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Chapter 4, Part 1
Newton’s Laws of Motion
Copyright © 2010 Pearson Education, Inc.
Force
Force: push or pull
Force is a vector – it has magnitude and
direction
Copyright © 2010 Pearson Education, Inc.
Newton’s First Law of Motion
If you stop pushing an object, does it stop
moving?
Only if there is friction! In the absence of any
net external force, a moving object will keep
moving at a constant speed in a straight line.
This is also known as the law of inertia:
objects in motion tend to stay in motion;
objects at rest tend to stay at rest.
In order to change the velocity of an object –
magnitude or direction – a net force is required.
Copyright © 2010 Pearson Education, Inc.
Newton’s Second Law of Motion
Acceleration is directly proportional to force:
Copyright © 2010 Pearson Education, Inc.
“Forces and Motion”
Copyright © 2010 Pearson Education, Inc.
Newton’s Second Law of Motion
An object may have several forces acting on it;
the acceleration is due to the net force:
Copyright © 2010 Pearson Education, Inc.
Forces and the Laws of
Motion
•  Problem
–  A 873-kg (1930-lb) dragster, starting from
rest, attains a speed of 26.3m/s (58.9 mph)
in 0.59 s. What is the magnitude of the
average net force on the dragster during
this time?
Copyright © 2010 Pearson Education, Inc.
Forces and the Laws of
Motion
•  Solution
m = 873kg
v f − vi 26.3m / s − 0m / s
vf = 26.3m/s a =
=
= 44.58m / s 2
t
0.59s
vi = 0m/s
t = 0.59s
2
4
F = ma = (873kg )(44.58m / s ) = 3.9 x10 N
Copyright © 2010 Pearson Education, Inc.
Newton’s Third Law of Motion
For every action, there is an equal and
opposite reaction.
Forces always come in pairs, acting on
different objects. These forces are called
action-reaction pairs.
Copyright © 2010 Pearson Education, Inc.
Newton’s Third Law of Motion
Some action-reaction pairs:
Copyright © 2010 Pearson Education, Inc.
Free-body diagrams:
A free-body diagram shows every force
acting on an object.
•  Isolate the object of interest
•  Sketch the forces as vectors
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Example of a free-body diagram:
Copyright © 2010 Pearson Education, Inc.
Weight
The weight of an object on the Earth’s surface
is the gravitational force exerted on it by the
Earth.
Fw
Fw
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(g is the absolute value of Earth’s
gravitational acceleration, –9.81 m/s2)
Normal Forces
The normal force is
the force exerted
by a surface on an
object.
Fw
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Normal Forces
The normal force may be equal to, greater than,
or less than the weight.
Fw
Fw
Copyright © 2010 Pearson Education, Inc.
Fw
Copyright © 2010 Pearson Education, Inc.
Fw
Normal Forces
The normal force is always perpendicular to the
surface.
Copyright © 2010 Pearson Education, Inc.
Forces and the Laws of
Motion
•  Example Problem
If two horizontal forces of 225N and 165N are
exerted in opposite directions on a crate,
what is the net force on the crate?
Copyright © 2010 Pearson Education, Inc.
Forces and the Laws of
Motion
•  Solution
225N
Copyright © 2010 Pearson Education, Inc.
165N
Forces and the Laws of
Motion
•  Solution
60N
Copyright © 2010 Pearson Education, Inc.
Forces and the Laws of
Motion
•  Problem
On Earth, a scale shows that Ben weighs
712N. What is Ben’s mass?
Copyright © 2010 Pearson Education, Inc.
Forces and the Laws of
Motion
•  Solution
Fw = 712N
g = a = 9.81m/s2
Fw
712N
m=
=
= 72.6kg(160lb)
2
g 9.81m / s
Copyright © 2010 Pearson Education, Inc.
Forces and the Laws of
Motion
•  Problem
The acceleration of gravity on the moon is
–1.60m/s2. What would the scale indicate
that Ben weighs if he were on the moon
and his mass is the same as in the
previous question? (72.6kg)
Copyright © 2010 Pearson Education, Inc.
Forces and the Laws of
Motion
•  Solution
m = 72.6kg
gmoon = 1.60m/s2
2
Fw = mgmoon = (72.6kg)(1.60m / s ) = 116N
Copyright © 2010 Pearson Education, Inc.
Forces and the Laws of
Motion
Male lions and human sprinters can both
accelerate at about 10.0 m/s2. If a typical
lion’s mass is 170 kg and a typical sprinter’s
mass is 75.0 kg, what is the difference in the
force exerted on the ground during a race
between these two species?
Copyright © 2010 Pearson Education, Inc.
Forces and the Laws of
Motion
•  Solution
ml = 170 kg
mh = 75.0kg
a = 10.0m/s2
Flion − Fhuman
mlion a − mhumana
(170kg)(10.0m / s 2 ) − (75.0kg)(10.0m / s 2 ) = 950N
Copyright © 2010 Pearson Education, Inc.
Homework
pp. 126-127 Multiple Choice 1-11 (odd)
pp. 128-132 9, 11, 17, 29, 39, 43
Copyright © 2010 Pearson Education, Inc.
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