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```					            Lecture 6
Ch5. Newton’s Law of Motion

University Physics: Mechanics
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
Announcement
Next week
1st half  : Lecture 7
2nd half  : Quiz 2
Material: Lecture 5 & 6
Over next week
1st half  : Lecture 8, Make-up Quiz 2
2nd half  : Discussion Quiz 2

Erwin Sitompul                         University Physics: Mechanics   6/2
What Causes an Acceleration?
 Out of common experience, we know that any change in
velocity must be due to an interaction between an object
(a body) and something in its surroundings.
 An interaction that can cause an acceleration of a body is
called a force. Force can be loosely defined as a push or pull
on the body.
 The relation between a force and
the acceleration it causes was first
understood by Isaac Newton.
 The study of that relationship is
called Newtonian mechanics.
 We shall now focus on its three
primary laws of motion.

Sir Isaac Newton
(1642—1727)
Erwin Sitompul                           University Physics: Mechanics   6/3
Newton’s First Law
 Newton’s First Law:
“If no force acts on a body, then the body’s velocity cannot
change, that is the body cannot accelerate.”
 In other words, if the body is at rest, it stays at rest. If the body
is moving, it will continue to move with the same velocity
(same magnitude and same direction).

Erwin Sitompul                           University Physics: Mechanics   6/4
Force

 A force can cause the acceleration of a body.
 As the standard body, we shall use the standard kilogram. It
is assigned, exactly and by definition, a mass of 1 kg.
 We put the standard body on a horizontal frictionless surface
and pull the body to the right, so that it eventually
experiences an acceleration of 1 m/s2.
 We can now declare, as a matter of definition, that the force
we are exerting on the standard body has a magnitude of
1 newton (1 N).

Erwin Sitompul                       University Physics: Mechanics   6/5
Force
 Forces are vector quantities. They have magnitudes and
directions.
 Principle of Superposition for Forces:
A single force with the magnitude and direction of the net
force acting on a body has the same effect as all the
individual forces acting together.
 Newton’s First Law: (proper statement)
→
“If no net force acts on a body (Fnet = 0), then the body’s
velocity cannot change, that is the body cannot accelerate.”

Erwin Sitompul                        University Physics: Mechanics   6/6
Net Force Calculation using Vector Sum

Fnet  ma
FA  FB  FC  0

FA, x  FB , x  FC , x  0
FA, y  FB , y  0

FC  FC , x ˆ   ( FA, x  FB, x ) ˆ
i                       i

Erwin Sitompul                 University Physics: Mechanics   6/7
Checkpoint
Which of the following six arrangements correctly show the
→      →
vector addition of forces F1 and F2 to yield the third vector,
→
which is meant to represent their net force Fnet?

Erwin Sitompul                          University Physics: Mechanics   6/8
Mass
 Mass is a scalar quantity.
 Mass is an intrinsic characteristic of a body.
 The mass of a body is the characteristic that relates a force
on the body to the resulting acceleration.
 A physical sensation of a mass can only be obtained when
we attempt to accelerate the body.

Erwin Sitompul                        University Physics: Mechanics   6/9
Newton’s Second Law
 Newton’s Second Law:
“The net force on a body is equal to the product of the body’s
mass and its acceleration.”
 The Newton’s second law in equation form
F net  ma
→
 It the net             on a body is zero, then the body’s
force Fnet
→
acceleration a is zero
 If the body is at rest, it stays at rest.
 If it is moving, it continues to move at constant velocity.

 1 N  (1 kg)(1 m s 2 )  1 kg  m s
2

Erwin Sitompul                            University Physics: Mechanics   6/10
Newton’s Second Law
→
 The vector equation Fnet = ma is equivalent to three
component equation, one written for each axis of an xyz
coordinate system:
Fnet,x  ma x , Fnet,y  ma y ,   Fnet,z  maz .

 The acceleration component along a given axis is caused
only by the sum of the force components along that same
axis, and not by force components along any other axis.

Erwin Sitompul                       University Physics: Mechanics   6/11
Free-Body Diagram
 The most important step in solving problems involving
Newton’s Laws is to draw the free-body diagram.
 Only the forces acting on the object of interest should be
included in a free-body diagram.

Fon book from hand

Fon book from earth

Erwin Sitompul                        University Physics: Mechanics   6/12
Free-Body Diagram

The system of interest is the cart

The free-body diagram of the cart

Erwin Sitompul                           University Physics: Mechanics   6/13
Example: Puck (Ice Hockey “Ball”)
Three situations in which one or two forces act on a puck that
moves over frictionless ice along an x axis, in one-dimensional
motion, are presented here.
→       →
The puck’s mass is m = 0.2 kg. Forces F1 and F2 are directed
along the axis and have magnitudes F1 = 4 N and F2 = 2 N.
→
Force F3 is directed at angle θ = 30° and has magnitude
F3 = 1 N.
In each situation, what is the acceleration of the puck?

F1  ma x

F1   4N
ax           20 m s 2
m 0.2 kg

Erwin Sitompul                        University Physics: Mechanics   6/14
Example: Puck (Ice Hockey “Ball”)
The puck’s mass is m = 0.2 kg. Forces F1 and F2 are directed
along the axis and have magnitudes F1 = 4 N and F2 = 2 N.
Force F3 is directed at angle θ = 30° and has magnitude
F3 = 1 N.

F3,x  F2  max
F3,x  F2
ax 
m
F3 cos   F2
F1  F2  max                          
m
4 N2 N
ax                                      (1 N)(cos 30)  2 N
0.2 kg                          
0.2 kg
 10 m s2                           5.67 m s 2

Erwin Sitompul                      University Physics: Mechanics   6/15
Some Particular Forces
 The Gravitational Force →
The gravitational force Fg on a body is a force that pulls on the
body, directly toward the center of Earth (that is, directly down
toward the ground.
F g  mg
Fg  mg

 The Weight
The weight W of a body is equal to the
magnitude Fg of the gravitational force
on the body
W  mg

Erwin Sitompul                         University Physics: Mechanics   6/16
Some Particular Forces
 The Normal Force
When a body presses against a surface, the surface (even a
seemingly rigid one)→deforms and pushes back on the body
with a normal force FN that is perpendicular to the surface.
 In mathematics, normal means perpendicular.

FN                According to Newton’s
second law,
FN  Fg  ma y
FN
FN  mg  m(0)
• Why?
FN  mg

Forces on a Body,
Resting on a Table
Erwin Sitompul                       University Physics: Mechanics   6/17
Some Particular Forces
 Friction                                          →
The frictional force or simply friction is a force f that resists
the motion when we slide or attempt to slide a body over a
surface.
 Friction is directed along the surface, opposite the direction of
the intended motion.

Erwin Sitompul                         University Physics: Mechanics   6/18
Some Particular Forces
 Tension
When a cord (or a rope, cable, or other such object) is
attached to a body and tensed, the cord pulls on the body with
→
a force T directed away from the body.
 The force is often called a tension force. The tension in the
→
cord is the magnitude T of the force on the body.

• A cord is considered as
massless and unstretchable
• A pulley is considered as
massless and frictionless
Erwin Sitompul                          University Physics: Mechanics   6/19
Beware of High Energy Concentration

Erwin Sitompul                 University Physics: Mechanics   6/20
Forms of Energy

Erwin Sitompul   University Physics: Mechanics   6/21
Newton’s Third Law
 Newton’s Third Law:
“When two bodies interact, the forces on the bodies from each
other are always equal in magnitude an opposite in direction.”
FBC  FCB
(Equal Magnitudes)

FBC  FCB
(Equal Magnitudes and
Opposite Directions)

FBC : The force on the book B from the box C
FCB : The force on the box C from the book B

Erwin Sitompul                       University Physics: Mechanics   6/22
Applying Newton’s Law: Problem 1
A block S (the sliding block) with mass M =3.3 kg is free to
move along a horizontal frictionless surface. It is connected by a
cord that wraps over a frictionless pulley, to a second block H
(the hanging block) with mass m = 2.1 kg. The cord and pulley
are considered to be “massless”.
The hanging block H falls as the sliding block S accelerate to
the right. Find:
(a) the acceleration of block S
(b) the acceleration of block H
(c) the tension in the cord

Erwin Sitompul                        University Physics: Mechanics   6/23
Applying Newton’s Law: Problem 1
FN

FN

The Forces Acting        Free-Body Diagram for
On The Two Blocks         Block S and Block H

Erwin Sitompul                     University Physics: Mechanics   6/24
Applying Newton’s Law: Problem 1
 The cord does not stretch,
Fnet,x  Max        so ax of M and ay of m have
T  Ma x         the same magnitude.
FN                           2        ax of M  ay of m  a
Fnet,y  Ma y
 The tension at M and the
FN  Fg S  0           tension at m also have the
FN  Fg S         same magnitude.
1      T  FgH  ma
• Why?
T  mg  ma
2    Ma  mg  ma
Fnet,y  ma y                       m
a       g
T  FgH  ma y                        M m
1

Erwin Sitompul                                  University Physics: Mechanics   6/25
Applying Newton’s Law: Problem 1
m          2.1
a        g            9.8  3.81m s
2

M m     3.3  2.1

(a) the acceleration of block S
ˆ
aS  3.81i m s 2

(b) the acceleration of block H
ˆ
aH  3.81j m s 2

(c) the tension in the cord
T  Ma  (3.3)(3.81)  12.573 N

Erwin Sitompul                               University Physics: Mechanics   6/26
Applying Newton’s Law: Problem 2
A cord pulls on a box of sea biscuits up
along a frictionless plane inclined at θ = 30°.
The box has mass m = 5 kg, and the force
from the cord has magnitude T = 25 N.
What is the box’s acceleration component a
along the inclined plane?

Free-Body Diagram              Free-Body Diagram,
→
of the Box                  Fg in components
Erwin Sitompul                          University Physics: Mechanics   6/27
Applying Newton’s Law: Problem 2
Fnet,x  max
T  mg sin   max
T  mg sin 
ax 
m
25  (5)(9.8)sin 30

5
 0.1 m s  2

• What is the meaning
of this value?
What is the force               Fnet,y  ma y
exerted by the plane    FN  mg cos   ma y
on the box?                                    No motion in
FN  mg cos   m(0) • y direction
FN  mg cos 
 (5)(9.8) cos30
 42.435 N
Erwin Sitompul                        University Physics: Mechanics   6/28
Homework 5: Two Boxes and A Pulley
A block of mass m1 = 3.7 kg on a frictionless plane inclined at
angle θ = 30° is connected by a cord over a massless,
frictionless pulley to a second block of mass m2 = 2.3 kg.
What are:
(a) the magnitude of the acceleration of each block,
(b) the direction of the acceleration of the hanging block, and
(c) the tension in the cord?
Hint: Draw the free-body diagram of m1 and m2 first.

Erwin Sitompul                        University Physics: Mechanics   6/29

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