Basic Physics
Introduction
1. What is Physics?
2. Give a few relations between physics
and daily living experience
3. Review of measurement and units SI,
METRIC, ENGLISH
VECTOR AND SCALAR
Scalar is a quantity which only signifies its
magnitude without its direction. (+ / - )
Ex. 1kg of apple, 273 degrees centigrade,
etc.
Vector is a quantity with magnitude and
direction. (+ / - )
Ex. Velocity of a moving object – a car
with a velocity of 100 km/hr due to North
West, etc.
VECTOR AND SCALAR
Writing conformity
F
F
Bold font
F
Normal Font with an arrow
head on top of it (Use this)
Italic font signifying its magnitude
VECTOR AND SCALAR
Defining a Vector by:
1. Cartesian Vector
Ex. F = 59i + 59j + 29k
N
the magnitude is F = (592 + 592 + 292)
F = 88.33 N
Due to which is the vector ??
VECTOR AND SCALAR
Z(k)
F = 59i + 59j + 29k
O
Y(j)
X (i)
N
VECTOR AND SCALAR
Defining a Vector by:
2. Unit Vector
Ex. F = F u (use the previous example)
u =
F
F
F for magnitude (F2 = Fx2 + Fy2 + Fz2)
u for direction (dimensionless and unity)
VECTOR AND SCALAR
Magnitude
F = (592 + 592 + 292)
F = 88.33 N
Direction
u =
59i + 59j + 29k
88.33
u = 0.67i + 0.67j + 0.33k



= cos-1 0.67 = 47.9 0 (angle from x-axis)
= cos-1 0.67 = 47.9 0 (angle from y-axis)
= cos-1 0.33 = 70.7 0 (angle from z-axis)
VECTOR AND SCALAR
Z (k)
F = 88.33 N
U
U = 0.67i + 0.67j + 0.33k
F
 = 47.9 0

 = 47.9 0
O
 = 70.7 0
X (i)


Y (j)
VECTOR AND SCALAR
Defining a Vector by:
3. Position Vector
Similar to unit vector, it differs on how to
locate the vector’s direction which is
using the point coordinate.
Ex. F = F u (see next example)
u=
r
r
(position vector)
(position vector magnitude)
VECTOR AND SCALAR
Z (k)
Given:
A
F = 150 N
F
Required:

a. F ?
b. , ,  ?
O

X (i)
U

6m
Y (j)
VECTOR AND SCALAR
Solution:
F=F u
u =
=
r
r
2i + 4j + 6k
7.48
u = 0.27i + 0.53j +0.80k
VECTOR AND SCALAR
Solution:
a. F = F u
= 150 (0.27i + 0.53j +0.80k)
F = 40.5i + 79.5j + 120k
b.
 = cos-1 0.27 = 74.3 0 (angle from x-axis)
 = cos-1 0.53 = 58.0 0 (angle from y-axis)
 = cos-1 0.80 = 36.9 0 (angle from z-axis)
VECTOR AND SCALAR
Operations of Vector
1. Addition
2. Subtraction
3. Dot Product
4. Cross Product
VECTOR AND SCALAR
1. Addition
F2
R
O

F1
R = F1 + F2
 = Tan
-1
Ry
Rx
R = (F1x + F2x) i + (F1y + F2y) j + (F1z+F2z) k
VECTOR AND SCALAR
1. Addition
F2
R
O
Resultant is directed
from initial tail towards
final arrow head

F1
R = F1 + F2
 = Tan
-1
Ry
Rx
R = (F1x + F2x) i + (F1y + F2y) j + (F1z+F2z) k
VECTOR AND SCALAR
2. Subtraction
O
F1

F2
R
R = F1 - F2
 = Tan
-1
Ry
Rx
R = (F1x - F2x) i + (F1y - F2y) j + (F1z - F2z) k
VECTOR AND SCALAR
2. Subtraction
O
F1

 = Tan
-1
Ry
Rx
F2
R
R = F1 - F2
Take note and watch
out !!!
(the sense is
opposite to the given
diagram)
R = (F1x - F2x) i + (F1y - F2y) j + (F1z - F2z) k
VECTOR AND SCALAR
3. Dot Product
Z (k)
F

F
X (i)
Y (j)
VECTOR AND SCALAR
Vector
(General Formula)



A . B = AB cos 
Magnitude
The angle between
vectors (between
their tails)
Cartesian Unit vector dot product
i.i=1
j.j=1
k.k=1
i.j=0
i.k=0
k.j=0
VECTOR AND SCALAR
From Example:
F . d = Fd cos 
(Using Vectors’ magnitude)
= (Fxi + Fyj + Fzk) . (dxi + dyj + dzk)
= Fx dx + Fy dy + Fz dz
(Using Component Vector)
The dot product of two vectors is called scalar product
since the result is a scalar and not a vector
VECTOR AND SCALAR
The dot product is used to determine:
1. The angle between the tails of the vectors.
 = cos
-1
A.B
AB
2.
The projected component of a vector V onto an axis
defined by its unit vector u
VECTOR AND SCALAR
Example:
Given : Figure 1
Z (k)
Required:
1. 
A
O
Y (j)
C
2. FBA (Magnitude)

X (i)
Fig.1
F = 100
N
B
VECTOR AND SCALAR
Solution :
1. Angle 
Find position vectors from B to A and B to C
rBA
= -200i – 200j + 100k
r BC
= -0i – 300j + 100k
= – 300j + 100k
rBA . rBC
0 + 60000 + 10000
70000
cos  =
=
=
= 0.738
rBA rBC
(300)(316.23)
94869
 = Cos -1 0.738 = 42.45 o
(answer)
VECTOR AND SCALAR
Solution :
2. FBA
rBA
uBA = r
BA
-200i – 200j + 100k
=
= -0.667i – 0.667j + 0.33k
300
rBC
-0i – 300j + 100k
uBC = r
=
BC
316.2
= – 0.949j + 0.316k
FBC = FBC . uBC = 100 . (– 0.949j + 0.316k) = -94.9j + 31.6k
FBA = FBC . uBA = (-94.9i + 31.6j) . (-0.667i – 0.667j + 0.33k)
= 63.3 + 10.5 = 73.8 N
(answer)
VECTOR AND SCALAR
Solution :
Alternative Solution
FBA = (100 N) (cos 42.45o)
= 73.79 N
FBA = FBA uBA
= 73.79 (-0.667i - 0.667j + 0.33k)
= -49.2i – 49.2j + 24.35k
VECTOR AND SCALAR
4. Cross Product
Z (k)
O
A
X (i)
B
F
Y (j)
VECTOR AND SCALAR
A = B x C
A is equal to B cross C
Apply the right hand rule
i
j
+
k
ixj=k
j x i = -k
ixi=0
jxk=i
k x j = -i
jxj=0
kxi=j
i x k = -j
kxk=0
VECTOR AND SCALAR
Right Hand Rule
VECTOR AND SCALAR
Right Hand Rule
VECTOR AND SCALAR
Right Hand Rule
VECTOR AND SCALAR
Right Hand Rule
……. (answer for
yourself)
VECTOR AND SCALAR
A = B x C
= (Bx i + By j + Bz k) x (Cx i + Cy j + Cz k)
i
=
j
k
Bx
By
Bz
Cx
Cy
Cz
=
i
j
k
i
j
Bx
By
Bz
Bx
By
Cx
Cy
Cz
Cx
Cy
-
+
A = (By Cz – Bz Cy)i + (Bz Cy – Bx Cz)j + (Bx Cy – By Cx)z
= (By Cz – Bz Cy)i – (Bx Cz – Bz Cx)j + (Bx Cy – By Cx)z
VECTOR AND SCALAR
A = B x C
= (Bx i + By j + Bz k) x (Cx i + Cy j + Cz k)
i
=
j
k
Bx
By
Bz
Cx
Cy
Cz
=
Full caution
for the +/- sign
and
subscripts
i
j
k
i
j
Bx
By
Bz
Bx
By
Cx
Cy
Cz
Cx
Cy
-
+
A = (By Cz – Bz Cy)i + (Bz Cy – Bx Cz)j + (Bx Cy – By Cx)z
= (By Cz – Bz Cy)i – (Bx Cz – Bz Cx)j + (Bx Cy – By Cx)z
VECTOR AND SCALAR
Example:
Given : Figure 2
Z (k)
Required :
O
A
Mo
1. Mo (Moment at
point O)
2. My (Moment
about y axis)
X (i)
F = 100N
Y (j)
B
VECTOR AND SCALAR
Solution:
Finding the vectors needed
F =F u
= 100
(
400i – 250j – 200k
(4002
+
2502
+
2002)
)
F = 78.07i – 48.79j – 39.04k
OA = 400j
OB = 400i + 150j – 200k
VECTOR AND SCALAR
Z (k)
O
A
400i + 150j – 200k
Y (j)
Mo
X (i)
B
F = 78.07i – 48.79j – 39.04k
F = 100 N
VECTOR AND SCALAR
Mo = OA x F
=
i
j
k
0
400
0
-48.79
-39.04
78.07
Mo = -15616i – 31228k
N.mm
Mo = 34914.86
N.mm
 = cos-1 (-0.447) = 116.55 0 (angle from x-axis)
 = cos-1 0 = 90.0 0 (angle from y-axis)
 = cos-1 0.894 = 26.57 0 (angle from z-axis)
VECTOR AND SCALAR
Mo = OB x F
=
i
j
k
400
150
-200
78.07
-48.79
-39.04
Mo = -15616i – 31228k
N.mm
Mo = 34914.86
N.mm
 = cos-1 (-0.447) = 116.55 0 (angle from x-axis)
 = cos-1 0 = 90.0 0 (angle from y-axis)
 = cos-1 0.894 = 26.57 0 (angle from z-axis)