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Mechanics
Mechanics
Kinematics
Dynamics
Force
Fundamental Forces
Comparing Contact and Field Forces
Concept Check – Newton’s 1st Law
A book is lying at rest on a table. The book will remain there at rest
because:
1. there is a net force but the book has too much inertia
2. there are no forces acting on it at all
3. it does move, but too slowly to be seen
4. there is no net force (SF= 0) on the book
5. there is a net force, but the book is too heavy to move
Concept Check – Newton’s 1st Law
A book is lying at rest on a table. The book will remain there at rest
because:
1. there is a net force but the book has too much inertia
2. there are no forces acting on it at all
3. it does move, but too slowly to be seen
4. there is no net force (SF= 0) on the book
5. there is a net force, but the book is too heavy to move
There are forces acting on the book, but the only forces acting are in
the y-direction. Gravity acts downward, but the table exerts an upward
force that is equally strong, so the two forces cancel, leaving no net
force.
Concept Check – Newton’s 1st Law (2)
A hockey puck slides on ice at constant velocity. What is the net force
acting on the puck?
1. more than its weight
2. equal to its weight
3. less than its weight but more than zero
4. depends on the speed of the puck
5. zero
Concept Check – Newton’s 1st Law (2)
A hockey puck slides on ice at constant velocity. What is the net force
acting on the puck?
1. more than its weight
2. equal to its weight
3. less than its weight but more than zero
4. depends on the speed of the puck
5. zero
The puck is moving at a constant velocity, and therefore it is not
accelerating. Thus, there must be no net force acting on the puck.
Follow-up: Are there any forces acting on the puck? What are they?
Concept Check – Newton’s 1st Law (3)
You put your book on the bus seat next to you. When the bus stops
suddenly, the book slides forward off the seat. Why?
1. a net force acted on it
2. no net force acted on it
3. it remained at rest
4. it did not move, but only seemed to
5. gravity briefly stopped acting on it
Concept Check – Newton’s 1st Law (3)
You put your book on the bus seat next to you. When the bus stops
suddenly, the book slides forward off the seat. Why?
1. a net force acted on it
2. no net force acted on it
3. it remained at rest
4. it did not move, but only seemed to
5. gravity briefly stopped acting on it
The book was initially moving forward (since it was on a moving bus).
When the bus stopped, the book continued moving forward, which was
its initial state of motion, and therefore it slid forward off the seat.
Follow-up: What is the force that usually keeps the book on the seat?
Concept Check – Newton’s 1st Law (4)
You kick a smooth flat stone out on a frozen pond. The stone slides,
slows down and eventually stops. You conclude that:
1. the force pushing the stone forward finally stopped pushing on it
2. no net force acted on the stone
3. a net force acted on it all along
4. the stone simply “ran out of steam”
5. the stone has a natural tendency to be at rest
Concept Check – Newton’s 1st Law (4)
You kick a smooth flat stone out on a frozen pond. The stone slides,
slows down and eventually stops. You conclude that:
1. the force pushing the stone forward finally stopped pushing on it
2. no net force acted on the stone
3. a net force acted on it all along
4. the stone simply “ran out of steam”
5. the stone has a natural tendency to be at rest
After the stone was kicked, no force was pushing it along! However,
there must have been some force acting on the stone to slow it down
and stop it. This would be friction.
Newton’s First Law
Newton’s First Law
Concept Check – Newton’s First Law
Consider a cart on a horizontal frictionless table. Once the cart has
been given a push and released, what will happen to the cart?
1. slowly come to a stop
2. continue with constant acceleration
3. continue with decreasing acceleration
4. continue with constant velocity
5. immediately come to a stop
Concept Check – Newton’s First Law
Consider a cart on a horizontal frictionless table. Once the cart has
been given a push and released, what will happen to the cart?
1. slowly come to a stop
2. continue with constant acceleration
3. continue with decreasing acceleration
4. continue with constant velocity
5. immediately come to a stop
After the cart is released, there is no longer a force in the x-direction.
This does not mean that the cart stops moving. It simply means that the
cart will continue moving with the same velocity it had at the moment of
release. The initial push got the cart moving, but that force is not
needed to keep the cart in motion.
Concept Check – Newton’s Second Law
We just decided that the cart continues with constant velocity. What
would have to be done in order to have the cart continue with constant
acceleration?
1. push the cart harder before release
2. push the cart longer before release
3. push the cart continuously
4. change the mass of the cart
5. it is impossible to do that
Concept Check – Newton’s Second Law
We just decided that the cart continues with constant velocity. What
would have to be done in order to have the cart continue with constant
acceleration?
1. push the cart harder before release
2. push the cart longer before release
3. push the cart continuously
4. change the mass of the cart
5. it is impossible to do that
In order to achieve a non-zero acceleration, it is necessary to maintain
the applied force. The only way to do this would be to continue pushing
the cart as it moves down the track. This will lead us to a discussion of
Newton’s Second Law.
Newton’s 2nd Law
a  SF
SF  ma
1
a
m
SF
a
m
SFx  max
SFy  ma y
SF represents the vector sum of all external forces
acting on the object, or the net force.
Concept Check – Newton’s 3rd Law
When you climb up a rope, the first thing you do is pull down on the rope.
How do you manage to go up the rope by doing that??
1. this slows your initial velocity, which is
already upward
2. you don’t go up, you’re too heavy
3. you’re not really pulling down – it just
seems that way
4. the rope actually pulls you up
5. you are pulling the ceiling down
Concept Check – Newton’s 3rd Law
When you climb up a rope, the first thing you do is pull down on the rope.
How do you manage to go up the rope by doing that??
1. this slows your initial velocity, which is
already upward
2. you don’t go up, you’re too heavy
3. you’re not really pulling down – it just
seems that way
4. the rope actually pulls you up
5. you are pulling the ceiling down
When you pull down on the rope, the rope pulls up on you!! It
is actually this upward force by the rope that makes you move
up! This is the “reaction” force (by the rope on you) to the
force that you exerted on the rope. And voilá, this is Newton’s
3rd Law.
Newton’s Third Law
Action and Reaction Forces
If A acts on B, then B acts back on A
Concept Check – Newton’s 3rd Law
A small car collides with a large 1. the car
truck. Which experiences the
2. the truck
greater impact force?
3. both the same
4. it depends on the velocity of each
5. it depends on the mass of each
Concept Check – Newton’s 3rd Law
A small car collides with a large 1. the car
truck. Which experiences the
2. the truck
greater impact force?
3. both the same
4. it depends on the velocity of each
5. it depends on the mass of each
Concept Check – Newton’s 3rd Law
A small car collides with a large 1. the car
truck. Which experiences the
2. the truck
greater acceleration?
3. both the same
4. it depends on the velocity of each
5. it depends on the mass of each
Concept Check – Newton’s 3rd Law
A small car collides with a large 1. the car
truck. Which experiences the
2. the truck
greater acceleration?
3. both the same
4. it depends on the velocity of each
5. it depends on the mass of each
We have seen that both vehicles experience the same magnitude
of force. But the acceleration is given by F/m so the car has the
larger acceleration, since it has the smaller mass.
Concept Check – Newton’s 3rd Law
In outer space, a bowling ball and a ping-pong ball attract each other
due to gravitational forces. How do the magnitudes of these attractive
forces compare?
1. the bowling ball exerts a greater force on the ping-pong ball
2. the ping-pong ball exerts a greater force on the bowling ball
3. the forces are equal
4. the forces are zero because they cancel out
5. there are actually no forces at all
F12
F21
Concept Check – Newton’s 3rd Law
In outer space, a bowling ball and a ping-pong ball attract each other
due to gravitational forces. How do the magnitudes of these attractive
forces compare?
1. the bowling ball exerts a greater force on the ping-pong ball
2. the ping-pong ball exerts a greater force on the bowling ball
3. the forces are equal
4. the forces are zero because they cancel out
5. there are actually no forces at all
The forces are equal and
opposite by Newton’s 3rd
Law!
F12
F21
Concept Check – Newton’s 3rd Law
In outer space, a bowling ball and a ping-pong ball attract each other
due to gravitational forces. How do the accelerations of the objects
compare?
1. they do not accelerate because they are weightless
2. accelerations are equal, but not opposite
3. accelerations are opposite, but bigger for the bowling ball
4. accelerations are opposite, but bigger for the ping-pong ball
5. accelerations are equal and opposite
F12
F21
Concept Check – Newton’s 3rd Law
In outer space, a bowling ball and a ping-pong ball attract each other
due to gravitational forces. How do the accelerations of the objects
compare?
1. they do not accelerate because they are weightless
2. accelerations are equal, but not opposite
3. accelerations are opposite, but bigger for the bowling ball
4. accelerations are opposite, but bigger for the ping-pong ball
5. accelerations are equal and opposite
The forces are equal and opposite
-- this is Newton’s 3rd Law!! But
acceleration is F/m and so the
smaller mass has the bigger
acceleration.
F12
F21
Follow-up: Where will the balls meet if they are released from this
position?
Newton’s 2nd and 3rd Laws
2nd Law
a=
SF
m
2nd Law
SF
a=
m
Car
a=
SF
m
Truck
a=
SF
m
You
Earth
SF
a=
m
SF
a=
m
Free Body Diagrams
•
•
•
•
•
Represent object with a dot
Draw arrows to represent forces
Label forces based on nature
Draw vectors proportionately
Show ONLY forces on ONE object
Fn (normal force)
Fk (force of kinetic friction)
Fa (applied force = pull)
Fg (force of gravity)
Common Forces to Consider
Fg = mg
Gravity
Fg (force of gravity)
Normal
Fn (force of surface on object)
Friction
Fk or s (force of friction - kinetic or static)
Push and Pull
Fa (applied force = push or pull)
Tension
FT (force due to rope or chain)
g = 9.81 m/s2
Free Body Diagrams
FT
FT,1
FT,2
Fg
Fg
FT,1
Fg
FT,2
Fg
Newton’s 2nd Law
a  SF
SF  ma
1
a
m
SF
a
m
SFx  max
SFy  ma y
SF represents the vector sum of all external forces
acting on the object, or the net force.
Solving Problems involving Force and Motion
1. Given
• List Variables
• Draw a Free Body Diagram (FBD)
2. Unknown
• Identify the unknown variable
3. Equation
• Newton’s Laws (2nd or 1st)
• Write out equation from the FBD
• Solve for the Unknown
Net Force equations from Free Body Diagrams
Description
FBD
Net Force Equations
SFx  0
Falling, no air
resistance
    SF
y
 Fg  ma y
Fg (force of gravity)
Fr (force of air resistance)
Falling through air,
speeding up
SFx  0
    SF
y
Fg (force of gravity)
 Fg  Fr  ma y
Air Resistance
Fair resistance  v 2
Net Force equations from Free Body Diagrams
Description
FBD
Net Force Equations
Fr (force of air resistance)
Falling at Terminal
Velocity
SFx  0
    SF
y
 Fg  Fr  ma y  0
Fg (force of gravity)
Fr (force of air resistance)
Falling, but slowing
SFx  0
    SF
y
Fg (force of gravity)
 Fg  Fr  ma y
Net Force equations from Free Body Diagrams
Description
FBD
Net Force Equations
FT (tension in rope)
Elevator at
rest
SFx  0
    SF
y
 FT  Fg  0
Fg (force of gravity)
Elevator moving
up or down at
constant velocity
FT (tension in rope)
SFx  0
    SF
y
Fg (force of gravity)
 FT  Fg  0
Net Force equations from Free Body Diagrams
Description
FBD
Net Force Equations
FT (tension in rope)
Elevator moving
up, speeding up
SFx  0
    SF
y
 FT  Fg  ma y
Fg (force of gravity)
Elevator moving
up, slowing down
FT (tension in rope)
SFx  0
    SF
y
Fg (force of gravity)
 FT  Fg  ma y
Net Force equations from Free Body Diagrams
Description
FBD
Elevator moving
down, speeding
up
FT
Net Force Equations
SFx  0
    SF
y
 Fg  FT  ma y
Fg
FT
Elevator moving
down, slowing
SFx  0
Fg
    SF
y
 Fg  FT  ma y
Sample
An elevator of mass _____ kg is slowing as it nears its final destination going up. If the tension in the
rope is ______ N, determine the elevator’s acceleration. (2 force vectors)
G:
m
FT
g
FT
Fg
U:
a ?
E:
    SF
y
ay 
S:
 FT  Fg  ma y
FT  Fg
m
FT  mg

m
(since Fg  mg )
S:
All of the problems in this section are without numbers because the focus is on the skill of
constructing a well-reasoned solution rather than simply obtaining a correct numerical answer. Once
you solve some of the problems, we will include some numbers in class.
Problems – 2
The ball from Q1 is falling straight down, but now consider that there is a ____N force or air
resistance. Draw an FBD and determine the mass’s acceleration. (2 force vectors)
G:
U:
E:
S:
Fg
a ?
    SF
y
ay 
S:
Fr
m
g
Fr
 Fg  Fr  ma y
Fg  Fr
m

mg  Fr
m
Problems – 3
An elevator (mass = ____kg) is held at rest from a vertical cable. Determine the tension in the cable.
(2 force vectors)
G:
m
g
FT
ay  0 m s2
Fg
U:
FT  ?
E:
    SF
y
 FT  Fg  ma y  0
FT  Fg  0
S:
S:
 FT  Fg  mg
Net Force equations from Free Body Diagrams
Description
FBD
Net Force Equations
Fn
SFx  0
At rest on a
surface or
sliding with
no friction
    SF
y
 Fn  Fg  0
Fg
Fn
Pushed on
a surface
Fa
    SFx  Fa  max
    SF
y
Fg
 Fn  Fg  0
Friction
Sliding friction is electrostatic.
The force of friction depends on only 2 factors:
• The texture of the surfaces in contact (coefficient of friction - 𝜇)
• The force pushing the surfaces together (normal force - 𝐹𝑛 )
Friction can be:
• Static (prevents an object from sliding)
• Kinetic (acts while an object is sliding)
In general, Static > Kinetic
Friction
The force of friction is simply the product of the coefficient of
friction and the normal force.
F f   Fn
Static friction ranges in value up to a maximum:
FS ,max   s Fn
Kinetic friction has a constant value of:
Fk  k Fn
Coefficients of Friction
You will find this table in your text on p. 138.
Friction Forces in Free-Body Diagrams
In free-body diagrams, the force of friction is always parallel to the
surface of contact.
Kinetic friction is always opposite the direction of motion.
Static friction must point in the direction that results in a net force of
zero.
Not moving
Fn
Fn
Fa
Fk (kinetic friction)
Fg
If moving, kinetic friction
opposes motion
Fa
FS (static friction)
Fg
If not moving, static
friction is in a direction
that provides equilibrium
Net Force Equations From Free Body Diagrams
Description
Sliding and
slowing due
to friction
FBD
Net Force Equations
Fn
    SFx   Fk  max
Fk (kinetic friction)
Fg
 Fk  max
 k Fn  max
    SF
y
 Fn  Fg  0
Fn  Fg
Net Force equations from Free Body Diagrams
Description
Pushed on
a surface
with friction
FBD
Net Force Equations
Fn
Fk or Fs
    SFx  Fa  Fk or s  max
Fa
Fg
    SF
y
 Fn  Fg  0
Fn  Fg
Sample Problem
m = 55 kg
Fa = 19 N to set in motion
g = 9.81 m/s2
ax = 0
ay = 0
s = ?
s 
Fs ,max
Fn
Fa
Fs ,max
Fg
Fn
    SFx  Fa  Fs ,max  0
    SF
Fs ,max  Fa
s 
Fs ,max
Fn
 s  0.035
19 N
Fa 

 55 kg   9.81m s 2 
mg
 Fn  Fg  0
Fn  Fg  mg (since Fg  mg )
y
Problem – 10
A box of mass ______ kg is at rest on a horizontal floor. If the coefficient of static friction is ____, find
the force needed to make the box begin moving. (4 force vectors)
G:
Fn
m
g
s
ax  0
ay  0
Fa
Fs
Fg
U:
Fa  ?
E:
    SFx  Fa  Fs  0
Fa  Fs
Fa  s Fn (since Fs  s Fn )
Fa  s mg
    SF
y
 Fn  Fg  0
Fn  Fg  mg (since Fg  mg )
Net Force equations from Free Body Diagrams
Description
Pushed
down at an
angle on a
surface (no
friction)
FBD
Net Force Equations
Fn

Fa
Fg
    SFx  Fa , x  max
Fn
Fg
SFx  Fa cos   max
Fa cos 

Fa sin 
Fa
    SF
y
 Fn  Fa , y  Fg  0
Fn  Fa sin   Fg  0
Net Force equations from Free Body Diagrams
Description
Pulled up at
an angle no
friction
FBD
Fn
Net Force Equations
Fa

Fg
Fn
Fa

    SFx  Fa , x  max
Fa sin 
Fa cos 
Fg
SFx  Fa cos   max
    SF
y
 Fn  Fa , y  Fg  0
Fn  Fa sin   Fg  0
Net Force equations from Free Body Diagrams
Description
FBD
Net Force Equations
Fn
Pulled up at
an angle
with friction
Fa

Fk
Fa sin 
Fa cos 
    SFx  Fa , x  Fk  max
Fa cos   Fk  max
    SF
y
Fg
 Fn  Fa , y  Fg  0
Fn  Fa sin   Fg  0
Fn
Pushed
down at an
angle with
friction
Fk
Fg
Fa cos 

Fa sin 
Fa
    SFx  Fa cos   Fk  max
    SF
y
 Fn  Fa sin   Fg  0
Solving Problems using Newton’s Laws
Given
Include Free Body Diagram
Choose + direction
Unknown
Equations
ΣF = ma (or ΣF = 0 )
Fk = μkFn (or Fs = μsFn)
Fg = mg
Use components if necessary
Solve system of equations if necessary
Substitute only at the end
Significant Figures
Fn
Fa
Fk
Fg
+→ Σ𝐹𝑥 = 𝐹𝑎 − 𝐹𝑘 = 𝑚𝑎𝑥
+↑ Σ𝐹𝑦 = 𝐹𝑛 − 𝐹𝑔 = 𝑚𝑎𝑦
𝐹𝑘 = 𝜇𝑘 𝐹𝑛
𝐹𝑔 = 𝑚𝑔
Chapter
Sample
Problem4
Fn
m
Fa
Fn
Fa

k

Fk
g
ay = 0
ax = ?
Fg
    SFx  Fa cos   Fk  max
ax 
Fa cos   Fk
m
Fk
Fa sin 
Fa cos 
Fa , y  Fa sin  (upward)
Fg
Fa , x  Fa cos  (to the right)
    SF
F cos    k Fn (since Fk  k Fn )
ax  a
m
Fa cos   k  mg  Fa sin  
ax 
m
y
 Fn  Fa sin   Fg  0
Fn  Fg  Fa sin 
Fn  mg  Fa sin  (since Fg  mg )
Problem – 12
An already-sliding hockey puck of mass _______ grams is pushed horizontally by a hockey stick with
a force of _______ N in the direction of its motion. The coefficient of kinetic friction is ______. Find
the acceleration of the puck. (4 force vectors)
G:
Fn
m
g
Fa
k
ay  0
U:
E:
Fa
Fk
Fg
ax  ?
    SFx  Fa  Fk  max
Fa  Fk Fa  k Fn (since F   F )
k
k n
ax 

m
m
ax 
Fa  k mg
m
    SF
y
 Fn  Fg  0
Fn  Fg  mg (since Fg  mg )
Problem – 18
A box of mass ______ kg is moving along a horizontal floor at a constant velocity of ______ m/s to
the right. A girl is pushing down-and-to-the-right on the box with a force of ______ N @ ______
degrees below the horizontal. Find the coefficient of kinetic friction. (4 force vectors)
G:
m
g
Fa

ax  0
ay  0
U:
E:
Fn
Fk
Fa cos 

Fg
Fa sin 
Fa
k  ?
    SFx  Fa cos   Fk  0
    SF
y
 Fn  Fa sin   Fg  0
Fk  Fa cos 
Fn  Fa sin   Fg
k Fn  Fa cos  (since Fk  k Fn )
Fa cos 
F cos 

k  a
Fa sin   mg
Fn
Fn  Fa sin   mg (since Fg  mg )
Objects on an Incline
y
y
y
Fn
Fn
Fn
Fk
Fk
𝜃
𝜃
𝜃
Fg
Fk
x
Fg
x
x
𝜃
Fg
X-axis is parallel to surface; y-axis is  to surface.
Force of gravity is straight down
Normal force is  to the surface (and will no longer be opposite to gravity)
The angle of the incline is equal to the angle between gravity and y-axis
• Find and use the components of gravity
Friction (when present) will always be along x-axis
Net Force equations from Free Body Diagrams
Description
FBD
Net Force Equations
y
Fn
Sliding down an
incline (no friction)
    SFx  Fg , x  max
SFx  Fg sin   max
Fg , x
Fg , y

x
Fg
    SF
y
 Fn  Fg , y  0
SFy  Fn  Fg cos   0
y
Fn  Fg cos 
Fg cos   F
g
Fg sin 
x
Net Force equations from Free Body Diagrams
Description
FBD
Net Force Equations
y
Sliding down an
incline with
friction
Fn
Fk
    SFx  Fg , x  Fk  max
SFx  Fg sin   Fk  max
Fg , x
Fg , y

x
Fg
    SF
y
 Fn  Fg , y  0
SFy  Fn  Fg cos   0
y
Fn  Fg cos 
Fg cos  
Fg
Fg sin 
x
Sample
A _____ kg child-and-sled combination is sliding down a snowy slope at ____ degrees below the
horizontal. If the coefficient of kinetic friction is ____, what is the acceleration? (3 force vectors)
G:
y
m
g
Fk
k

y
Fn
Fk
Fg sin 
ay  0
x
Fg
U:
ax  ?
E:
    SFx  Fg sin   Fk  max
ax 
Fg sin   Fk
m (since F
Fg sin   k Fn
m
Fg cos 
    SF
y
Fg
x
 Fn  Fg cos   0
Fn  Fg cos   mg cos 
k
ax 
Fn
 k Fn )
(since Fg  mg )
mg sin   k mg cos 
 g sin   k g cos 

m