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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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The Four Fundamental
Interactions
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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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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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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
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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
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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
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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
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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
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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
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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
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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 ma y
N mg ma
a=2 m/s2
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
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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:
F3, y
q tan
F
3
,
x
1 7.35 N
60
tan
4.24 N
F1=9.8 N
1
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Does it
work?
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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
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
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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
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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
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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°
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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
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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
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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
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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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