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Chapter 5
Forces (Part the First)
Forces
“A push or a pull”
 Can cause acceleration
 Aristotelian (wrong) vs Newtonian (mostly
right) mechanics

 Limitations
of Newtonian mechanics
Objects moving very fast (special relativity)
 Very small objects (quantum mechanics)
 Very small objects moving very fast (quantum field
theory)

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2
Forces

Unit: kg m/s2 [Newtons]

Force is a vector quantity
 Direction

and magnitude
The sum of the forces on an object is Fnet
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3
The Four Fundamental
Interactions
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4
The Practical Forces of Nature
Gravity
 Restoring forces

 Springs,

Constraining forces
 Tension,

etc.
normal force
Dissipative forces
 Friction,
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air resistance
5
Newton’s 1st Law


If there are no forces acting on an
object, the velocity of the object will
not change
The “law of inertia”
 An
object in motion will remain in
motion, and object at rest will remain at
rest, until acted upon by an outside
force

Then why do things seem to slow
down when we stop pushing them?
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Newton’s First Law vs Hollywood
Is this possible?
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7
Inertial Reference Frame

A reference frame in which Newton’s laws
hold
 Non-accelerating

The earth is approximately an inertial
reference frame
 Rotation,
motion around the
sun/galaxy/universe
 Coriolis effect
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8
Newton’s 2nd Law

The net force on a body is equal to the
product of the body’s mass and the
acceleration of the body


Fnet  ma

Can break force into components
 Treat
each direction independently
Fnet , x  max
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Fnet , y  may
Fnet , z  maz
9
What is Mass?
Intrinsic property
 NOT dependent on an objects size, weight
or density
 Proportionality constant between force and
acceleration

 Difference
between pushing a tennis ball and
a bowling ball
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Newton’s 3rd Law
When object A exerts a force on object B,
object B exerts a force on object A of equal
magnitude and opposite direction
 “Action and reaction”

A
B


FA on B   FB on A
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Forces: Gravity

An object of mass m in freefall will experience a
force:


Fg  m g

Weight is equal to magnitude of Fg
 A peculiarity
in Halliday, Resnick, and Walker
W mg

Note: Mass is intrinsic, while weight depends on
where you are. They are NOT the same thing.
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Demo: Newton’s Carts

Two people push off each other
 The
force is the same in magnitude, opposite in
direction


FA on B   FB on A


m A a A   mB a B
FB on A
FA on B
A
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B
13
Demo: Newton’s Carts

Double the mass on one cart. What happens?


FA on B   FB on A


m A a A   mB a B
FB on A
A
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FA on B
B
14
Forces: Normal


“Normal” in the mathematical sense meaning
perpendicular
When an object pushes against a surface, the
surface pushes back perpendicular to its surface
with a force N
N
Fg
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Forces: Friction



Force that resists motion when an object slides
over a surface
Always opposite to the direction of motion
We’ll discuss more next chapter
v
Ffr
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Forces: Tension



Force exerted by a rope
Always points in the same direction as the rope
Ropes are assumed to be massless unless
otherwise specified
v
T
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Ffr
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Tension (Continued)
T
Frope
Frope
1. A rope can only pull (cannot push)
2. Objects at the ends of a taut rope have the same
magnitude velocity and acceleration
3. Frictionless and massless
pulleys only change direction
of rope, not the tension.
T
F
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T
F
18
Free Body Diagrams
The first step in solving any force problem:
1.
2.
3.
4.
Sketch the object in question
Draw an arrow for each force acting on the
object
Label each force
Indicate the direction of acceleration off to the
side (acceleration is NOT a force)
F2
F1
F3
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a
19
Net Force
A free body diagram is used to calculate the net
force on one object.
Fnet= m a
Note: The two equal forces in Newton’s Third
Law are on different objects. They don’t appear
on the same free body diagram.
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How to Solve a Force Problem
1.
2.
3.
4.
5.
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Draw a free body diagram
Guess at forces of unknown magnitude or
direction
Break forces into components
Sum of all the forces in one direction = mass *
acceleration in that direction
i.e. Fnet,x=m ax
Repeat step 3 for each direction
Solve for unknown quantities
21
Example: A Standing Person
Say a man has a mass of 60 kg. Find the normal
force exerted on him by the ground.
1) Draw a diagram
N
2) Sum the forces in the y direction
a=0
Fnet , y  may
N  mg  0
N  mg  (60 kg)(9.8 m / s 2 )  588 N
Fg=mg
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Note: We know the direction of the
normal force from the diagram
22
Example: A Standing Person
Now say our 60 kg man is jumping with a = 2 m/s2
1) Draw a diagram
N
2) Sum the forces in the y direction
Fy  may
a=2 m/s2
N  mg  ma
N  mg  ma  m( g  a)


N  (60 kg) (9.8 m / s 2 )  (2 m / s 2 )  708 N
Fg=mg
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Notice: The normal force is variable
23
Example: Ring on a Table
Ring centered on round table and strings with
weights are attached
0.5 kg
How much force is
needed to balance
forces so the ring
stays centered (zero
acceleration)?
30°
1 kg
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Solution: Draw a Free-Body Diagram
A free body diagram
shows all the forces
operating on a single
object
F3,y
F2
4.9 N
We had to make an
educated guess for where
to put F3. If we guessed
wrong, that’s OK. We will
just pick up a minus sign
F3
q
30°
F3,x
F1
9.8 N
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Solution: Break Forces Into Components
F3,y
F1 :
F3
F1, x  0
F1, y  9.8 N
F2
4.9 N
30°
F3,x
F2 :
F2, x  (4.9 N ) cos 30
 4.24 N
F2, y  (4.9 N ) sin 30
 2.45 N
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F1
9.8 N
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Solution: Summing Forces
In the y-direction:
In the x-direction:
Fnet , x  max
Fnet , y  may
F1, x  F2, x  F3, x  0
F1, y  F2, y  F3, y  0
0  4.24 N  F3, x  0
 9.8 N  2.45 N  F3, y  0
F3, x  4.24 N
F3, y  7.35 N
F3,y
F3
F2
4.9 N
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30°
F3,x
F1=9.8 N
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Solution:
The magnitude of F3 is:
F3  F3, x  F3, y
2
F3,y
F3
2
 (4.24 N ) 2  (7.35 N ) 2
F2
4.9 N
30°
q
F3,x
 8.5 N  (0.87 kg) g
The angle of F3 is:
F1=9.8 N
 F3, y 

q  tan 

F
3
,
x


1
 7.35 N 
  60
 tan 
 4.24 N 
1
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Does it
work?
28
Pig on Frictionless Inclined Plane
N
a
Fg
q
Assume the pig has a mass of m and the angle of
the ramp is q. Find N and a.
Need to consider forces || and  to plane…
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Pig Forces: Break Into Components
Acceleration:
y
ax  a
ay  0
N
x
a
Normal Force:
Nx  0
Ny  N
θ
Fg,y
θ Fg
Gravity:
Fg , x  Fg sin q  mg sin q
Fg,x
Fg , y  Fg cosq  mg cosq
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Pig Forces: Sum Forces
y-direction:
Fnet , y  may
N y  Fg , y  0
y
N
x
a
N  mg cos q  0
N  mg cosq
x-direction:
Fnet , x  max
θ
Fg,y
θ Fg
Fg,x
Fg , x  max
mg sin q  ma
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a  g sin q
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Pig on a Plane: Sanity Check
N
a
q
Fg
N = mg cos θ
θ= 0
θ = 90°
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

and
a = g sin θ
a=0
and N = mg
a=g
and N = 0
32
Cart on a Plane
Now let’s add friction. A cart of mass m is
stationary on an inclined plane of angle q.
Find Ffr and N.
N
Ffr
Fg
q
a=0
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Cart on a Plane: Components
Normal Force:
N
Nx  0
Ny  N
Gravity:
Ffr
Fg,y
q
Fg
Fg,x
Fg , x  Fg sin q  mg sin q
Fg , y  Fg cosq  mg cosq
Friction:
F fr , x  F fr
Once again, use a “tilted”
coordinate system
F fr , y  0
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Cart: Summing the Forces
y-direction:
Fnet , y  may
N
Ffr
N y  Fg , y  0
N  mg cos q  0
N  mg cosq
Fg,y
q
Fg
Fg,x
x-direction:
Fnet , x  max
Fg , x  F fr , x  max
F fr  mg sin q  0
F fr  mg sin q
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Does it Work?
If we replace the normal
force and friction with another
force, the cart should stay
stationary after the ramp is
removed.
The mass of the cart is
1 kg and the angle of
the ramp is 30°
N  mg cosq  (1 kg) g cos 30  (0.87 kg) g
Ffr  mg sin q  (1 kg) g sin 30  (0.5 kg) g
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Tension
A 16 lb. bowling ball is hung from a rope attached to a scale
which is attached to a wall. The scale reads 16 lbs.
It is now attached to another bowling ball. What does it read?
A) 0 B) 16 lbs C) 32 lbs
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Tension in Two Dimensions
T3
30o
60o
T2
30o
60o
T1
Consider the mass:
T1
m
Summing forces:
Fnet , y  ma y  0
m
mg
T1  mg  0
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T1  mg
38
Tension in 2-D (Continued)
Consider the knot:
T3
T3
60o
30o
T2
30o
60o
m
30o
60o
T1
y-direction:
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T2
T1
x-direction:
Fnet , y  may  0
Fnet , x  max  0
T2 sin 60  T3 sin 30  T1  0
T2 cos 60  T3 cos 30  0
T2 sin 60  T1
T3 
sin 30
cos 30
T2  T3
cos 60
39
Tension in 2-D (Continued)
What we know:
T1  mg
cos 30
T2  T3
cos 60
T2 sin 60  T1
T3 
sin 30
Jumping through the algebraic
hoops (left as an exercise for the
ambitious student):
mg
T3 
 0.5 mg



cos 30 tan 60  sin 30
T2  0.866 mg
T1  mg
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Example: Fuzzy Dice
With no acceleration
Frope
Fnet,y = may = 0
Fg
Frope- Fg = 0
Frope= -Fg = mg
What happens when we accelerate?
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Fuzzy Dice…
Find q:
q
Frope
x direction:
Frope sin q  ma
y direction:
ma

Frope cosq mg
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Fnet , y  ma y  0
Frope cos q  mg  0
Fg
Frope sin q
Fnet , x  max
Frope cos q  mg
a
tan q 
g
e.g. a=g
 θ=45°
42
Another Example:
Find the acceleration of this
system:
Start with free body
diagrams of both blocks
a1
N
T
T
m2
q
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a2
x
m2
FG
Recall: a1 = a2 = a
y
FG
43
Solution:
Summing forces in the x direction:
a1
N
T
Fnet , x  ma1, x
T  m1 g sin q  m1a
T  m1 (a  g sin q )
Summing forces in the y direction:
FG
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N  m1 g cosq  0
N  m1 g cosq
44
Solution
y
T
a2
x
Summing forces in the y direction:
Fnet , y  m2 a2, y
T  m2 g  m2 a
m2
T  m2 (a  g )
FG
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Solution
m2
What we have:
N  m1 g cosq
T  m1 (a  g sin q )
T  m2 (a  g )
q
The tensions must be equal to
each other
m2 (a  g )  m1 (a  g sin q )
m2 a  m2 g  m1a  m1 g sin q
a (m1  m2 )  m2 g  m1 g sin q
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m2 g  m1 g sin q
a
m1  m2
46
Fan Cart With a Sail: Will It Move?
A) Yes, in the same direction as without the sail
B) Yes, in the opposite direction
C) No
D) No Clue
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