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Chapter 4. Force and Motion
Chapter Goal: To establish
a connection between force
and motion.
Student Learning Objectives
• To recognize what does and does not constitute a
force.
• To identify the specific forces acting on an object.
• To draw an accurate free-body diagram of an object.
• To begin the process of understanding the connection
between force and motion.
A force is an interaction between two objects
• A force is a push or a pull on an object.
• If I push a book across a table, the book pushes
me back (inanimate objects can exert force!)
• A force is a vector. It has both a magnitude and a
direction.
• The force (interaction) has the same magnitude
for both me and the book. However the direction of
the force on me is opposite to the direction of the
force on the book.
A force is an interaction between two objects
•A force on an object requires an agent. The agent is
another object.
•This is another way of confirming that a force is an
interaction between objects. It takes two to tango!
• A force is either a contact force or a long-range
(noncontact) force.
•Gravity is the only long-range force we will study
this semester. It is an interaction between two
objects, but they are not necessarily in contact.
•All other forces (this semester) only exist when the
two objects are in contact.
A Short Catalog of Forces - Gravity
• Gravity is a long-range
attractive force between two
objects.
•In this class, our emphasis is
on the interaction between the
Earth and objects on or near its
surface (the weight force).
•Your text uses W for the
weight force.
•Near the surface of the earth,
W = mg, where m is the object
mass in kilograms and g= 9.8
m/s2 is the acceleration due to
gravity.
Earth
A Short Catalog of Forces: Tension
Ropes, strings, cables and hooks exert a tensional force.
These agents always pull, they never push.
Although it seems strange, the sled exerts a tension force on
the rope.
A Short Catalog of Forces - Normal
The normal force is a contact force between 2
surfaces that is perpendicular to the surfaces. Your
book uses FN for the normal force
A Short Catalog of Forces - friction
The frictional force is a contact force between 2
surfaces that is parallel to the surfaces
A Short Catalog of Forces

Fp

Other forces are usually shown as an F with an identifying
subscript.
You’ve just kicked a rock, and it is now
sliding across the ground about 2 meters
in front of you. Which of these forces act
on the rock at this instant? Choose all
that apply
A. Gravity, acting downward
B. The normal force, acting upward
C. The force of the kick, acting in the
direction of motion
D. Friction, acting opposite the direction of
motion
E. All of the above
Drawing force vectors
Free-body diagrams
• Identify all forces acting on the
object.
• Draw a coordinate system. If
motion is along an incline, the
coordinates system should be
tilted relative to true horizontal
and true vertical.
• Represent the object as a dot at
the origin.
• Draw vectors representing each
of the identified forces and label
with the appropriate symbol.
truck not moving due to friction
Freebody diagrams
Draw a free body diagram
for the truck.
There is no friction, but a
cable attached to the
wall keeps the truck
from moving.
Freebody diagrams
Draw a free body diagram for the sled. Assume friction
between the snow and the sled.
4.3 Newton’s Second Law of Motion
Newton’s Second Law
An object of mass m, subject to forces F1 , F2 ,
etc. will undergo an acceleration with a
magnitude directly proportional to the net force
and inversely proportional to the mass:

a


F
m



F  ma
The direction of the acceleration is
the same as the direction of the net
force.
4.3 Newton’s Second Law of Motion
Newton’s First Law
This is a special case of the 2nd Law. If the net
force equals 0, there is no acceleration and the
velocity of the object will not change. If it is at
rest, it will stay at rest.

F  0
In this case the object is in equilibrium.
•Static equilibrium – object is at rest for a finite
period
•Dynamic equilibrium –object is moving at constant
velocity.
Graphical Interpretation of Newton’s Second
Law
• Newton’s 2nd Law is the equation of a line with a
0 value for the y-intercept (in this case the aintercept!).


F
a
m
a
F
The following graphs plot acceleration vs force for different objects.
Which object has the greatest mass?
A
B
a
a
F
F
C
D
a
a
F
A.
B.
C.
D.
F
Problem-Solving Strategy for Newton’s Law
Problems
Problem-Solving Strategy: Equilibrium
Problems

 Fx  0

 Fy  0
If the net force is equal to zero, must
the object be at rest? The next example
shows the answer to this:
Newton’s 1st Law: Towing a car up a hill
A car with a weight (not mass) of 15,000 N is being
towed up a 20° frictionless slope at constant velocity.
The tow rope is parallel to the slope surface. What is
the tension on the tow rope?
Why is this a Newton’s 1st Law problem (ΣF = 0) and not
a Newton’s 2nd Law problem (ΣF = ma)?
Draw a free body diagram for this problem.
Newton’s 1st Law: Towing a car up a hill
A car with a weight (not
mass) of 15,000 N is
being towed up a 20°
frictionless slope at
constant velocity.
The tow rope is
parallel to the slope
surface. What is the
tension on the tow
rope?
What part of the story
indicates that ΣF = 0?
Newton’s 1st Law: Towing a car up a hill
Once the pictorial representation is completed, use Newton’s
1st Law in component form:
T and n have only one nonzero component, but FG has
non-zero components in both x
and y directions. How do we
write the components of FG in
terms of θ ?
Newton’s 1st Law: Towing a car up a hill
Solving the 2nd equation tells us that n = FG
cos θ, which is not necessary information for
this problem.
Problem-Solving Strategy for Newton’s 2nd
Law Problems
1. Use the problem-solving strategy outlined for
Newton’s 1st Law problems to draw the free body
diagram and determine known quantities.
2. Use Newton’s Law in component form to find the
values for any individual forces and/or the
acceleration.
3. If necessary, the object’s trajectory (time, velocity,
position, acceleration) can be determined by using
the equations of kinematics.
4. Reverse # 2 and 3 if necessary.
Newton’s 2nd Law – Speed of a Towed Car
•Draw a free-body diagram. Find the acceleration using Newton’s
second law in component form.
•Draw a pictorial representation and Find the velocity using one of
the kinematic equations.
Speed of a towed car – pictorial representation of
motion, motion diagram and free body diagram
Speed of a towed car
Use the free body diagram to write
Newton’s 2nd Law in component form:
ΣFy = may = 0 (no change in speed up or
down)
ΣFy = n – FG , or n=FG (not pertinent in
this problem)
ΣFx = max
T – f = max
Speed of a towed car
Speed of a towed car
Apparent Weight – What the scale says (even if there
is no scale!)
• The value the scale reads when scale and object are
accelerating or being acted upon by some force other
than gravity (e.g the buoyant force in water)
• Not equal to the true weight (FG = mg)
• Apparent weight is either a normal force (step on
scale) or tension force (hanging scale)
• An object inside an accelerating elevator has the same
acceleration as the elevator.
• Acceleration is not a force so don’t include directly
on free body diagram.
Typical FBD for an apparent weight problem
n
FG = mg, but n does not!
Apparent Weight – What the scale would say,
even if there is no scale
It takes the elevator in a skyscraper 4.0 s to reach a
constant speed of 10.0 m/s. A 60.0 kg passenger gets
on at the ground floor. What is her apparent weight
during those 4 seconds?
Use the problem-solving strategy (repeated in next
slide).
Problem-Solving Strategy for Newton’s Law
Problems
Problem-Solving Strategy for Newton’s 2nd
Law Problems
1. Use Newton’s Law in component form to find the
values for any individual forces and/or the
acceleration.
2. If necessary, the object’s trajectory (time, velocity,
position, acceleration) can be determined by using
the equations of kinematics.
3. Reverse # 1 and 2 if necessary (Hint: it is!)
Mass – a quantitative measure of inertia
• An object’s mass (m) is a measure of the amount
of matter that it contains.
• The SI unit of mass is the kilogram (kg). The
English unit of mass is the slug, which is virtually
never used.
• The mass of an object does not change unless you
add or subtract matter from it.
• The mass of an object is the same on any planet
and in deep space.
• Scales do not directly measure mass, except for a
balance scale, which compares one mass to
another.
Weight
• The weight of an object on or above the earth is the
gravitational force that the earth exerts on the object
(FG or W).
• Since weight is a force, the SI unit of weight is the
Newton (N); the English unit is the pound.
• The weight of an object changes on other planets
since the gravitational force of other planets are
different than that of the earth.
• A scale does not directly measure weight; it
measures the normal force or (hanging scale) the
tension force. In most situations, this force
measured by the scale is numerically equal to the
weight force.
Apparent weight or “what the scale says”
• In accelerating reference frames (e.g. the ever-popular
elevator with a scale), the scale reading will differ
from the true weight. This is called apparent weight.
• Metric scales assume measurement on earth and are
calibrated to read in mass units (kg). American scales
read in English force units (lb). It’s a culture thing.
4.7 The Gravitational Force
Newton’s Law of Universal Gravitation
• Every particle in the universe exerts an attractive force on
every other particle.
• A particle is a piece of matter, small enough in size to be
regarded as having no volume. In practice we use the particle
model even for larger bodies
• The Law of Universal Gravitation is an example of an inverse
square law. The force between the two bodies is inversely
proportional to the square of the distance between them.
m1m2
FG  2
r
For two particles that have masses m1 and m2 and are
separated by a distance r, the force has a magnitude
given by:
F1 2
m1m2
 F21  G 2
r
G  6.673 10
11
N  m kg
2
2
F1-2 and F2-1 are equal in magnitude but point in
opposite directions
If m1 has a mass of 12 kg and m2 has a mass of 25 kg and the
two objects are 1.2 meters apart, what is the magnitude of the
gravitational force between them?
F1 2
m1m2
 F21  G 2
r
G  6.673 1011 N  m 2 kg 2
Ratio Reasoning with Inverse Square Laws
Two masses, separated by a distance of r experience
a gravitational attraction, F. Now the distance is
increased by a factor of 3. By what factor does
the magnitude of the force change (what is the
ratio F2 )?
F1
a. 1/3
c. 9
b. 3
d. 1/9
The figure shows the moon being attracted by the
earth. The earth is also attracted by the moon.
Compared to FEonM (the force shown), FMonE is:
A. much smaller and in the same
direction.
B. much smaller and in the opposite
direction.
C. the same size and same direction.
D. the same size and
opposite direction.
earth
moon
FEonM
4.7 The Gravitational Force
Definition of Weight
The weight of an object on or above the earth is the
gravitational force that the earth exerts on the object.
The weight always acts downwards, toward the center
of the earth.
On or above another astronomical body, the weight is the
gravitational force exerted on the object by that body.
SI Unit of Weight: newton (N)
W G
W G
M Em
r
2
M Em
RE
2
If the object is close (within 100 miles) to the
earth, this becomes:
Define the quantity g,
for all the unchanging
entities:
ME
g G 2
RE
Calculate g
G  6.673 1011 N  m 2 kg 2
ME = 5.98 x 1024 kg
RE = 6.38 x 106 m (don’t
forget to square!)
The Gravitational Force
W G
M Em
RE
2
ME
g G 2
RE
W  mg
g = 9.8 m/s2 on Earth
1 kg of mass x g = 9.8
N, where N = kg m/s2
Ratio Reasoning
A space traveler weighs 540.0 N on earth.
What will he weigh on another planet whose
radius is twice that of earth and whose
mass is 3 times that of earth?
Friction
Kinetic Friction
Experiments show that the
kinetic friction force is nearly
constant and proportional to the
magnitude of the normal force.
where the proportionality
constant μk is called the
coefficient of kinetic friction.
Static Friction
The box is in static
equilibrium, so the static
friction must exactly balance
the pushing force:
Static friction
• An object remains at rest as long
as fs < fs max
• The object slips when fs = fs max
• A static friction force fs > fs max is
not physically possible.
• fs max >fk for the same surfaces
where the proportionality constant μs is
called the coefficient of static friction.
A model of friction
• “ motion” indicates motion relative to the two surfaces
• the max value static friction, fs max occurs at the very instant
the object begins to move (which often means 1 nanosecond
before, for problem-solving purposes.
• for any given materials, μs > μk
Determine which of the frictional forces is the
largest. The box and the floor are identical in
all situations.
Which of these have the same frictional force?
A. none
B. a and d
C. c,d,e,
D. c,d
Example Problem w tilted axis
A 75-kg snowboarder starts down a 50-m high (not
long!), 100 slope with μk = 0.06. Assume he has just
started moving, but his starting velocity is essentially
zero. What is his speed at the bottom?
• Draw the free body diagram (fbd) to determine the
forces.
• Use Newton’s Law in component form.
• Determine the acceleration.
• Use kinematics to solve for the final speed.
• Start with the fbd.
Example Dynamics Problem w tilted axis
A 75-kg snowboarder starts
down a 50-m high, 100
slope, with μk = 0.06. What
is his speed at the bottom?
Write Newton’s Law for y-axis
values. Is this 2nd Law, or
1st Law (a=0)?
There is one force along the +
y axis and one force with a
negative y component.
n
fk
FG
Find:
v1
uk = .006
Example Dynamics Problem w tilted axis
A 75-kg snowboarder starts down
a 50-m high, 100 slope, with μk
= 0.06. What is his speed at the
bottom?
Write Newton’s Law for y-axis
values. Is this 2nd Law, or 1st
Law (a=0)?
n
fk
FG
ΣFy = n – mg cos θ = may
But no acceleration along y axis
means the net force is 0
therefore
n = mg cos θ
Find:
v1
uk = .006
Example Dynamics Problem w tilted axis
A 75-kg snowboarder starts down
a 50-m high, 100 slope, with μk
= 0.06. What is his speed at the
bottom?
n
ΣFx = mg sin θ – fk = max
where fk = uk n
fk
and we know that n = mg cos θ;
therefore
mg sin θ – ukmg cos θ = max
You didn’t even need to know the
mass; it cancels out!
plug n chug: a = 1.12 m/s2
FG
Find:
v1
uk = .006
n = mg cos θ
Example Dynamics Problem
A 75-kg snowboarder starts
down a 50-m high, 100 slope
on a frictionless board.
What is his speed at the
bottom? Couldn’t find a
picture of a snowboarder….
but the idea is the same
Find
v1
a, =1.12 m/s2
Example Dynamics Problem
A 75-kg snowboarder starts
down a 50-m high, 100 slope
on a frictionless board. What
is his speed at the bottom?
Time is not important.
Use the mantras of the Indian
Princess to calculate the
length (hypotenuse) of the
slope. That’s our x1.
Find
v1
Then use the appropriate
kinematic equation
v1 = 25.4 m/s
a, =1.12 m/s2
Reference Frames
A reference frame is a coordinate system that allows description
of time and position relative to some “observer”.
• The origin moves (or stays at rest) in the same way as the
observer does.
• If a person is sitting still in a moving train, the description of
the person's motion depends on the chosen frame of reference.
• If the frame of reference is the train (i.e origin on the train), the
person is considered to be not moving IN THAT FRAME OF
REFERENCE
• if the frame of reference is the Earth, (i.e. origin on the Earth),
the person is considered to be moving IN THAT FRAME OF
REFERENCE
• The “observer” doesn’t have to be a person
Inertial Reference Frames
• An inertial reference frame is one
where the origin is either at rest (the
Earth, for our purposes) or moving
at constant velocity.
• Newton’s 1st Laws are valid in an
inertial reference frame.
• Newton’s 1st Law appears to be
broken if one is looking at motion
from a non-inertial (i.e. accelerating)
reference frame.
• The truck is accelerating to the right.
Imagine 2 frames of reference; that
of the truck driver and that of a
pedestrian.
Reference frames, revisited
The truck is accelerating to
the right.
The truck driver thinks the
box in the back is not
moving, because its not
sliding toward the back
of the truck.
The ground-based observer
sees the box getting
faster, just like the truck.
Which observer is in an
inertial reference frame?
A. The truck driver
B. The ground-based guy
C. The box (if it could
observe)
D. None of the above.
Free body diagram of the box
In a free-body diagram of
the box, there is a:
a. static friction vector,
pointing right
b. static friction vector,
pointing left
c. kinetic friction vector,
pointing right
d. kinetic friction vector,
pointing left
e. no friction vector, the
box is stationary in the
truck
Make sure the cargo doesn’t slide
• A box is in the back of a flatbed
truck What is the maximum
acceleration the truck can have
without the box sliding?
Coefficients of friction between
truck and box are μs = 0.40 and
μk = 0.20. To solve this
problem:
• Is the object of interest the box,
truck, or both?
• Do the forces on the object of
interest include static friction,
kinetic friction, both, or neither?
EOC 115 – Hanging a picture
A person is trying to judge whether a picture of mass
1.10 kg is properly positioned by pressing it against a
wall. The pressing (normal) force is perpendicular to
the wall. The coefficient of static friction between
picture and wall is 0.660. What is the minimum
amount of pressing force required?
Draw a freebody diagram. In which direction is the
normal force in this problem? Does it have anything
to do with the weight of the picture?
Push the crate
A dockworker pushes a 50-kg crate with a
downward force of 300.0 N, at an
angle of 30 degrees below horizontal.
The coefficient of kinetic friction is
0.30.
A. What is the normal force?
B. What is the frictional force?
C. What is the acceleration?
• Draw a freebody diagram of the object
of interest (will it have tilted axes?)
• Apply Newton’s Law in component
form.
Push the crate - Freebody diagram
Knowns
m = 50 kg
Fpush = 300N
θ = 30˚
μk = 0.30
n
fk
θ
Find
n, fk ,a
Fpush
Fg
Push the crate (problem 1 from Exploration 4)
A. ΣFy = may
n - FG - Fp sin 30˚ = 0
this solves for n = 640N
B. fk = μk |n| = 192N
C. ΣFx = max
Fp cos 30˚ - μk |n| = max
This will give us ax
which is a = 1.36 m/s/s
Keep it up! Problem 3 from Exploration 4
A man pushes with a 100 N force on a 10- kg
crate, which is sliding down the wall
(despite his efforts). The pushing force is
directed 53 degrees above horizontal. The
coefficient of kinetic friction is 0.60.
Determine the magnitude of the friction
force between wall and crate.
Draw a freebody diagram. In which
direction is the normal force in this
problem?
Use Newton’s law in the appropriate
direction to determine the normal force.