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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
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.
Drawing force vectors
Combining Forces
What is the net force on the object? Your answer
should include both magnitude and direction (in the
form of an angle from the x-axis)
3N
4N
What is the net force on the object?
Individual Forces
Net Force
5N
37
3N
4N
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.
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
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?
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
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 ball? You may choose more than
one answer.
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
Ans: A, B, and D
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.
Extension 1 – Freebody diagrams
Mass
• 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 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.
• Scales do not measure weight, at least not directly.
Apparent weight or “what the scale says”
• Scales measure a normal force or a tension force
(i.e.hanging scales).
• In most situations, the force measured by the scale is
numerically equal to the weight force.
• 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. 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 a mathematical point.
The force that each exerts on the other is directed along the line
joining the particles.
4.7 The Gravitational Force
For two particles that have masses m1 and m2 and are
separated by a distance r, the force has a magnitude
given by:
m1m2
F G 2
r
G  6.673 1011 N  m 2 kg 2
4.7 The Gravitational Force
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 gravitational force
between them?
m1m2
F G 2
r
G  6.673 1011 N  m 2 kg 2
4.7 The Gravitational Force
m1m2
F G 2
r

 6.67 10
11
N  m kg
2
2

12 kg 25 kg 

2
1.2 m
8
 1.4 10 N
The gravitational force between 2 objects is usually pretty small,
unless one of the object is a planet!
If m1 has a mass of 12 kg and m2 has a mass of 25 kg and the
two objects are twice as far apart as before (i.e. 2.4 m), what is
the gravitational force between them? Twice as much? Half as
much? Or something else
m1m2
F G 2
r
G  6.673 1011 N  m 2 kg 2
The figure shows a binary star
system. The mass of star 2 is
twice the mass of star 1.
Compared to
, the
magnitude of the force
is
A. one quarter as big.
B. half as big.
C. the same size.
D. twice as big.
E. four times as big.
The figure shows a binary star
system. The mass of star 2 is
twice the mass of star 1.
Compared to
, the
magnitude of the force
is
A. one quarter as big.
B. half as big.
C. the same size.
D. twice as big.
E. four times as big.
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)
4.7 The Gravitational Force
W G
M Em
r
and W = mg
2
ME
g G 2
r
Calculate g
ME = 5.98 x 1024 kg
RE = 6.38 x 106 m (don’t
forget to square!)
4.7 The Gravitational Force
Relation Between Mass and Weight
W G
M Em
r
2
ME
g G 2
r
g = 9.80 m/s2 on Earth
W  mg
4.7 The Gravitational Force
Relation Between Mass and Weight
W G
M Em
r
ME
g G 2
r
W  mg
2
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.
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?
Is this a Newton’s 2nd Law problem (ΣF = ma) or
a Newton’s 1st Law problem (ΣF = 0)?
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
•Find the acceleration using Newton’s second law in component form.
•Find the velocity using one of the kinematic equations.
•Note that in this problem, the car’s mass, not the weight, is given.
Speed of a towed car – pictorial representation of
motion, motion diagram and free body diagram
Speed of a towed car
EXAMPLE 6.3 Speed of a towed car
Our pictorial analysis (and common sense) indicate that there is
no acceleration along the y-axis.
EXAMPLE 6.3 Speed of a towed car
Apparent Weight – What the scale says (even if there
is no scale!)
• The value the scale reads when the object is
accelerating or being acted upon by some other force
(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!)
Apparent Weight Problem
a = +2.5 m/s2
n = apparent weight = 738 N
n
FBD of elevator passenger
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 Fintercept!).



F  ma
The following graphs plot net force vs. acceleration for different
objects. Rank each situation according to mass. That is, order the
situation from largest to smallest mass, using Newton’s Second Law.
Ranking Order from greatest to least mass: F C (A D) B E
(Note this one is slightly different than the example in Exploration 3.
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
Rank order, from largest to smallest, the size
of the friction forces fa to fe in these five
different situations. The box and the floor are
made of the same materials in all situations.
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
so the net force is 0 there
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 = ma
and fk = uk n
fk
and n = mg cos θ
FG
mg sin θ – ukmg cos θ = ma
You didn’t really need the mass; it
cancels out.
plug n chug: a = 1.12 m/s2
now find v1
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?
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 Indian Princess to
calculate the length
(hypotenuse) of the slope.
That’s our x1.
Find
v1
v1 = 25.4 m/s
a, =1.12 m/s2
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?
Make sure the cargo doesn’t slide
fbd - box
n
fs
Fg
Knowns
μs = 0.40
μk = 0.20
FG = 9.8 m
find
amax
Make sure the cargo doesn’t slide
n
fs
max
Fg
ΣFy = may = 0
n – FG = 0, therefore n = mg and
fsmax = μs mg
Knowns
μs = 0.40
μk = 0.20
FG = mg
find
amax
Make sure the cargo doesn’t slide
n
fs
max
Fg
ΣFx = max friction is the only force!
μs mg = ma, and the m drops out
a = 3.9 m/s/s
Knowns
μs = 0.40
μk = 0.20
FG = mg
find
amax
EOC #107
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 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?
EOC # 107 Freebody diagram
fbd - picture
Knowns
m = 1.10 kg
μs = 0.660
fs
Fpush
n
Find
Fpush
Fg
Forces which are usually x are y in this problem and vice
versa (with the exception of gravity). Newton’s Laws
still work.
EOC # 107 Freebody diagram
Knowns
m = 1.10 kg
μs = 0.660
Find
Fpush
fbd - picture
fs
Fpush
n
Fg
Newton’s Law in the x direction
tells us that n = Fpush but nothing
else about the value of either.
Moving right along to the ydirection:
ΣFy = may = 0 = fs – FG or fs = mg. No matter how hard
you press, the picture will not levitate up. Fact. However, if
you don’t push hard enough….
EOC # 107 Freebody diagram
Knowns
m = 1.10 kg
μs = 0.660
Find
Fpush
fbd - picture
fs
Fpush
n
FG
The minimum value of n must be
the value that allows fsmax to be
equal to the weight of the picture:
ΣFy = 0 = fsmax – FG
or μs |n| = mg
n = Fpush = 16.3 N
Push the crate (problem 1 from Exploration 4)
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.
Keep it up! Freebody diagram
fbd - box
fk
n
Fpush
θ
Knowns
m = 10 kg
Fpush = 100N
θ = 53˚
μk = 0.60
Find
fk ,a
Fg
Forces which are usually x are y in this problem and vice
versa (with the exception of gravity). Newton’s Laws
still work.
Keep it up! Freebody diagram
fbd - box
fk
n
Fpush
θ
Knowns
m = 10 kg
Fpush = 100N
θ = 53˚
μk = 0.60
Find
fk ,a
Fg
ΣFx = Fp cos θ – n = 0
n = 60.2 N fk = 36.1 N
ΣFy = Fp sin θ + fk – FG = max
a = + 1.79m/s/s therefore box is slowing down.