Lecture 7
Ch5. Newton’s Law of Motion
University Physics: Mechanics
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
2012
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)
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Force in Various Forms
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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).
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Mass, Motion & Force
 Experimental Observations:
 An object acted on by a net force accelerates in the same
direction as the net force.
 Some objects accelerate slower or faster than others when
subjected to the same force  inertial mass.
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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).
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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.”
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Net Force Calculation using Vector Sum
Fnet  m a
FA  FB  FC  0
FA, x  FB , x  FC , x  0
FA, y  FB , y  0
FC  FC , x ˆi   ( FA, x  FB, x ) ˆi
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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?
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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.
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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
→
force
F
→ net
 It the net
on a body is zero, then the body’s
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
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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,z  maz .
Fnet,x  max , Fnet,y  ma y ,
 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.
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3 Astronauts and an Asteroid
Three astronauts, propelled by jet backpacks,
push and guide a 120 kg asteroid toward a
processing dock, exerting the forces shown in
the figure, with F1 = 32 N, F2 = 55 N, and F3 =
41 N. What is the (a) magnitude and (b)
angle of the asteroid's acceleration?
F1  (32 N)(cos30 ˆi  sin 30 ˆj)
F2  (55 N) ˆi
F  (41N)(cos(60) ˆi  sin(60) ˆj)
3
Fnet 103.21 ˆi  19.51 ˆj

 0.860 ˆi  0.163 ˆj m s 2
Fnet  m a  a 
120
m
1  0.16 
2
2
  tan 
a  (0.860)  ( 0.163)
  10.73
 0.86 
 0.88 m s 2
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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
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Free-Body Diagram
The system of interest is the cart
The free-body diagram of the cart
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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  max
F1
4N
ax 
 20 m s2

m 0.2 kg
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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
ax 
F1  F2  max
4 N2 N
ax 
0.2 kg
 10 m s 2
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F3,x  F2
m
F3 cos   F2

m
(1 N)(cos 30)  2 N

0.2 kg
 5.67 m s2
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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
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University Physics: Mechanics
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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.
According to Newton’s
second law,
FN
FN
FN  Fg  ma y
FN  mg  m(0)
FN  mg
• Why?
Forces on a Body,
Resting on a Table
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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.
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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
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Beware of High-Energy Tension
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Various Forms of Energy
• Energy can convert from one form to the other.
• Energy in higher level (more concentrated) tends to
convert to lower (less concentrated) condition.
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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
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University Physics: Mechanics
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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
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Applying Newton’s Law: Problem 1
FN
FN
The Forces Acting
On The Two Blocks
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Free-Body Diagram for
Block S and Block H
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Applying Newton’s Law: Problem 1
Fnet,x  Max
T  Max
FN
 The cord does not stretch,
so ax of M and ay of m have
the same magnitude.
ax of M  ay of m  a
2
Fnet,y  Ma y
FN  Fg S  0
FN  Fg S
 The tension at M and the
tension at m also have the
same magnitude.
1
T  FgH  ma
T  mg  ma
• Why?
2 Ma  mg  ma
Fnet,y  ma y
T  FgH  ma y
1
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a
m
g
M m
T
Mm
g
M m
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Applying Newton’s Law: Problem 1
a
m
2.1
g 
9.8  3.81m s 2
M m
3.3  2.1
(a) the acceleration of block S
aS  3.81iˆ m s 2
(b) the acceleration of block H
aH  3.81jˆ m s 2
(c) the tension in the cord
T  Ma  (3.3)(3.81)  12.573 N
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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
of the Box
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Free-Body
Diagram,
→
Fg in components
University Physics: Mechanics
7/29
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
2
 0.1 m s
• What is the meaning
of this value?
What is the force
exerted by the plane
on the box?
Fnet,y  ma y
FN  mg cos   ma y
No motion in
FN  mg cos   m(0) • y direction
FN  mg cos 
 (5)(9.8) cos 30
 42.435 N
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University Physics: Mechanics
7/30
Homework 6: 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.
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Homework 6A
Two blocks of mass 3.5 kg and 8.0 kg are connected by a
massless string that passes over a frictionless pulley (see
figure below). The inclines are frictionless.
When the assembly is released from rest, find
(a) the magnitude of the acceleration of each block; and
(b) the tension in the string.
(c) the increase of elevation of the climbing block (which one?)
after 0.4 s.
Hint: Draw the free-body diagram of the two blocks.
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Homework 6B
1. A 2.75-kg cat moves in a straight line (the
x-axis). The figure below shows a graph
of the x-component of this cat’s velocity
as a function of time. (a) Find the
maximum net force on this cat. When
does this force occur?; (b) When is the
net force on the cat equal to zero?; (c)
What is the net force at time 8.5 s?
2. Three rainbow boxes are attached by cords,
one of which wraps over a frictionless pulley
with negligible mass. The three masses are
mA = 30 kg, mB = 40 kg, and mC = 10 kg.
When the assembly is released from rest,
determine (a) the tension in the cord
connecting B and C, and (b) how far A moves
in the first 0.25 s (assuming it does not reach
the pulley)?
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