Lecture Outline Chapter 5 Physics, 4th Edition James S. Walker

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Lecture Outline
Chapter 5
Physics, 4th Edition
James S. Walker
Copyright © 2010 Pearson Education, Inc.
Chapter 5
Newton’s Laws of Motion
Copyright © 2010 Pearson Education, Inc.
Units of Chapter 5
•  Dynamics
•  Force and Mass
•  Newton’s 1st, 2nd and 3rd Laws of Motion
•  The Vector Nature of Forces: Forces in Two
Dimensions
•  Weight
•  Normal Forces
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Newton’s Concepts
1)  m (mass): How much matter something contains
2)  v (velocity): A body’s speed and direction
3)  a (acceleration): The change in a body’s velocity
per unit time
4)  F (force): any external agent that causes a
change in the motion of a free body
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5-1 Force and Mass
Force: push or pull
Force is a vector – it has magnitude and
direction
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5-1 Force and Mass
Mass is the measure of
how hard it is to
change an object’s
velocity.
Mass can also be
thought of as a
measure of the quantity
of matter in an object.
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5-2 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, an object will keep moving
at a constant speed in a straight line, or
remain at rest.
This is also known as the law of inertia.
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5-2 Newton’s First Law of Motion
In order to change the velocity of an
object – magnitude or direction – a net
force is required.
An inertial reference frame is one in
which the first law is true. Accelerating
reference frames are not inertial.
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5-3 Newton’s Second Law of Motion
Observations show that the acceleration of an object is
proportional to the net force exerted on it and inversely
proportional to its mass:
Or, more familiarly,
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5-3 Newton’s Second Law of Motion
An object may have several forces acting on it;
the acceleration is due to the net force:
(5-1)
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5-3 Newton’s Second Law of Motion
An object may have several forces acting on it;
the acceleration is due to the net force:
(5-1)
In terms of vector components :
∑F
x
= ma x
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∑F
y
= ma y
∑F
z
= ma z
5-3 Newton’s Second Law of Motion
Free-body diagrams:
A free-body diagram shows every force acting
on an object.
•  Sketch the forces
•  Isolate the object of interest
•  Choose a convenient coordinate system
•  Resolve the forces into components
•  Apply Newton’s second law to each
coordinate direction
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5-3 Newton’s Second Law of Motion
Example of a free-body diagram:
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5-3 Newton’s Second Law of Motion
Example 5-1: mboat = 753 kg, F = 80.5N. What are
magnitudes and directions of accelerations in (a) and (b)
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5-3 Newton’s Second Law of Motion
Example 5-1: mboat = 753 kg, F = 80.5N. What are
magnitudes and directions of accelerations in (a) and (b)
(a) Create free - body diagram, isolate source of interest
and sketch the external forces on this body and their components.
3(80.5N)
∑ Fx = ma x ⇒ FL + FM + FC = ma x ⇒ a x = 753kg = 0.321m/s2
(b) ∑ Fx = ma x ⇒ −FL + FM − FC = ma x ⇒ a x =
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€
(−80.5N)
= −0.107m/s2
753kg
5-3 Newton’s Second Law of Motion
Example 5-2: mball = 0.15 kg, v = 90 mi/h and Δx = 2m.
Estimate the force on the ball.
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5-3 Newton’s Second Law of Motion
Example 5-2: mball = 0.15 kg, v = 90 mi/h and Δx = 2m.
Estimate the force on the ball.
Equation of motion along x - direction :
⎛ mi ⎞ 2 ⎛ 1609.3m ⎞ 2
⎟
⎜ 90 ⎟
⎜90
2
v
⎠
⎠
⎝
⎝
h
3600s
=
=
= 400m/s2
v 2 = v 20 + 2a x Δx = 2aΔx ⇒ a =
2Δx
2(2m)
4m
Newton's Second Law :
∑F
x
€
= ma x ⇒ Fx = (0.15kg)(400m/s2 ) = 60N
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5-4 Newton’s Third Law of Motion
Forces always come in pairs, acting on
different objects:
If object 1 exerts a force on object 2, then
object 2 exerts a force – on object 1.
These forces are called action-reaction pairs.
NOTE: The action and reaction forces would
appear in the free-body diagrams of different
objects so they do not cancel out.
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5-4 Newton’s Third Law of Motion
Some action-reaction pairs:
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5-4 Newton’s Third Law of Motion
Example 5-3: m1 = 150 kg, m2 = 250 kg. The force of
person in canoe 1 on canoe 2 is F = 46N. (a) acceleration
of each canoe? (b) separation of canoes after t = 1.2s?
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5-4 Newton’s Third Law of Motion
Example 5-3: m1 = 150 kg, m2 = 250 kg. The force of
person in canoe 1 on canoe 2 is F = 46N. (a) acceleration
of each canoe? (b) separation of canoes after t = 1.2s?
Apply Newton's second law on each canoe :
canoe 1:
canoe 2 :
∑F
∑F
1x
= m1a1
2x
= m 2a 2
Apply Newton's third Law :


F1 = (- 46 N)xˆ F2 = (46 N)xˆ
F1 −46 N
F
46N
=
= −0.31m/s2 a 2 = 2 =
= 0.18m/s2
m1 150 kg
m2 250 kg
Canoe 1 with the smallest mass accelerates more.
a1 =
1 2 1
a1t = (−0.31m/s2 )(1.2s) 2 = −0.22m
2
2
Δx = x 2 - x1 = 0.35m
x1 =
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x2 =
1 2 1
a 2t = (0.18m/s2 )(1.2s) 2 = 0.13m
2
2
5-4 Newton’s Third Law of Motion
Contact forces:
The force exerted by
one box on the other is
different depending on
which one you push.
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5-4 Newton’s Third Law of Motion
Apply Newton's 2nd Law on both boxes :
F
F = (m1 + m2 )a ⇒ a =
(m1 + m2 )
case1: Force on small box 2. Box 2 exerts a force F1 on box 1 and box 1
exerts an equal and opposite force F2 on box 2.
Contact force = F1 = F2
Apply Newton's 2nd law on box 1
(only one external force in x direction on box 1) :
m1F
Contact Force : F1 = m1a =
(m1 + m2 )
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5-4 Newton’s Third Law of Motion
Apply Newton's 2nd Law on both boxes :
F = (m1 + m2 )a ⇒ a =
F
(m1 + m2 )
case2 : Force on large box 1. Box 2 exerts a force F1 on box 1 and box 1
exerts an equal and opposite force F2 on box 2.
Contact force = F1 = F2
Apply Newton's 2nd law on box 2
(only one external force in x direction on box 2) :
m2F
Contact Force : F2 = m2a =
(m1 + m2 )
The contact force is larger when you push on the smaller box!
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Newton’s Laws of Motion
1)  An object remains at rest, or moves in a straight line at
a constant speed, unless acted upon by a net outside
force.
2)  A body’s acceleration is proportional to the net outside
force acting on the object, and will be in the direction of
that force. Its resistance to acceleration depends on its
mass. In equation form, this is
F=ma
m = mass of object (kg)
a = acceleration of object (m/s2)
F = net external force (kg m/s2 = Newton)
3)  Whenever one object exerts a force on a second object,
the second object exerts an equal and opposite force on
the first object.
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5-5 The Vector Nature of Forces: Forces in
Two Dimensions
The easiest way to handle forces in two
dimensions is to treat each dimension
separately, as we did for kinematics.
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5-5 Forces in Two Dimensions
Example 5-5: Jack and Jill. mp = 1.30 kg
F1 = 7.0 N, F2 = 11 N, W = 12.8 N and φ=28°.
(a) θ = ? For ax = 0 (b) ay =?
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5-5 Forces in Two Dimensions
Example 5-5: Jack and Jill. mp = 1.30 kg
F1 = 7.0 N, F2 = 11 N, W = 12.8 N and φ=28°.
(a) θ = ? For ax = 0 (b) ay =?
Use component form of Newton's seconds law :
∑F
x
= ma x
∑F
y
= ma y
a x = 0 ⇒ ∑ Fx = 0 ⇒ −F1sinθ + F2sin28  = 0 ⇒ sinθ =
F2
sin28 
F1
θ = 48 
F1cosθ + F2cos28  - W
∑ Fy = F1cosθ + F2cos28 - W ⇒ a y =
mp

(7.0 N)cos48  + (11.0 N)cos28  -12.8 N
ay =
= 1.2m/s2
1.30 kg
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5-6 Weight
The weight of an object on the Earth’s surface
is the gravitational force exerted on it by the
Earth.
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5-6 Weight
Example 5-6 : Where’s the fire? mf = 97 kg, h = 3.0m, t =
1.2s. What is the upward force from the pole on the
fireman?
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5-6 Weight
Example 5-6 : Where’s the fire? mf = 97 kg, h = 3.0m, t =
1.2s. What is the upward force from the pole on the
fireman?
Apply Newtons' Second Law :
∑F
y
= mf a y ⇒ F − W = mf a y ⇒ F - mf g = mf a y
1
2h
2(3.0m)
2
y = h + a yt 2 ⇒ a y = − 2 = −
2 = −4.2m/s
2
t
(1.2s)
Note that the total acceleration of the fireman is negative
because it is pointed in the - y direction.
F = mf (g + a y ) = 97 kg(9.81m/s2 − 4.2m/s2 ) = 540N
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5-6 Weight
Apparent weight:
Your perception of your weight is based on the
contact forces between your body and your
surroundings.
If your surroundings
are accelerating,
your apparent weight
may be more or less
than your actual
weight.
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5-6 Weight
Apparent weight : The force from the floor on person.
Newton's Second Law on person :
Wa - W = ma y ⇒ Wa = m(g + a y )
Case 1: At rest a y = 0 and the apparent weight is equal to our weight
Case 2 : For elevator accelerating upward ay > 0 and Wa > W
Case 3 : For elevator accelerating downward ay < 0 and Wa < W
€
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5-6 Weight
Example 5-7 : Don’t buy Salmon in an upward
accelerating elevator. What is the scale weight (a)
at rest (b) upward acceleration of 2.5m/s2 (c)
downward acceleration of 3.2m/s2.
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5-6 Weight
Example 5-7 : Don’t buy Salmon in an upward
accelerating elevator
∑F
y
= ma y
Fy = Wa − W = Wa − mg

Wa = m(g + a y )yˆ

(a) At rest a y = 0 and Wa = mgyˆ = 5.0kg(9.81m/s2 )yˆ = 49Nyˆ

(b) Wa = 5.0kg(9.81m/s2 + 2.5m/s2 )yˆ = 62Nyˆ (your're paying too much)

(c) Wa = 5.0kg(9.81m/s2 - 3.2m/s2 )yˆ = 33Nyˆ
€
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Wa = m(g + a)
For a = - g one has zero apparent weight
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5-7 Normal Forces
The normal force is
the force exerted by
a surface on an
object.
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5-7 Normal Forces
The normal force may be equal to, greater than,
or less than the weight.
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5-7 Normal Forces
Example 5-8: Ice Block F1 = 13 N, F2 = 11 N
(a)  What is the acceleration of the ice?
(b) What is the normal force on the ice?
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5-7 Normal Forces
Example 5-8: Ice Block F1 = 13 N, F2 = 11 N, m = 6.0 kg
(a)  What is the acceleration of the ice?
(b) What is the normal force on the ice?
∑F
= ma x
∑F
= F1x + F2x = F1 cos60  − F2 cos 30  = (13N)0.5 - (11N)0.87 = 6.5N - 9.5N = -3N
x
x
∑F
y
= ma y
(a) ma x = - 3 N ⇒ a x =
∑F
y
-3 N

= −0.5m/s2 a = - 0.5 m/s2 xˆ
6.0kg
= 0 ⇒ FN − W − F1y − F2y = 0 ⇒ FN − mg − F1 sin60  − F2 sin 30  = 0
FN = mg + F1 sin60  + F2 sin 30  = (6.0kg)(9.81m/s2 ) + (13 N)(0.87) + (11 N)(0.5)

FN = 75N FN = 75N yˆ
€
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5-7 Normal Forces
The normal force is always perpendicular to the
surface.
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5-7 Normal Forces
Example 5-9: Toboggan, m and θ known
(a)  What is the acceleration of the child?
(b)  What is the normal force on the child?
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5-7 Normal Forces
Example 5-9: Toboggan, m and θ known
(a)  What is the acceleration of the child?
(b)  What is the normal force on the child?
∑F
x
= ma x
∑F
y
= ma y
Wx mgsin θ
∑ Fx = Wx = ma x ⇒ a x = m = m = gsinθ

∑ Fy = FN − Wcosθ = ma y = 0 ⇒ FN = mgcosθ FN = mgcosθyˆ
€
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Summary of Chapter 5
•  Force: a push or pull
•  Mass: measures the difficulty in
accelerating an object
•  Newton’s first law: if the net force on an
object is zero, its velocity is constant
•  Inertial frame of reference: one in which the
first law holds
•  Newton’s second law:
•  Free-body diagram: a sketch showing all the
forces on an object
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Summary of Chapter 5
•  Newton’s third law: If object 1 exerts a force
on object 2, then object 2 exerts a force – on
object 1.
•  Contact forces: an action-reaction pair of
forces produced by two objects in physical
contact
•  Forces are vectors
•  Newton’s laws can be applied to each
component of the forces independently
•  Weight: gravitational force exerted by the Earth
on an object
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Summary of Chapter 5
•  On the surface of the Earth, W = mg
•  Apparent weight: force felt from contact with a
floor or scale
•  Normal force: force exerted perpendicular to a
surface by that surface
•  Normal force may be equal to, lesser than, or
greater than the object’s weight
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