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Basic Physics
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Introduction
1. What is Physics?
2. Give a few relations between physicsand daily living experience
3. Review of measurement and units SI,METRIC, ENGLISH
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VECTOR AND SCALAR
Scalar is a quantity which only signifies itsmagnitude 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 carwith a velocity of 100 km/hr due to NorthWest, etc.
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VECTOR AND SCALAR
Writing conformity
F Bold fontF Italic font signifying its magnitude F Normal Font with an arrow
head on top of it (Use t h is )
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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 ??
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VECTOR AND SCALAR
X (i)
Z(k)
Y(j)
F = 59i + 59j + 29k N
O
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VECTOR AND SCALARDefining a Vector by:2. Unit Vector
Ex. F = F u (use the previous example)
=
F for magnitude (F 2 = F x2 + F y2 + F z2)
u for direction (dimensionless and unity)
uF F
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VECTOR AND SCALARMagnitude F = (59 2 + 59 2 + 29 2)
F = 88.33 N
Direction =
= 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)
59i + 59j + 29k88.33u
u
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VECTOR AND SCALAR
X (i )
Z (k )
Y ( j )
F = 88.33 N
U
FU = 0.67i + 0.67j + 0.33k
= 47.9 0
= 47.9 0
= 70.7 0
O
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VECTOR AND SCALAR
Defining a Vector by:3. Position Vector
Similar to unit vector, it differs on how tolocate the vectors direction which isusing the point coordinate.
Ex . F = F u (see next example)r (position vector)r (position vector magnitude)u =
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VECTOR AND SCALAR
U
6 m
Given:
F = 150 N
Required:
a. F ?
b. , , ?
X (i )
Z (k )
Y ( j )
F
O
A
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VECTOR AND SCALAR
Solution:
r r
u =
F = F u
=2i + 4j + 6k
7.48 = 0.27i + 0.53j +0.80ku
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VECTOR AND SCALAR
Solution:
F = F u
= 150 (0.27i + 0.53j +0.80k)
F = 40.5i + 79.5j + 120k
= 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)
a.
b.
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VECTOR AND SCALAR
Operations of Vector1. Addition
2. Subtraction3. Dot Product4. Cross Product
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VECTOR AND SCALAR
1. Addition
F1
RF2
R = F1 F2+
R = (F 1x + F 2x) i + (F 1y + F 2y) j + (F 1z+F 2z) k
=R y
RxTan -1
O
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VECTOR AND SCALAR
1. Addition
F1
RF2
R = F1 F2+
R = (F 1x + F 2x) i + (F 1y + F 2y) j + (F 1z+F 2z) k
=R y
RxTan -1
O
Resultant is directedfrom initial tail towardsfinal arrow head
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VECTOR AND SCALAR
2. SubtractionF1
RF2
R = F1 F2-
R = (F 1x - F 2x) i + (F 1y - F 2y) j + (F 1z - F 2z) k
= RyRx
Tan -1
O
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VECTOR AND SCALAR
2. SubtractionF1
RF2
R = F1 F2-
R = (F 1x - F 2x) i + (F 1y - F 2y) j + (F 1z - F 2z) k
=
Ry
RxTan -1O
Take note and watchout !!!
(the sense isopposite to the given
diagram)
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VECTOR AND SCALAR
3. Dot Product
F
X (i )
Z (k )
Y ( j )
F
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VECTOR AND SCALAR
A . B = AB cos (General Formula)
Vector Magni tude
The ang le betw eenvectors (be tweenth eir tai ls )
Cartesian Unit vector dot product
i . i = 1
i . j = 0 i . k = 0 k . j = 0
j . j = 1 k . k = 1
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VECTOR AND SCALAR
From Example: F . d = Fd cos (Using Vectors magnitude)
= (F xi + F y j + F zk) . (d xi + d y j + d zk)
= F x dx + F y d y + F z d z (Using Component Vector)
The dot product of two vectors is called scalar productsince the result is a scalar and not a vec tor
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The dot produ ct is used to dete rmine:1. The angle between the tails o f the vectors .
VECTOR AND SCALAR
= cos-1
A . B
AB2. The projecte d component of a vector V onto an axis
defined by it s unit vector u
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VECTOR AND SCALAR
X (i )
Z (k )
Y ( j )O
B
A
C
F = 100N
Given : Figure 1
Required:
1.
2. F BA (Magnitude)
Fig.1
Example:
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VECTOR AND SCALAR
Solution :
1. Angle
Find position vectors from B to A and B to C
r BA = -200i 200j + 100k
r BC = -0i 300j + 100k = 300j + 100k
cos =r BA . r BC r BA r BC
=0 + 60000 + 10000
(300)(316.23)=
7000094869
= 0.738
= Cos -1 0.738 = 42.45 o (answer)
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VECTOR AND SCALAR
Solution :
=r BAuBA =
r BA
2. F BA
-200i 200j + 100k300 = -0.667i 0.667j + 0.33k
r BCuBC =
r BC=
-0i 300j + 100k316.2
= 0.949j + 0.316k
FBC = F BC . u BC = 100 . ( 0.949j + 0.316k) = -94.9j + 31.6k
FBA = F BC . u BA = (-94.9i + 31.6j) . (-0.667i 0.667j + 0.33k)
= 63.3 + 10.5 = 73.8 N (answer)
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VECTOR AND SCALAR
Solution : Alternative Solution
FBA = (100 N) (cos 42.45o)
= 73.79 N
FBA = F BA u BA = 73.79 (-0.667i - 0.667j + 0.33k)
= -49.2i 49.2j + 24.35k
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VECTOR AND SCALAR
4. Cross Product
B
A
F
X (i )
Z (k )
Y ( j )O
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VECTOR AND SCALAR
A = B x C A is equal to B cross C
Apply the right hand rule
i
j k
i x j = k
j x k = ik x i = j
j x i = -k
k x j = -ii x k = -j
i x i = 0
j x j = 0k x k = 0+
-
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VECTOR AND SCALAR
Right Hand Rule
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VECTOR AND SCALAR
Right Hand Rule
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VECTOR AND SCALAR
Right Hand Rule
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VECTOR AND SCALAR
Right Hand Rule
. answer for yourself)
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VECTOR AND SCALAR
A = B x C
= (B x i + B y j + B z k ) x (C x i + C y j + C z k )
i j kBx By Bz
Cx C y C z
=
= (B y C z Bz C y)i + (B z C y Bx C z) j + (B x C y By C x)z A
i j kBx By Bz
Cx C y C z
=i jBx By
Cx C y
+-
= (B y C z Bz C y)i (B x C z Bz C x) j + (B x C y By C x)z
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VECTOR AND SCALAR
A = B x C
= (B x i + B y j + B z k ) x (C x i + C y j + C z k )
i j kBx By Bz
Cx C y C z
=
= (B y C z Bz C y)i + (B z C y Bx C z) j + (B x C y By C x)z A
i j kBx By Bz
Cx C y C z
=i jBx By
Cx C y
+-
= (B y C z Bz C y)i (B x C z Bz C x) j + (B x C y By C x)z
Full cautionfor the +/- signand
subscripts
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VECTOR AND SCALAR
Example: Given : Figure 2
Required :
1. M o (Moment atpoint O)
2. M y (Moment
about y axis)
B
A
F = 100N
X (i )
Z (k )
Y ( j )O
Mo
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VECTOR AND SCALAR
Solution:Finding the vectors neededF = F u
= 100400i 250j 200k
(400 2 + 250 2 + 200 2)( )
F = 78.07i 48.79j 39.04k
OA = 400j
OB = 400i + 150j 200k
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VECTOR AND SCALAR
B
A
F = 100 N
X (i )
Z (k )
Y ( j )
O
Mo400i + 150j 200k
F = 78.07i 48.79j 39.04k
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VECTOR AND SCALAR
Mo =
=
Mo = -15616i 31228k N.mm
F OA x
i j k
0 400 0
78.07 -48.79 -39.04
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)
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VECTOR AND SCALAR
Mo =
Mo = -15616i 31228k N.mm
F OB x
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)
i j k
400 150 -200
78.07 -48.79 -39.04
=