 j & B 2i 3 j
Solution:
A B (3i 5 j ) (2i 3 j ) 3i 2i 5 j 3 j
5i 2 j
2
Dr. Sabah AL-Naimi
Qatar University
Physics Program
FALL 2009
PHYS191
Example (2):
Find the sum of three vectors A , B and C lying in x - y plane and given
by:
A 2i 2 j ,
B 20i 6 j , C i j
Solution:
A B C ( 2i 2 j ) ( 20i 6 j ) (i j )
2i 20i i 2 j 6 j j
23i 7 j
Example (3):
Find the following:
a ) A 3B 5C
,
b) 2 A 0.5 B 0C
where A 3i j k , B 2i 4 j 3k , C i j 2k
Solution:
a)
A 3B 5C (3i j k ) 3(2i 4 j 3k ) 5(i j 2k )
3i j k 6i 12 j 9k 5i 5 j 10k
(3i 6i 5i ) j 12 j 5 j k 9k 10k
14i 8 j 18k
b) 2 A 0.5B 0C 2(3i j k ) 0.5(2i 4 j 3k ) 0(i j 2k )
6i 2 j 2k i 2 j 1.5k 0
6i i 2 j 2 j 2k 1.5k
7i 4 j 0.5k
3
Dr. Sabah AL-Naimi
Qatar University
Physics Program
FALL 2009
PHYS191
Example (4):
Find the following:
5 R S 0.8 K
where R i j , S 2i 3 j , K j k
Example (5):
Find the following:
1
3M 1 N O
2
3
M 3i k , N 2 j , O 3i 2k
4
1.4 Components of a vector and unit vectors
Any vector A lying in the x-y Plane and
making any arbitrary angle with the
positive x-axis, we can represent this vector
by the following expression:
A Ax A y
or
A Ax i Ay j
where
A x Ax i A cos i
and
A y Ay j A sin j
The magnitude of vector A is :
A
2
2
Ax Ay
and the direction of vector A is :
Ay
tan
Ax
tan 1 (
Ay
Ax
)
At this expression
A Ax i Ay j
we call i and j unit vectors , where i is a
unit vector parallel to the x-axis and j
is a unit vector parallel to the y-axis.
5
Dr. Sabah AL-Naimi
Qatar University
Physics Program
FALL 2009
PHYS191
Dr. Sabah AL-Naimi
Qatar University
Physics Program
FALL 2009
PHYS191
1.5 Vector Analysis
A Ax A y
where
A x Ax i A cos i
and
A y Ay j A sin j
A x A sin (i )
A y A cos ( j )
A x A sin (i )
A y A cos ( j )
A x A cos (i )
A y A sin ( j )
And So on
6
Dr. Sabah AL-Naimi
Qatar University
Physics Program
FALL 2009
PHYS191
1.6 Analysis of Forces at Equilibrium
Foreces at Equilibrium :
x axis :
F
x
0
....... .......... ... .......... . 0
y axis :
F
y
0
......... .......... . .......... .... 0
7
Dr. Sabah AL-Naimi
Qatar University
Physics Program
FALL 2009
PHYS191
Example
F3
F3 cos
F2 sin
F2
F1
F3 sin
F2 cos
F4
Solution:
F3 cos
x axis :
F
x
F2 sin
0
F1 F2 cos F3 sin 0
y axis :
F
y
0
F1
F3 sin
F2 cos
F4
F2 sin F3 cos F4 0
8
Dr. Sabah AL-Naimi
Qatar University
Physics Program
FALL 2009
PHYS191
1.7 Examples
Example (1):
Find the magnitude of force F1 and F4 .The system is at equilibrium.
F2 150 N
F3 180N
30
F3 sin 50
F2 cos 30
50
F1
F3 sin
0
F1 F2 sin 30 F3 cos 50 0
F1 F3 cos 50 F2 sin 30
180 cos 50 150 sin 30
115.7 75
40.7 N
F
y
0
F2 cos 30 F3 sin 50 F4 0
F4 F2 cos 30 F3 sin 50
150 cos 30 180 sin 50
129.9 137.9
267.8 N
Therefore,
F1 = 40.7 i
F2 sin 30
F4
Solution:
x
F1
F3 cos 50
F4
F
Graph:
N
F4 = 267.8 j N
9
Dr. Sabah AL-Naimi
Qatar University
Physics Program
FALL 2009
PHYS191
Example (2):
Find the tension in each for cord this system. The system is at
equilibrium.
40
Graph:
70
T2
T1
T2
T1
40
70
T1 sin 70
T2 sin 40
T1 cos 70
m 5kg
T2 cos 40
mg
Solution:
F
x
0
T1 cos 70 T2 cos 40 0
T1 cos 70 T2 cos 40
T1 T2
F
y
cos 40
cos 70
0
T1 sin 70 T2 sin 40 mg 0
cos 40
sin 70 T2 sin 40 mg
cos 70
T2 ( 2.12) T2 (0.64) 5 X 9.8 49
T2
T2 ( 2.76) 49 T2
T1 17.8
49
17.8 N
2.76
cos 40
40.3 N
cos 70
Therefore,
Magnitude of T1 is
T1= 40.3 N
Magnitude of T2 is
T2= 17.8 N
10
Dr. Sabah AL-Naimi
Qatar University
Physics Program
FALL 2009
PHYS191
Example (3):
Find the magnitude of the force F1 and F2 where F1=F2 . The system
is at equilibrium.
F2
Graph:
F1
20
F1 sin 20
20
F2 sin 20
200N
F1 cos 20
F2 cos 20
200
Solution:
F
x
0
F1 cos 20 F2 cos 20 0
F cos 20 F cos 20 0
F
y
0
F1 sin 20 F2 sin 20 200 0
F sin 20 F sin 20 200 0
F (sin 20)(1 1) 200
2 F sin 20 200
200
200
F
294.1 N
2 sin 20 2 X 0.34
F1 F2 F 294.1 N
11
Dr. Sabah AL-Naimi
Qatar University
Physics Program
FALL 2009
PHYS191
Example (4):
Find magnitude of F2 and the weight (w) .The system is at equilibrium.
F3 50N
35
Solution:
F2 ?
F3 sin 25
F2 cos 35
25
10
F1 cos10
F1 30 N
w
F2 sin 35
F3 cos 25
w
F1 sin10
Solution:
F
x
0
F1 cos 10 F2 sin 35 F3 cos 25 0
F2 sin 35 F3 cos 25 F1 cos 10
F3 cos 25 F1 cos 10 45.3 29.5
27.72 N
sin 35
0.57
F2
F
y
0
F2 cos 35 F3 sin 25 F1 sin 10 w 0
w F2 cos 35 F3 sin 25 F1 sin 10
(27.72) cos 35 50 sin 25 30 sin 10
22.7 21.1 5.2 38.6 N
12
Dr. Sabah AL-Naimi
Qatar University
Physics Program
FALL 2009
PHYS191
Example (5):
A man walking 25km southeast from his camp, as shown to you.
Determine the components of the mans displacement.
north
west
30
Graph:
east
r cos 30
25km
r sin 30
south
Solution:
x component : r cos 30i (25km) cos 30 (i )
25 (0.87) i 21.75i Km east
y component : r sin 30( j ) (25km) sin 30 ( j )
25 (0.5) ( j ) 12.5 j Km south
13
Dr. Sabah AL-Naimi
Qatar University
Physics Program
FALL 2009
PHYS191
Example (6):
On the first day, a girl walking 20m north from her house. On the
second day, she walks 40m in a direction 60 northwest. Determine the
components of her displacement for each day.
north
day 2
r2 40m
day1
20m r1
60
east
west
south
Solution:
for day 1 :
X component : 0
Y component : r1 ( j ) 20m j
day1
20m r1
for day 2 :
X component : r2 cos 60(i) 40 cos 60(i)
40(0.5)(i)
20 i m
Y component : r2 sin 60( j ) 40 sin 60( j )
40(0.87)( j )
34.8 j m
14
r2 40m
60
r2 cos 60
r2 sin 60
Dr. Sabah AL-Naimi
Qatar University
Physics Program
FALL 2009
PHYS191
Example (7):
An Airplane starts from an airport takes the route as shown to you. It
flies to city A located 175 km in a direction 30 north of east. Next, it
files to city B located 150 km 20 west of north . Finally, it flies 190 km
west to city C. Find the location of the plane regarding to the starting
point O.
north
c
Graph:
b
b
20
20
30
a
a
east
30
west
c
south
Solution:
x component Rx
Rx a cos 30(i ) b sin 20(i ) c (i )
175 cos 30 i 150 sin 20 i 190 i
151.6 i 51.3 i 190 i
89.7 i
a sin 30 j
b cos 20 j
c
y component R y
b sin 20 i
R y a sin 30( j ) b cos 20( j )
175 sin 30( j ) 150 cos 20( j )
87.5 ( j ) 140.95 ( j )
228.5 j
The location of the plane :
R R x R y 89.7 i 228.5 j
15
a cos 30 i
Dr. Sabah AL-Naimi
Qatar University
Physics Program
FALL 2009
PHYS191
Example (8):
Find the resultant vector, its magnitude and its direction.
C
5m
Graph:
B
40
B
A
4m
4m
5m
3 sin 20
3m
40 c
20
3 cos 20
20
A
B
5 sin 40
C
5 cos 40
3m
Solution:
Solution:
The resultant vector :
x component :
R x 5 cos 40 i 3 cos 20 i
5(0.77) i 3(0.94) i
3.38 i 2.28 i
1.01 i
y component :
R y 4 j 5 sin 40 j 3 sin 20 j
4 j 5(0.64) j 3(0.34) j
3.21 j 1.03 j
8.24 j
The magnitude :
R
2
2
Rx R y
(1.01) 2 (8.24) 2
The direction :
R
8.24
tan 1 ( y ) tan 1 (
) 83.01
Rx
1.01
16
68.92 8.3 m
Whe
.. F
n
Con
necti
ng 2
or
mor
e
capa
citor
in
serie
s
con
necti
on ,
We
have
the
follo
wing
:
1
2
1
Z
C3
T
h
e
E
q
Dr. Sabah AL-Naimi
Qatar University
Physics Program
FALL 2009
PHYS191
NOTE :
The resultant vector of R is :
R Rx i R y j
The magnitude of R :
R
2
2
Rx R y
The Direction of R :
Ry
Ry
tan
tan 1 ( )
Rx
Rx
H.W. # 1
P. 71 76
3, 7, 10, 15, 18, 27, 29,
33, 35, 49, 59
17