Introduction Chapter
Dr. Yousef Abou-Ali
yabouali@iust.edu.sy
Introduction Chapter
1.3
Standards and Units
1.7 Vectors and Vector Addition
1.10 Products of Vectors
4.2 Newton's First Law
4.3 Newton's Second Law
4.5 Newton's Third Law
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
Goals for Introduction Chapter
 To know standards and units and be able to do unit
conversions.
To be able to add vectors.
 To be able to break down vectors into x and y components.
 To be able to calculate dot and cross products.
 To understand force – either directly or as the net force of
multiple components.
 To study and apply Newton’s First Law.
 To study and apply the concept of mass and acceleration as
components of Newton’s Second Law.
 To differentiate between mass and weight.
 To study and apply Newton’s Third Law.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
Why We Study Physics?
Two Reasons:
• Physics is one of the most fundamental of the science.
• The study of physics is an adventure. If you wondered: why the
sky is blue, why radio waves travel through empty space and how a
satellite stays in orbit? You can not find the answers without first
understanding the basic laws of physics.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
Introduction
• Physics
is an experimental science. Physicists observe the
phenomena of nature and try to find patterns and principles that
relate these phenomena.
• Example: Galileo dropped light and heavy objects from the top
of the Leaning Tower of Pisa to find out whether their rates of fall
were the same or different. Galileo recognized that only experimental investigation could answer this question. From examining the
results of his experiments he made the inductive leap to the
principle, or theory, that the acceleration of a falling body is
independent of its weight.
•Every physical theory has a range of validity outside of which it is not
applicable.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
Introduction
• Getting back to Galileo, suppose we drop a
feather and a cannonball. They certainly do not
fall at the same rate. This does not mean that
Galileo was wrong; it means that his theory was
incomplete. If we drop the feather and the
cannonball in a vacuum to eliminate the effects
of the air, then they do fall at the same rate.
Galileo's theory has a range of validity .
Introduction
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
Introduction
• At
some points in their studies, almost all physics students find
themselves thinking “ I understand the concepts but I just cant solve
the problems”.
You do not know physics unless you can do physics.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
1.3 Standards and Units
• Physics
is an
measurements.
experimental
science,
Experiments
require
• We
use numbers to describe the results of measurements (any
number used to describe a physical phenomena quantitatively is a
called physical quantity).
Example: two physical quantities that describe you, are your weight
and height .
• Some physical quantities are so fundamental that we can define
them only by describing how to measure them. Such a definition is
called an operational definition.
Examples: measuring a distance by using a rule, and measuring a
time interval by using a stopwatch.
Dr. Y. Abou-Ali,
Abou-Ali IUST
University Physics, Chapter 1
1.3 Standards and Units
• When
we measure a quantity, we always compare it with some
reference standard. Such a standard defines a unit of the quantity.
• To
make accurate, reliable measurements, we need units of
measurements that do not change. The system used by scientists and
engineers around the world called the “metric system” but since
1960 known as International System or SI.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
1.3 Standards and Units
• There are three fundamental SI Units:
• Time
 Second (s)
• Length
 Meter (m)
• Mass
 Kilogram (kg)
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
1.3 Standards and Units
The Second
• Originally tied to the length of a day.
• Now, exceptionally accurate.
 Atomic clock
 9,192,631,770 oscillations of a low-energy transition in Cs
 In the microwave region
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University Physics, Chapter 1
3
1.3 Standards and Units
The meter
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
3
1.3 Standards and Units
The meter
• Now
tied to Kr discharge and counting a certain number of
wavelengths.
• Exceptionally accurate, in fact redefining c, speed of light.
• New definition is the distance that light can travel in a vacuum in
1/299,792,458 s.
• So accurate that it loses only 1 second in 30 million years.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
3
1.3 Standards and Units
The kilogram
• The kilogram, defined to be the mass of a particular cylinder of
platinum-iridium alloy.
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University Physics, Chapter 1
3
1.3 Standards and Units
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
3
1.3 Standards and Units
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
1.3 Standards and Units
The British System:
• Length: 1 inch = 2.54 cm.
• Force: 1 pound = 4.448221615260 newtons.
• For time, the British unit is second.
• British units are used only in mechanics and thermodynamics. No
BU for electrical units.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
Dimensional analysis
The word dimension has a special meaning in physics. It
denotes the physical nature of a quantity.
Whether a distance is measured in units of feet or meters
or fathoms, it is still a distance. We say its dimension is
length.
• The symbols we use in this book to specify the
dimensions of length, mass, and
• time are L, M, and T, respectively. We shall often use
brackets [ ] to denote the dimensions of a physical
quantity. For example, the symbol we use for speed in
this book is v, and in our notation the dimensions of
speed are written [v] = L/T.
Dimensional analysis
• Dimensional analysis makes use of the fact that
dimensions can be treated as algebraic quantities. For
example, quantities can be added or subtracted only if
they have the same dimensions.
• For example we can Show that the expression v = at is
dimensionally correct, where v represents speed, a
acceleration, and t an instant of time. So by used
dimensional analysis we write
• [V]= L/T
• And
Dimensional analysis
• Example : Analysis of a Power Law
• Suppose we are told that the acceleration a of a particle
• moving with uniform speed v in a circle of radius r is
proportional to some power of r, say rn, and some power
of v, say vm. Determine the values of n and m and write the
simplest form of an equation for the acceleration.
• Solution :
• Let us take a to be
Dimensional analysis
• where k is a dimensionless constant of proportionality.
• Knowing the dimensions of a, r, and v, we see that the
dimensional equation must be
This dimensional equation is balanced under the conditions
n+m = 1
and
m=2
Therefore n=-1 , and we can write the acceleration
expression as
Dimensional analysis
• which of the following equations are
dimensionally correct ?
• (a)
• (b)
• Solution :
• (a) This is incorrect since the units of [ax] are
m2/s2 , while the units of [v] are m/s .
• (b) This is correct since the units of [y] are m, and
cos (kx) is dimensionless if [k] is in m−1.
Dimensional analysis
• Newton’s law of universal gravitation is
represented by
• Here F is the magnitude of the gravitational force
exerted by one small object on another, M and m
are the masses of the objects, and r is a distance.
Force has the SI units kg·m/s2.
• What are the SI units of the proportionality
constant G?
Dimensional analysis
1.7 Vectors and Vectors Addition
• Time, temperature, mass, density…. Can be describe by a single
number with unit.
• Some
important physical quantities have a direction associated
with them and cant be describe by a single number.
• Example: motion of airplane.
• Physical quantity is described
by a single number we call it a
scalar quantity. For scalar number 6 kg + 3 kg = 9 kg.
• Vector quantity has both a magnitude (how much or how many)
and direction in space. To combine vectors require different set of a
operation.
• Simplest vector quantity is Displacement (change in position of a
point).
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
1.7 Vectors and Vectors Addition
P2
• Change from P1 to P2 represent by line from P1 to P2
P1
• We write vector quantity such as displacement:
A
• In the book vector symbols are boldface italic type with an arrow
above them.
• Displacement is not related directly to the total distance travelled.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
1.7 Vectors and Vectors Addition
• In the “world of vectors” 1+1 does not necessarily equal 2.
• Graphically?
Anti
parallel vectors
Parallel vectors
• Magnitude of
Dr. Y. Abou-Ali, IUST
A A A
1.2
University Physics, Chapter 1
Components of Vectors
A  Ax  Ay
(1.6)
where Ax & Ay :
are called the components vectors of
vector A
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
Components of Vectors
• We need a single number to describe Ax & Ay
• When the component vector Ax points in the positive x-direction,
we define the number Ax to be equal to the magnitude of Ax .
• When the component vector Ax points in the negative x-direction,
we define the number Ax to be equal to the negative of that
magnitude.
• The magnitude of a vector quantity always positive (never negative).
• We can calculate the components of A if we know its magnitude
and direction, we will describe the direction of a vector by its angle.
• In last figure θ (theta) is the angle between
Dr. Y. Abou-Ali, IUST
A and +x axis.
University Physics, Chapter 1
Components of Vectors
Ax
cos  
A
 Ax  A cos
& sin  
Ay
A
& Ay  Asin 
(1.7)
• We measured θ from +x
axis toward +y axis.
y
θ
cx(-)

B
Bx(-)
By(+)
θ
0
x

C
0
x
cy(-)
y
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
Components of Vectors
Example 1.6 (Finding Components):
a. What are the x- and y- components of vectorD In the figure (a) below?
The magnitude of the vector is D = 3.00 m and the angle α = 45o.
b. What are the x- and y- components of vectorE In the figure (b) below?
The magnitude of the vector is E = 4.50 m and β = 37.0o.
y
α
Dx(+)
x

D
Dy(-)
(a)
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
Components of Vectors
Solution:
a) Identify and step: all we need Eqs (1.7), the angle is not measured
from +x toward +y axis.
Execute: the angle between D and +x axis is α (alpha)
θ = - α = - 45o
Dx = D cos θ = (3m)(cos(-45o)) = 2.1 m
Dy = D sin θ = (3m)(sin(-45o)) = - 2.1 m
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
Components of Vectors
b) x is not horizontal & y is not vertical. The angle β (beta) is between E
and +y axis, we cant use Eqs 1.7.
E is the hypotenuse of a right triangle and the other side are Ex & Ey.
Ex = E sin β = (4.50 m)(sin(37.0o)) = 2.71 m
Ey = E cos β = (4.50 m)(cos(37.0o)) = 3.59 m
Evaluate: if you want to use Eqs 1.7 you have to find the angle between
+x and vector E.
θ = 90o – 37o = 53o
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
Unit Vectors
• A unit vector that has magnitude of 1 with no unit.
y

A

Aˆj
Ax  Ax iˆ
A  A ˆj
ˆj
y
0
iˆ

Aiˆ
y
x
A  Ax iˆ  Ay ˆj
Dr. Y. Abou-Ali, IUST
(1.13)
(1.14)
University Physics, Chapter 1
Unit Vectors
• When we have two vectors
A& B
A  Ax iˆ  Ay ˆj
B  Bx iˆ  By ˆj
R  A B
 ( Ax iˆ  Ay ˆj )  ( Bx iˆ  B y ˆj )
 ( Ax  Bx )iˆ  ( Ay  By ) ˆj
 Rx iˆ  Ry ˆj
Dr. Y. Abou-Ali, IUST
(1.15)
University Physics, Chapter 1
Unit Vectors
• If we have third unit k:
A  Ax iˆ  Ay ˆj  Az kˆ
B  Bx iˆ  By ˆj  Bz kˆ
R  Rx iˆ  Ry ˆj  Rz kˆ
Dr. Y. Abou-Ali, IUST
(1.16)
(1.17)
University Physics, Chapter 1
Unit Vectors
Example 1.9 (Using Unit Vectors):
Given the two displacement: D  (6iˆ  3 ˆj  kˆ ) m
E  (4iˆ  5 ˆj  8kˆ) m
Find the magnitude of the displacement:
2D  E
Solution:
Identify, step and execute: Letting: F
 2D  E
F  2(6iˆ  3 ˆj  kˆ ) m  (4iˆ  5 ˆj  8kˆ) m
 [(12  4)iˆ  (6  5) ˆj  (2  8)kˆ] m
 (8iˆ  11 ˆj 10kˆ ) m
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
Unit Vectors
• The unit of D, E & F
are m
F
the components also are in m.
2
Fx
2
 Fy
2
 Fz
 (8)  (11)  (10)
2
2
2
 17 m
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
1.10 Products of Vectors
Scalar Product:
• The scalar product of A & B is donated by A.B
 Also called dot product.

B


A
A.B  AB cos   A B cos 
(1.18)
Definition of the scalar (dot) product
Work
Dr. Y. Abou-Ali, IUST
W  F .S
Force
Displacement
University Physics, Chapter 1
1.10 Products of Vectors
A.B  B. A
Commutative law
iˆ.iˆ  ˆj. ˆj  kˆ.kˆ  (1) (1) cos 0  1
iˆ. ˆj  iˆ.kˆ  ˆj.kˆ  (1) (1) cos 90o  0
(1.19)
A.B  ( Ax iˆ  Ay ˆj  Az kˆ ).( Bx iˆ  By ˆj  Bz kˆ )
A.B  Ax Bx  Ay By  Az Bz
(1.21)
Scalar (dot) product in terms of components
• Scalar product, find the angle Φ between any two vectors whose
components are known.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
1.10 Products of Vectors
Example 1.10 (Calculating a Scalar Product):
Find the scalar product A.B Of the two vectors in the figure below.
The magnitude of the vectors are A = 4.00 and B = 5.00
y

B

A
130.0o
Φ
ˆj
iˆ
Dr. Y. Abou-Ali, IUST
53.0o
x
University Physics, Chapter 1
1.10 Products of Vectors
Solution:
Identify and step:
• Two ways to calculate:
 Eq (1.18) using magnitudes of the vectors and Φ.
 Eq (1.21) using the components of the two vectors.
Execute:
1. Φ = 130.0o – 53.0o = 77.0o
A.B  AB cos  (4.00)(5.00)cos77.0  4.50
o
Positive because Φ between 0 – 90o
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
1.10 Products of Vectors
2. Find the components of vectors A & B
Ax  (4.00) cos 53.0o  2.407
Ay  (4.00) sin 53.0o  3.195
Az  0
Bx  (5.00) cos130.0o  3.214
By  (5.00) sin130.0o  3.830
Bz  0
A.B  Ax Bx  Ay By  Az Bz
 (2.407)(3.214)  (3.195)(3.830)  (0)(0)  4.50
Dr. Y. Abou-Ali, IUST
Read example 1.11
University Physics, Chapter 1
1.10 Products of Vectors
Vector Product:
• The vector product of A & B is donated by A  B
 Also called cross product.
• Vector product is a vector quantity with direction perpendicular
to the plane which contain two vectors A and B and a magnitude
equal to AB sinΦ.
 
A B
If C  A B, then
C  AB sin 
(1.22)
Magnitude of the vector (cross)
product of A & B
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Φ

A

B
University Physics, Chapter 1
1.10 Products of Vectors
Vector Product:
• We measure Φ from vector A toward B
• Φ from 0o – 180o
C  A
C  B
• When A & B are parallel or anti-parallel  C  A  B  0
• The vector product of any vector with itself is zero A  A  0
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
1.10 Products of Vectors
Contrast between the scalar & vector product:
Scalar product
Vector product
A \ \ B  C = maximum
A \ \B  C  0
AB C 0
A  B  C  maximum
A.B  B. A
iˆ.iˆ  ˆj. ˆj  kˆ.kˆ  1
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A  B  B  A (1.23)
iˆ  ˆj   ˆj  iˆ  kˆ
ˆj  kˆ  kˆ  ˆj  iˆ (1.24)
kˆ  iˆ  iˆ  kˆ  ˆj
University Physics, Chapter 1
1.10 Products of Vectors
A  B  ( Ax iˆ  Ay ˆj  Az kˆ )  ( Bx iˆ  By ˆj  Bz kˆ )
 Ax iˆ  Bx iˆ  Ax iˆ  B y ˆj  Ax iˆ  Bz kˆ
 Ay ˆj  Bx iˆ  Ay ˆj  B y ˆj  Ay ˆj  Bz kˆ
(1.25)
 Az kˆ  Bx iˆ  Az kˆ  By ˆj  Az kˆ  Bz kˆ
• We can rewrite
Ax iˆ  By ˆj  ( Ax By )iˆ  ˆj 
A  B  ( Ay Bz  Az By )iˆ  ( Az Bx  Ax Bz ) ˆj  ( Ax By  Ay Bx )kˆ
Dr. Y. Abou-Ali, IUST
(1.26)
University Physics, Chapter 1
1.10 Products of Vectors
C  A  B  Components:
Cx= AyBz - AzBy
Cy= AzBx – AxBz
(1.27)
Cz= AxBy - AyBx
iˆ
A B 
Dr. Y. Abou-Ali, IUST
ˆj kˆ
Ax
Ay
Az
Bx
By
Bz
University Physics, Chapter 1
1.10 Products of Vectors
Example 1.12 (Calculating a Vector Product):
Vector A has magnitude 6 units and is in direction of the +x axis.Vector
B has magnitude 4 units and lies in +xy plane, making an angle of 30o
with the +x axis (figure below). Find the vector product A  B
Solution:
Identify and step:
 Eq (1.22), determine the magnitude and right rule for direction.
 Using the components of vectors A & B to find vector C
components using Eq (1.27).
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
1.10 Products of Vectors
Execute: 1. from Eq (1.22): AB sin   (6)(4)(sin30o )  12
• From the right hand rule, the direction of A  B

is along + z
A B  12 kˆ
Execute: 2. Write the components of A & B
Ax = 6
Ay = 0
Bx = 4 cos30o
By = 4 sin30o
2 3
Dr. Y. Abou-Ali, IUST
Az = 0
Bz = 0
2
University Physics, Chapter 1
1.10 Products of Vectors
C = A  B, Using Eqs (1.27)
cx  (0)(0)  (0)(2)  0
c y  (0)(2 3)  (6)(0)  0
cz  (6)(2)  (0)(2 3)  12
C has only z component and lies along z-axis.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 1
Force and Interactions
• Force is push or pull.
• The concept of force gives us the interaction between 2 bodies or
between body and its environment.
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University Physics, Chapter 4
Force and Interactions
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 4
Force and Interactions
• Force is vector quantity (magnitude & direction)
• When two forces F1 & F2 act at the same time at a point A of a
body:

F2
A
R  F1  F2

R

F1
Superposition of forces
• Forces combine according to vector addition.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 4
Force and Interactions
• We need to find the vector sum (resultant) of all the forces acting
on a body:
R  F1  F2  F3  .......   F
(4.1)
the vector sum of the forces
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 4
Force and Interactions
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 4
Force and Interactions
• The components version of (4.1):
Rx   Fx &
Ry   Fy
• Each component may be positive or negative.
• The magnitude and direction of the net force are:
R
2
Rx

2
Ry
&
tan  
(4.2)
Ry
Rx
where θ the angle between vector R and the + x-axis.
• In 3-D:
Rz   Fz
Dr. Y. Abou-Ali, IUST
&
R
2
Rx
2
 Ry
2
 Rz
University Physics, Chapter 4
Force and Interactions
Example 4.1 ( Superposition of forces):
Three professional wrestlers are fighting over the same champion’s
belt. As viewed from above, they apply the three forces to the belt
that are shown in Fig 4.5a, where the belt is located at the origin.
The magnitude of the three forces are F1= 250 N, F2= 50 N and
F3= 120 N. Find the x- and y-components of the net force on the belt,
and find the magnitude and the direction of the net force.
y

F1
53o
y

F2
F1x
(a)
Dr. Y. Abou-Ali, IUST


R  F
F1y

F3
x
Ry
x
Rx
(b)
University Physics, Chapter 4
Force and Interactions
Solution:
Identify and Set Up: Vector addition, will find the magnitude and
direction of the net force as well its x- and y-components using
Eq.(4.2).
Execute: From Fig. 4.5a, the angles of forces F1, F2, F3 and the
+xaxis are:
θ1 = 180o – 53o = 127o
θ2 = 0
θ3 = 270o
• The x- and y- components of forces F1, F2 and F3 are:
F1x = (250 N) cos 127o = - 150 N
F1y = (250 N) sin 127o = 200 N
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 4
Force and Interactions
F2x = (50 N) cos 0o = 50 N
F2y = (50 N) sin 0o = 0 N
F3x = (120 N) cos 270o = 0 N
F3y = (120 N) sin 2700 = - 120 N
• The net force is:
R  F
Rx  F1x  F2 x  F3x  (150N)  50N  0N  100 N
Ry  F1y  F2 y  F3 y  200N  0N  (120N)  80 N
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 4
Force and Interactions
R
Rx2  Ry2 
(100N)2  (80N)2  128 N
tan 
Ry
Rx
 80N 
  arctan
 arctan 
  arctan(0.80)
Rx
 100N 
Ry
θ = – 39o or 141o
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 4
4.2 Newton’s First Law
Newton’s First Law: A body acted on by no net force moves with
constant velocity (which may be zero) and zero acceleration.
• The tendency of a body to keep moving once it is set in motion
results from a property called Inertia.
push of surface force
net force = 0
gravity force
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 4
4.2 Newton’s First Law
• When a force is perpendicular to the contact surface, we called:
Normal force.
• Zero net force is equivalent to no force at all.
• When a body is acted on by no forces or by several forces that
vector sum is zero, body is in equilibrium.
F  0
 Fx  0
Dr. Y. Abou-Ali, IUST
&
body in equilibrium
 Fy  0
(4.3)
body in equilibrium
(4.4)
University Physics, Chapter 4
4.3 Newton’s Second Law
• In
presence of a net force acting a body causes the body to
accelerate. The direction of the acceleration is the same net force.
• When:
 F  constant  a  constant
 F  ma
• We call the ratio
(4.5)
 F / a  m , the inertial mass or mass.
• The inertia of a body tends to cause the body to resist any change
in its motion.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 4
4.3 Newton’s Second Law
Mass and Force:
• Mass is a quantitative measure of inertia.
• When the mass is high, the body resists more to being accelerated.
• When hit a basketball and a table-tennis ball with the same force,
the acceleration of basketball << the acceleration of table-tennis ball.
• SI unit of mass is the kilogram.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 4
4.3 Newton’s Second Law
Newton: one newton is the amount of net force that gives an
acceleration of one meter per second square to a body with a mass of
one kilometer.
1 N = 1 kg . m/s2
m1a1  m2 a2
Dr. Y. Abou-Ali, IUST
m2
a1


m1
a2
(the same net force) (4.6)
University Physics, Chapter 4
4.3 Newton’s Second Law
 F  ma
(4.7)
• If a net external force acts on a body, the body accelerates. The
direction of acceleration is the same as the direction of net force. The
net for vector is equal to the mass of the body times the acceleration
of the body.
 F The acceleration is in the same direction as the net
a 
m force.
• Use components:
 Fx  max
 Fy  may
 Fz  maz
(4.8)
(Newton’s Second Law)
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 4
4.3 Newton’s Second Law
• Eqs. (4.7) and (4.8) are valid only:
 External force.
 m = constant.
 In inertial frames of reference.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 4
4.3 Newton’s Second Law
Mass and Weight
• The
weight of a body is a familiar force, it is the earth’s
gravitational attraction for the body.
• Mass characterise the inertial properties of a body.
Large mass = Large weight
• What the relationship between mass and weight?
• A freely falling body has an acceleration = g
• F = ma = (1kg)(9.8 m/s2) = 9.8 kg.m/s2 = 9.8 N.
• w = mg magnitude of the weight of a body of mass m.
• The weight of a body is a force, a vector quantity.
w  mg 4.10
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 4
4.3 Newton’s Second Law
Mass and Weight
Notice: force weight is acting on the body all the time.
Variation of g with Location:
• g varies from point to point on the earth’s surface:
g ~ 9.78 – 9.82 m/s2
due to the earth is not perfectly spherical and the effect due to its
relation and orbital motion.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 4
4.3 Newton’s Second Law
Example 4.4 ( Determining acceleration from force):
A worker applies a constant horizontal force with magnitude 20 N to
a box with mass 40 kg resting on a level floor with negligible friction.
What is the acceleration of the box?
Solution:
Identify: This problem involves force and acceleration, using
Newton’s second law.
Set Up: (i) choose a coordinate system (ii) identify all the forces
y
acting on the body in question.
n
F =20 N
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w
x
University Physics, Chapter 4
4.3 Newton’s Second Law
• The forces applied:
(i) horizontal force by the worker (force F).
(ii) force w weight of the box (downward).
(iii) upward supporting force n (normal force) exerted by the floor.
• Friction force = 0
Execute:
 Fx  F  20N
 Fx 20 N 20kg.m/s 2
ax 


 0.50 m/s2
m
40kg
40kg
Read example 4.5/133.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 4
4.5 Newton’s Third Law
FA on B  FB on A
• If body A exerts a force on a body B (an “action”), than
the body B exerts a force on body A (a “reaction”). These
two forces have the same magnitude but are opposite in
direction. These two forces act on different bodies.
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 4
Motion with Constant Acceleration
vx  vox  ax t
(constant acceleration only)
2
1
x  xo  vox t  ax t
2
(constant acceleration only)
2
2
vx  vox  2 a x ( x  xo )
(constant acceleration only)
 vox  vx 
x  xo  
t

 2 
(constant acceleration only)
Dr. Y. Abou-Ali, IUST
University Physics, Chapter 4
Equilibrium
• We say an object is in equilibrium if
all the forces on it balance (add up to
zero). Figure 1.40 shows a beam
weighing 124 N that is supported in
equilibrium by a lOO.O-N pull and a
force F at the floor. The third force on
the beam is the l24-N weight that acts
vertically downward. (a) Use vector
components to find the magnitude and
direction of F. (b) Check the
reasonableness of your answer in part
(a) by doing a graphical solution
approximately to scale.
Solution
Newton’s law
• Two adults and a child want to posh a
wheeled cart in the direction marked x
in Fig. The two adults push with
horizontal forces F 1 and F 2 as shown
in the figure. (a) Find the magnitude
and direction of the smallest force that
the child should exert. You can ignore
the effects of friction. (b) If the child
exerts the minimum force found in
part (a), the cart accelerates at 2.0
m/S2 in the + x-direction. What is the
weight of the cart?
Solution
Solution