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UNIT 2 - DYNAMICS
Unit 2A – Newton’s 3 Laws of Motion
Isaac Newton and the Principia Mathematica
Dynamics is the study of why things move. Why things move the
way they do is because of forces and Newton’s 3 Laws of Motion.
In 1687, Newton outlined his monumental 3 Laws of Motion:
Law I.
Every body continues in its state of rest, or of uniform motion in
a right line, unless it is compelled to change that state by forces
impressed upon it.
Law II.
The change of motion is proportional to the motive force
impressed; and it is made in the direction of the right line in
which that force is impressed.
Law III.
To every action there is always opposed an equal reaction: or,
the mutual actions of two bodies upon each other are always
equal, and directed to contrary parts.
Historical Background to the Principia – The Copernican Revolution
“If I have seen far, it is because I have stood on the shoulders of
giants.”
- Isaac Newton
All moving objects on earth tend to slow down to a stop. Prior to
Newton, then, from the days of Aristotle (350 B.C.) and the ancient Greeks,
the belief was that the natural tendency for motion of objects, from intuition,
was to move towards rest or no motion. There was also a viewpoint since
the time of Ptolemy (A.D.127) and the Romans that the earth was the center
of the universe, a view held dominant throughout the middle ages by the
Catholic Church.
Then, in the early1500’s, a reluctant Polish cleric by the name of
Nicolas Copernicus (b. 1473) put forward the heretical idea that the earth
revolved around the sun. Published in 1543 on his deathbed, De
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revolutionibus orbium coelestium (On the Revolutions of the Heavenly
Spheres), revolutionized not just our understanding of the solar system, but
our understanding of nature in general, as future attempts to understand the
universe - which were before the domain of God and religion - instead
switched to a belief in simple and demonstrable mathematical relationships
or laws between objects, in other words, the realm of science. The two
principle champions of Copernicus’ heliocentric (sun-centered) system were
Johannes Kepler and Galileo Galilei.
Born in 1571 in what is now Germany, Kepler, by good fortune, as a
young man went to work for the Royal Astronomer Tycho Brahe, who died
soon after, leaving Kepler with the title and a wealth of astronomical data to
sift through and disseminate. In 1609, Kepler published the Astronomia
Nova (New Astronomy), outlining his first two laws of planetary motion. In
1618, Kepler discovered his third law. What his three laws showed was that
all planets behaved according to identical mathematical relationships- that
there was an underlying common pattern in the motion of the heavenly
bodies, irrespective of which or where they were1..
Galileo was born in Pisa, Italy, in 1564. Considered by no less than
Einstein to be the father of experimental science, Galileo performed
experiments on balls rolling down inclined planes and swinging pendulums
(and perhaps cannonballs dropped from the leaning tower of Pisa) that
refuted Aristotle’s ideas of motion. It was Galileo who refined the
refracting telescope and first used it to gaze at the stars. What he saw
convinced him of the correctness of Copernicus’ heliocentric theory,
something for which he was punished by the Inquisition in 1633 - forced to
recant his views publicly and spend the last 9 years of his life under house
arrest. Just as Kepler was the forerunner for the formulation of
mathematical relationships about nature, Galileo’s experiments, use of
evidence from observations and general theorizing became the forerunner to
what is today the modern scientific method.
Guided by Copernicus, both Kepler and Galileo looked to the stars,
for the cosmic answer to nature. Planets do not slow down or come to rest in
the new sun-centred universe. Kepler’s third law became the basis of
Newton’s future Law of Universal Gravitation. From the consideration of
the motion of planets and the reduced air resistance or friction in space,
Galileo was able to formulate his principle of inertia that Newton later
adopted to formulate his first and second law.
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Johannes Kepler, following the theories of Copernicus and using the
astronomical data of Tycho Brahe, mapped the motion of the planetary
bodies around the sun. As Copernicus had predicted, the motion very
closely resembled circular motion, where the circling planet traveled at
constant speed but constantly changed direction as it orbited the sun. From
his data, Kepler was able to form his three laws that all heavenly bodies
seemed to follow, irrespective of their identity.
Yet planets do not orbit exactly at constant speed, nor exactly in a
circular path. Kepler's 1st Law states that planetary orbits are ellipses.
Ellipses are egg-shaped orbits that differ from circular motion in one very
fundamental aspect: Instead of one center that a circle has, ellipses have two
centers -or more properly termed foci. If you take a string from the centre of
a circle to a point on the circle itself, then run that taut string around you will
describe a circular orbit. If you run a taut string from the two foci of an
ellipse (one end of the string connected to each focus) to a point on the
ellipse, you will transcribe an ellipse's egg-shaped orbit.
Kepler's 2nd Law states that equal areas subtended by a string
from the sun around a planet's orbital path will correspond to equal
periods of time. This means that planets move at different speeds along
different parts of the elliptical orbit.
Example 17: From Kepler’s 2nd Law, deduce whether a planet is moving
faster when it’s at a point nearer to the sun, or when it’s farther
from the sun in its elliptical orbit.
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(nearer)
Kepler's 3rd Law, which states the mathematical relationship
between the periods of orbit, T, and the radius of orbit, R, is given by:
Kepler's Constant, K, holds for all planetary bodies orbiting around
the same central body or mass, so if you work out K for one planet in the
solar system, you can apply this K to all other planets in the solar system to
derive their periods of orbit or radius.
Example 18: Calculate the orbital radius of Mercury given that the time it
takes Mercury to travel around the sun is equal to 88.4 earth
days. (Note: earth's orbital radius around the sun Rearth = 1.49
x 1011 m)
(5.79 x 1010 m)
For planets in our solar system, one of the two principal foci in the
elliptical orbits is, of course, the sun. The second focus is presumed to be
the center of mass of the other planetary bodies in the solar system, a
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shifting focus because of the different orbital periods. In reality, most
planetary orbits follow very close to a circle with the two foci very near each
other. For most problems, we can assume the two elliptical foci are close
enough to be identical, and that planetary orbits approximate a circular
orbit.
Example 19: Calculate Kepler’s constant from the moon’s orbital radius
around the earth (3.84 x 108 m) and its orbital period (1 lunar
month = 27.32 days), and use this value to determine the
orbital period around the earth of a satellite in orbit 3.59 x 107
m above the surface of the earth. What is the speed of this
satellite in orbit?
(1.00 days, 3.07
km/s)
Astronomical Unit:
To make the mathematics a little easier, astronomers used Earth’s numbers
in Kepler’s third law:
T2/R3= 1 Earth year2/ 1 AU3 = a constant
Where an AU(astronomical unit) is the mean orbital radius of Earth to the
sun. The constant for this is therefore 1!
Any other planet in our solar system can be compared to Earth and therefore
its constant is also 1! (However, the units must match Earth’s)
Eg. Pg 272 #1
Do pg 272 #2,3 and pg 275#1,3
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Newton’s First Law – The Principle of Inertia
Law I.
Every body continues in its state of rest, or of uniform motion in
a right line, unless it is compelled to change that state by forces
impressed upon it.
There are basically two parts to the 1st Law:
1) The natural tendency of all moving objects is uniform motion or
constant velocity (this includes moving at the same speed and
direction forever!).
2) if the object is not moving at constant velocity (non-uniform
motion) there must be unbalanced forces or a net force acting on it.
Law I- Part 1:
But moving objects on earth naturally slow down to a stop.
It is important to remember that Newton (and Galileo) was referring
to an object in the idealized system of space, without friction. On earth,
moving objects have to deal with friction and air-resistance which slows
them to a stop.
Activity:
Slide a chalkduster along the blackboard ledge. What happens?
Why?
___________________________________________________
_____
Now roll a marble along the floor. Again, what happens and
why?
___________________________________________________
_____
Which of the two motions – the chalkduster or the marble,
more closely represents planets moving in space?
________________________________________________________
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Example 1: Imagine you are a bus driver on a crowded bus. You spot a
boy/girl (of your dreams) who gets on and stands in the middle
of the aisle. Of course, you can’t help but keep looking at
him/her. By Newton’s first law, predict what will happen to the
girl when:
a) the bus starts moving forward.
________________________________________________________
__________
________________________________________________________
_________
b) the bus turns left
______________________________________________________
_________
______________________________________________________
_________
c) the bus slows to a stop.
______________________________________________________
_________
______________________________________________________
_________
In the absence of any forces like friction, all objects try to keep
moving at the same constant velocity – the same speed and same direction.
Note that the case of an object at rest is just a special case of uniform
motion, where the constant velocity is zero.
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The tendency of an object to move at constant velocity is called an
object’s inertia. (resistance to the change in motion) The first law is often
referred to as the principle of inertia. The amount of inertia depends on
the amount of mass of an object, ie. The greater the mass, the greater the
inertia.
Question:
speed?
Which has more inertia, a bus or a car moving at the same
___________________________________________________
________
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Law I – Part 2
The second part of Newton’s 1st Law is the idea of forces causing changes in
motion or velocity. Here, Newton is referring to the net force or sum of all the forces.
Net force equation: Fnet = F1 + F2 + F3 +…
The unit for force is the Newton, N, and forces are vector quantities.
What the first law is saying is there are basically only two cases for motion:
Uniform Motion (constant velocity) case, when Fnet = 0
Non-Uniform Motion (changing velocity/acceleration), when Fnet  0
***Note that a net force of zero does not mean no forces are acting on an object.
There can be forces acting on an object where the net force is zero, as long as the forces
balance each other. Conversely, the presence of a net force is an indication of an
unbalanced set of forces.
Example 2:
Indicate whether the forces are balanced or unbalanced acting on a
chalkduster on a blackboard ledge that is:
1) At rest? __________
2) Pushed by a hand and then let go? ___________
3) Pushed by a hand at constant velocity across the ledge? __________
Since forces are vectors, if two or more forces act on an object, the forces must be added
vectorally to produce the net balanced or unbalanced force.
Example 3:
Three forces of equal magnitude 10.0 N act on an object. What is the net
force (magnitude and direction) if the direction of the forces is given by:
1) F1 is east, F2 is west, and F3 is west
2) F1 is east, F2 is north, and F3 is east
3) F1 is east, F2 is north, and F3 is 35.0 0 S of W
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Newton’s 2nd Law
Law II.
Fnet = ma
The change of motion is proportional to the motive force impressed; and it
is made in the direction of the right line in which that force is impressed.
The second law states the direct mathematical relationship between force and the
acceleration (change in motion). The force acting on the object is directly proportional to
the rate of acceleration: doubling the force doubles the acceleration. This gives a linear
relationship between a plot of force F versus acceleration a: a straight line graph through
the origin.
The constant of proportionality – the slope of this graph - is the mass m. From the
equation of a straight line y = mx + b, where m is the slope and b is the y-intercept, the
equation here is F = ma (+0). This yields Newton’s 2nd Law:
Fnet = ma
Fnet – net or sum of the forces
m – mass
a – acceleration
The unit of force is a product of mass times acceleration units: 1 N = (1 kg)(1 m/s2)
Example 4:
A man pulls a stationary 10 kg box along a floor with 10 N of force
applied. What is the acceleration of the box when:
A) The force of friction opposing the motion is 10 N?
B) The force of friction opposing the motion is 5 N?
C) The force of friction opposing the motion is 0 N?
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Free-Body Diagrams and Newton’s 2nd Law
A diagram of all the forces of interest acting on an object is called a
free-body diagram. All forces are drawn as vector arrows leading away from
the center of the object (forces are drawn as pulls). Usually the forces of
interest are only those forces influencing the object’s motion (or lack of
motion). These forces are then used to write a net force equation.
For typical dynamics problems we are only concerned with the four forces the force of gravity Fg, the normal force FN, the tension FT (or applied forceFA) and the force of friction Ff .
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The Force of Gravity (Weight)
Fg = m g
The force of gravity Fg –also known as the weight of an object - is a
product of the object’s mass m and its acceleration due to gravity g.
An object’s weight varies based on the location of the object (another
planet!) since the acceleration of gravity will change, but an object’s mass
remains constant.
On the surface of the earth, the acceleration due to gravity is relatively
constant at g = 9.81 m/s2.
Example 5: Sketch the free-body diagram of a 56.0 kg parachutist when
she just steps out of an airplane at 2000 m and calculate her
weight.
The formula for weight is derivable from Newton’s laws.
Begin by looking for all the forces; here the net force is the force of
gravity
Force equation:
Fnet = Fg
Next, invoke Newton’s 2nd Law
if
Fnet = m a then m a = m g
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Dynamics Problems
After sketching a free-body diagram showing all the forces in the
direction of interest (usually the direction of motion) the general format to
follow - in any order - is to
1) sum up all the forces in the direction of motion to get the net force Fnet
Fnet = F1 + F2 + F3 …
2) invoke Newton’s 2nd Law and equate the net force to the mass times
acceleration
Fnet = F1 + F2 + F3 … = m a
3) if a = 0, then the forces are balanced and we have a constant velocity
problem
If a ≠ 0, then we have an acceleration problem
Example 6: Sketch the free-body diagram of a 25.0 kg boy sitting on a level
floor.
Since there is a force of gravity going downward, because the boy is
not moving (not accelerating), there must be a balancing force upward. Any
object at rest on a surface will have a normal force being exerted by the
surface on the object to keep it at rest (relative to the surface).
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Normal Force (FN): the perpendicular force a surface exerts on an object
to keep it at rest relative to the surface.
By Newton’s 2nd Law, calculate the normal force acting on the boy
above.
This allows us to generalize a formula for the normal force acting on an
object on a horizontal surface.
FN = mg
Applied Force (Tension) FT or FA
When a force is applied on an object, either a push or a pull, from a
second body or person, it is drawn as a pull from the center of the first object
and labelled as the tension or applied force FT.
Applied Force or Tension : any non-natural or man-made applied force
applied to an
object (originally referring to the Tension on a
rope)
When the applied force or any other force does not all fall in the same line
(linear), then we have a 2 dimensional problem that must be resolved into
components.
Example 7: Sketch the free-body diagram of a 25.0 kg wagon being pulled
across a floor with a horizontal force of 50.0 N.
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a)
What is the net force acting on the wagon in the perpendicular
direction to the surface? Describe the motion of the wagon in
the up/down direction.
Fnet =
b)
What is the net force acting on the wagon in the parallel
direction to the surface? Describe the motion of the wagon in
the horizontal direction.
Fnet
We are usually only concerned with the components in the direction of
motion, and can ignore the components in the direction perpendicular to the
motion. In the above example, we would normally only consider the tension
as the force of interest.
Sometimes individual forces have to be broken into components, one
component in the direction of motion and a second component perpendicular
to the direction of motion. Again, the forces in the direction of motion are
the forces of interest.
Example 8: Calculate the final speed of a 25.0 kg wagon initially at rest
being pulled by its handle at an angle of 35.00 to the horizontal
with an applied force of 50.0 N over a distance of 5.00 m.
(Note: since the motion is along the horizontal, only the forces
along the horizontal component are of interest. We can ignore
the vertical forces)
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The first part of this problem is to write a force equation:
Secondly, use Newton’s second law to calculate the acceleration in
the horizontal direction from the net force.
The third part of this problem is to use the acceleration calculated
above to determine the final speed from one of the five acceleration
formulas.
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The Force of Friction
So far, only the ideal cases – without friction – have been considered.
In real cases, friction does exist from contact with surfaces. Usually, the
direction of friction is opposite the direction of motion.
Force of Friction Ff:
The natural contact force that is usually in direct
opposition to motion (in the direction opposite to
motion).
Example 9: Calculate the final speed of the 25.0 kg wagon being pulled at
an angle of 35.00 to the horizontal by a force of 50.0 N over a
distance of 5.00 m. This time there is a force of friction of 20.0
N.
Start with a free-body diagram labelling the forces of interest (in the
direction parallel to the motion).
Determine the force equation in the direction of motion.
From Newton’s 2nd Law, calculate the acceleration.
Finally, from the acceleration, using one of the five acceleration
formulas
calculate the final speed
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Elevator Problems
In elevator problems, the tension T in the cable above the elevator is
opposing the force of gravity Fg. Depending on whether this tension is
greater than, equal to, or less than the force of gravity, the elevator is –
respectively – accelerating upward, not accelerating, or accelerating
downward.
Note that when the elevator is not moving, or moving at constant velocity
either up or down, the tension T is equal to the force of gravity Fg.
Example 17:
A high speed elevator moves up a downtown skyscraper
from the first
floor to the 12th floor. What is the tension in the cable when the
1000 kg elevator (9,810 N) is: (sketch the values obtained in the
diagrams above)
a) accelerating upwards at 5.00 m/s2?
b) moving up at a constant velocity of 10.0 m/s?
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c) decelerating to a stop at –5.00 m/s2?
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Weigh Scale Readings on Elevators
A person standing on a weigh scale in an elevator will see the scale
reading change as the accelerator is accelerating. The scale reading, or
weight, is the tension T opposing the force of gravity.
Example 18: What is the weigh scale reading in Newtons of a 60.0 kg woman
(589 N) in a high speed elevator that is: (Sketch free-body
diagrams with values)
a) accelerating upwards at 5.00 m/s2?
b) moving up at a constant velocity of 10.0 m/s?
c) decelerating to a stop at –5.00 m/s2?
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Atwood Machine (Pulley) Problems
Atwood machine problems involve changes in the direction of force
through the tension in a rope strung across a pulley.
Example 19: An Atwood machine is connected as diagrammed below. Given
that mass 1 is 500 g and mass 2 is 1.00 kg. Assuming the surface
is frictionless, what is the acceleration of mass 1 and mass 2?
The problems can be simplified by straightening out the rope and laying
the entire apparatus along a horizontal. The weight (Fg) of any suspended
mass acts like an "engine" pulling the train of all the cars attached by the rope
forward along a track. It is important to remember that while only one car's
weight is doing the pulling, the total mass of all the cars must be considered to
calculate the acceleration.
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Sometimes two masses are suspended, in which case it becomes a two
headed train with engines pulling in opposite directions.
Example 20: a) Given the same masses as above, calculate the acceleration of
the two
masses below.
b) Assuming they both start at rest, what is the final maximum
velocity
reached?
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Incline Problems
One special type of motion problem in physics involves motion along a
ramp or incline. In incline problems, since motion is along the incline
parallel to the surface, the forces must be broken up into parallel // and
perpendicular  components where
 The forces in the perpendicular components are balanced as the object
is stationary (not accelerating) relative to the surface
 The forces in the parallel direction determine the motion along the
incline surface.
In particular, the force of gravity Fg must be resolved into parallel Fg// and
perpendicular Fg components relative to the surface of the incline.
Example 10:
A 10.0 kg box is on a frictionless incline of 35.00 to the horizontal.
Calculate the force of gravity parallel Fg// and perpendicular Fg
to the surface of the incline.
By similar triangles, a new triangle with the force of gravity Fg as the
hypotenuse allows us to calculate its parallel component Fg// and
perpendicular component Fg from the angle .
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Fg// = mg sin  = 10.0 kg (9.81 m/s2) sin 350 = 56.3 N
Fg = mg cos  = 10.0 kg (9.81 m/s2) cos 350 = 80.4 N
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Motion Along the Incline
The parallel component is the usual direction of motion (if any) along
an incline. In all cases there will be the force of gravity in the parallel
direction Fg//. Depending on the problem, there may or may not be
additional forces such as an applied force T or a force of friction Ff opposing
this force of gravity.
Example 11:
What is the acceleration of the 10.0 kg box on a
frictionless incline at 35.00 to the horizontal?
Here, the only force in the parallel direction is the force of
gravity Fg//.
By Newton’s 2nd Law:
Example 12:
What force of friction would hold the 10.0 kg box
motionless on the incline in the problem above? (Note: these
forces will be opposite in direction to motion down the incline)
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The Normal Force FN on an Incline
In general, since there is no motion (no acceleration) in the
perpendicular direction (the object is not rising above or dropping below the
surface of the incline), the net force in the perpendicular direction to the
incline surface will be balanced by a normal force FN opposing the Fg . This
balancing normal force FN opposing Fg = [Fg cos , gives a general
equation for the normal force on an incline at angle 
FN = Fg cos 
The Normal Force on an Incline
Example 13:
A 25.0 kg girl sits on the side of a hill at 30.00 to the
horizontal. Complete
the free-body diagram below and calculate the normal force FN
as well as the force of friction Ff acting on the girl.
Note: Here there is no motion or - more importantly - no
acceleration, so in both the parallel and perpendicular direction
the forces must be balanced.
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Acceleration On an Incline
When an object is accelerating up or down an incline, the forces in the
parallel direction to the incline cannot be balanced. If there are two forces,
one force must be bigger than the other, and the difference- the net force –
produces the acceleration.
Since the object is not accelerating in the perpendicular component to
the incline, the net force is zero in this component, meaning the normal force
still balances the force of gravity in the perpendicular direction Fg (FN =
Fg)
Example 14:
A 25.0 kg boy starts from rest at the top of a slide, then
slides down the 3.00 m long incline at 25.00 to the horizontal.
The force of friction is 30 N.
Calculate the boy’s final speed at the bottom.
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The Coefficient of Friction
Depending on the nature of the surface of the incline, the force of friction
may be relatively low or high. A characteristic of any surface’s force of
friction is given by its coefficient of friction .
Coefficient of Friction : the ratio of the force of friction to the force
normal is given by
the coefficient of friction, a constant for any given surface
 = Ff
FN
The higher the coefficient, the greater the force of friction for any given
surface.
Example 15: Calculate the coefficient of friction for the sliding 25.0 kg boy
above.
Static versus Kinetic Friction
Experimentally, it is determined that the force of friction is different
for static (non-moving) objects versus kinetic (moving) objects. In general,
the force of friction decreases as an object begins moving, yielding a slightly
lower value of the coefficient of friction for moving objects.
Static friction is the force of friction that opposes an applied force. It
prevents the object from moving. Therefore, the net force is zero and
static force is equal and opposite to the applied force. The amount of
static friction depends on the force applied. In other words, it varies!
Kinetic friction is the force of friction that acts on a moving object. It has a
set value determined by the surfaces of the two materials in contact and does
not vary.
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The following is from http://www.school-forchampions.com/science/frictioncoeff.htm
(Since the quality of the surfaces is not mentioned, you
should only use these readings as a guide. It is best to
measure the coefficients for your specific materials and
conditions of use to obtain accurate values)
Coefficient of Sliding Friction (clean
surfaces)
Material 1
Material 2
Static
Kinetic
Aluminum
Aluminum
1.05 1.35
1.4
Brake
Material
Cast Iron
0.4
-
Brake
Material
Cast Iron (wet) 0.2
-
Cast Iron
Cast Iron
1.1
0.15
Leather
Oak (parallel
grain)
0.61
0.52
Nickel
Nickel
0.7 - 1.1 0.53
Nickel
Mild Steel
-
Solids
Rubber
1.0 - 4.0 -
Teflon
Steel
0.04
-
Teflon
Teflon
0.04
-
Zinc
Cast Iron
0.85
0.21
0.64
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Example 16:
A 25.0 kg boy starts from rest down a slide at an incline
0
of 25.0 to the horizontal. The slide is 3.00 m long down the incline.
Calculate the coefficient of kinetic friction of the slide if the boy is
sliding at constant velocity down the incline.
(0.466)
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Lab 5 - An Incline Problem
Background:
Incline problems are special 2-dimensional problems where the x and y
dimensions are parallel and perpendicular to the slope of the ramp. Each dimension is
treated independently. Any vectors not directly along either dimension like the force of
gravity must be resolved into these two components first.
In a frictionless incline, the parallel component of the force of gravity would be
the only force in the parallel direction. As such it is the net force, responsible for the
acceleration, a, in the parallel direction. If we released a cart (vi = 0) along a ramp of
some length, d, then a theoretical final speed, vf, at the end of the ramp could be
calculated from one of the constant acceleration formulae (which one?).
The real final speed might be somewhat less than the theoretical final speed,
which can be blamed on friction in a real incline. Working backwards, if we can measure
the real final speed at the end of the ramp (for instance, analyzing a ticker tape as before),
then we can determine the real acceleration in the parallel direction. This real
acceleration can be translated back to a net force by Newton's 2nd Law. Since the
parallel component of the force of gravity and the opposing force of friction are the only
two forces in the parallel direction, their vector sum will equal this net force; so knowing
the parallel component of the force of gravity and the net force, we can calculate the
force of friction.
Problem:
Using the cart, ticker tape ramp setup as below, calculate the force of friction
Write-up:
Procedure
- Tabulate and explain your collection of data
Data
- Show measured values needed to solve problem
Analysis
- Show calculations for values not directly measured
Challenge Question: Calculate the coefficient of kinetic friction, (1 mark)
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Newton’s 3rd Law: Action and Reaction Quick lab 3-8
“To every action there is always opposed an equal reaction: or, the mutual
actions of two bodies upon each other are always equal, and directed to
contrary parts.”
The 3rd law restated says that whatever force one object exerts on a
second object, F1 on 2, the second object is exerting and exactly equal force
back on the first object, F2on 1, in the opposite direction of the first force.
F1 on 2 = - F2 on 1
Example :
A 2.00 kg gun (object 1) fires a 2.00 g bullet (object 2),
accelerating the bullet to a maximum speed of 200 m/s over the
gun barrel’s length of 10.0 cm. What is maximum speed of the
gun?
Fgun on bullet = - Fbullet on gun
mbulletAbullet = - mgunAgun
The force of the bullet on the gun is the “recoil” felt upon firing. The
negative refers to the opposite direction from the bullet.
On earth, we can only move forward on very similar lines as the gun and the
bullet: we push backward against the earth which allows the earth to push us
forward. The reason we do not notice the movement of the earth backward
is that we are a very small bullet compared to the earth as a very big gun.
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Example 23: A 59.8 kg boy accelerates as he walks at 1.0 m/s2. The earth’s
mass is 5.98 x 10 24 kg. What is the resultant acceleration of
the earth?
In space or on frictionless surfaces there is nothing to push against, so
motion across a distance becomes quite a problem. Spaceships use rockets
and the principle of Newton’s third law to cause forward motion. The action
is burning fuel propelled at high speed out the back end of the rocket, while
the reaction is the rocket propelled forward. Here, the fuel is the bullet,
while the rocket is the gun.
Example 24: Imagine you are placed in a frictionless suit at rest on centre
ice in the Coliseum, in a game of Survivor where there is a
water bottle and food on the side bench. You are left there for
the week. How will you survive?
Example 25: If a horse tries to pull a cart forward with the exact same force
that the cart pulls on the horse backward, since both forces are
equal and opposite, how is it possible for the horse to pull the
cart forward?
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Unit 2B- Gravitational Forces and Fields
Newton's Law of Universal Gravitation
Kepler had devised a mathematical relationship describing the motion
of the planets. From this, Newton developed a law that not only covered all
planetary objects and their motion in space, but a law meant to cover all
objects in the universe. The power of this law cannot be overstated: from
planets and stars to atoms and molecules, everything in this universe with
mass is covered by this law, hence its name the law of universal gravitation.
The force of gravity Fg is a fundamental force, one of four
fundamental forces in nature.
 very weak force
 only becomes significant for very large masses, like planetary bodies.
 For planets, the force of gravity supplies a centripetal (centre-seeking)
force to all objects near it. Eg. It keeps planets in circular orbits
around the sun, and the moon in a circular orbit around the earth.
Newton's universal law of gravitation relates the force of gravity, Fg ,
between any two objects. There are three quantities involved:
 the mass of one object (m1)
 the mass of a second object (m2)
 the distance or radius of the circular orbit between them (R).
The effect of all three quantities on the force of gravity was tested by
Cavendish's experiment with lead spheres. Cavendish suspended a rod on a
string. A lead ball (m1) on one end of the rod was balanced with a
counterweight on the other end. A second lead ball (m2) was brought a
measured distance away (R) from the first lead ball. The gravitational
attraction between the two caused the rod to twist on its own axis. The angle
of twist corresponded to the force of gravity (Fg).
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Example 26. In a Cavendish type experiment with two lead balls m1 and m2
a distance R, the angle is measured to be 1.5o. Assuming the force of
gravity is directly correlated with the angle, what will be the new angle if the
first lead ball is halved in size, the second tripled, and the distance between
them is reduced by one half?
(9.0o)
By varying the masses (m1, m2) and the distance between r, he was
able to see the three relationships Fg m1 , Fg m2 , Fg 1/R2, leading to
the general equation:
Newton's Law of Universal Gravitation
Fg  G
m1m2
R2
Cavendish used this experiment to determine the universal gravitational
constant, G
G = 6.67 x 10-11 Nm2/kg2
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Example 27: Calculate the force of gravity on a 100 kg man 1000 km above
the surface of the earth. (Data for the radius and mass of the
earth can be found on your data sheet). Compare this weight to
his weight on earth’s surface.
With the universal gravitational constant known, the mass of the earth
could be calculated from the weight of a known mass of an object on the
earth’s surface, and the distance of that object from the center of the earth
(the earth’s radius)
Example 28: What is the mass of the earth if a 1.00 kg mass weighs 9.83 N
on the surface of the earth?
For three object systems:
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Gravitational Fields
By Newton's Law of Universal Gravitation, we know the force of
gravity takes two masses to interact (eg. the earth and the moon).
Sometimes it is convenient to identify the gravitational "force" associated
with only one mass. For instance, we say the earth by itself exerts a force of
gravity, and this force exists even when there is no moon, or person, or any
other second mass present.
To describe how only one object or mass exerts a force of gravity we
need to introduce the concept of field theory. A ‘field’ is the ‘sphere of
influence’ around an object – the area in which it will apply a force onto an
object.
A gravitational field assigns a number to all points in 3-dimensional
space around one mass, representing the gravitational "pull" exerted by that
one mass. We have used one field number already and probably were not
aware of it: the number 9.81, given the symbol g, which represents the
gravitational field strength on earth’s surface. This number, 9.81 m/s2, is
only one possible number for g - all points exactly the same distance from
the earth's center (the earth’s radius along the poles).
Field lines (vector arrows) represent the strength (spacing and length) and
direction (arrowheads). Like force, the gravitational field, g, is a vector, and
its direction is pointed towards m1’s center, for instance the earth's.
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Like gravitational force, the gravitational field of the earth exerts gets
weaker as we move away from earth's center, the earth's gravitational field
numbers will get lower than 9.81 for points farther away from the surface of
the earth, for example, 6.0.
To get the gravitational field number, g, for all points in space around
one mass m1, we use Newton's Law of Universal Gravitation, and divide out
the second mass m2.
The Gravitational Field, g
How far from Earth’s surface would you have to be to have a g of 6.0?
This tells us the g-field number for any one mass, at any point a
distance R from the center of the mass exerting the g-field, m1. For the
earth's g-field, the earth is m1. Any other mass affected by the earth's field
becomes m2.
Fields are not forces: to get back the gravitational force from a field, we
need only multiply the g-field number by the second mass m2.
Example 29:
Use the mass and radius of the earth from your data
tables to calculate the g-field at a point in space twice the radius of the earth
away from the earth's center.
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Gravitational Acceleration g on Earth and Other Planets
The earth is not a perfect sphere. As the earth revolves on its own axis, the central
equatorial region bulges outwards from the spinning, which tends to flatten out the pole regions.
The result is a (very slight) egg shaped ovoid instead of a sphere.
Since the gravitational field g is dependent on the planetary radius R, because the radius
of the earth is larger at equator and smaller at the poles, the value for g is slightly different from
equator to poles. The value for g on the surface of the equator is 9.79 N/kg –farther from the
earth’s centre - whereas the value for g is 9.83 N/kg on the surface of the poles – nearer to the
earth’s centre. The average value is 9.81 m/s2, and is very near the value in Canada just north of
the 49th parallel (49o latitude). For most instances, a value of 9.8 m/s2 is quite safe to assume for
anywhere on earth’s surface.
The value of g also depends on planetary mass m. This means different planets have
differing gravitational fields. The moon, for example, has a gravitational field approximately
one-sixth that of earth. This means that objects on the moon will have one-sixth the weight as
compared to the surface of the earth, or will accelerate at one-sixth the rate of g = 9.81 m/s2
towards the moon’s centre. This means that people jumping on the moon can jump much higher,
or an average golfer can hit a golf ball on the moon over 500 yards, much further than Tiger
Woods’ best drive!
Example 30: The mass of the moon is 7.349 x 1022 kg and its equatorial radius is 1737.4
km.
a) What is the gravitational field g on the moon’s equator?
b) Supposing a person can jump 1.00 vertical metres high on the surface of
the earth. How high could he jump on the surface of the moon? (the height and
gravitational field are inversely related)
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Weightlessness:
In what situations do you feel weightless?
1.no gravitational force is acting – in space
2.when normal force becomes zero – when an elevator hits the top,
when a ride changes direction from going up to coming down.
3.when gravitational force is balanced by another force – free fall at
terminal velocity – on a ride, skydiving.
True Weight – Fg
Apparent weight (weight felt) – is FN
In the above examples, are you truly weightless in each??
1. Fg is zero so you are weightless!
2. And 3. there is still gravity acting on the object so you are not truly
weightless. Gravity is still affecting you.
To be truly weightless, there has to be no force of gravity and no net force
on an object. A condition that occurs in outer space when not moving or
moving at a constant speed (a=0).
Pg 225 #1,2 on bottom
Pg229 #7-9
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GRAVITY AND EARTH’S TIDES
Newton explained that the tides are caused by the moon and the sun’s forces
of gravity on Earth. However, other factors have to be taken into effect as
well. eg. shape of coastline, ocean floor, see pg 211.
The height of tides varies with the location on Earth. In some areas they are
a low as 0.5 m difference and others vary by 18m between low and high
tides. The Bay of Fundy has the highest tides in the world. (Canada’s east
coast)
Read bottom of pg 211 (last half of the page)
Do practice problem pg 212 #1
Read pg 213-214
Do #4,5 on page 215
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Newton- The Great Thinker (*optional)
There is a story about Newton’s discovery of the universal law of gravitation
while sitting in his garden one day. Back in Newton’s early days at Cambridge
university, the Black Plague (bubonic plague carried by rats) swept over Europe, wiping
out a large percentage of the population. In 1466, the university was forced to shut down
and Newton went home, back to his mother’s farm in Lincolnshire. It was during this
time that he formulated most of his important contributions to mathematics and physics
including the binomial theorem, differential calculus, vector addition, the laws of motion,
centripetal acceleration, optics, and universal gravitation. Newton was not even 22 at the
time.
Newton’s biographer writes in his memoirs that an apple falling from a tree
provided Newton his inspiration for the law of universal gravitation. As Newton thought
about the apple and its fall, he began to see the relationship between the similarity
between the moon and the apple’s motions. Might not the motion of the moon and the
apple be caused from the same thing?
“ I began to think of gravity extending to the orb of the moon, and … from
Kepler's rule I deduced that the forces which keep the Planets in their orbs must be
reciprocally as the squares of their distances from the centres about which they revolve:
and thereby compared to the force required to keep the moon in her orb with the force of
gravity at the surface of the earth, and found them to answer pretty nearly. All this was in
the two plague years of 1665 and 1666, for in those days I was in the prime of my age for
invention, and minded mathematics and philosophy more than at any time since…”
We can attempt to recreate Newton’s derivation of the force relationships between
the apple on earth and the moon as an example of the use of proportions rather than
equations to solve a physics problem. Knowing only the radius of the earth (RE = 6.37 x
108 m) and approximating the orbital radius of the moon as 60 times RE, Newton was
able to arrive at a very close approximation to the period of the moon, verifying the
inverse square relationship.
For an object in circular motion due to the force of gravity, we start with the
centripetal force Fc (force causing circular motion- see next unit) being due to the
gravitational force Fg.
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Fg  Fc
v2
m
R
 2 R 
 T 
m 
R
4 m 2 R

T2
4 m 2 R

K R3

2
substituting (v 
2 R
)
T
substituting (T 2  K R 3 Kepler ' s Law)
4 m 2
K R2
 1 
 cons tan t  2 
R 
1
Fg  2
R
This inverse square relation gives an approximation to the force of gravity on the moon
relative to the force of gravity on an apple on the surface of the earth. If the distance R to
the moon is approximately 60 times the radius of the earth, then the force of gravity on
the moon is approximately 1/60 squared or 1/3600 as compared to the apple on earth.
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Again, starting with the first principle that the gravitational force is the centripetal force:
Fg  Fc
m g m
v2
R
 2 R 


T 
g
R
2
4 R
 2
T
2
or ,
4 2R
g
This equation allows Newton to estimate the period T of the moon’s orbit.
Substituting the new radius of the moon as 60 times the radius on earth, and the
gravitational field on the moon as 1/60 squared times the acceleration due to gravity on
earth, we can obtain a very close approximation of the period of the moon’s orbit.
4 2R
Tmoon 2 
g
T2 

4  2 (60 RE )
60 g E
4  2 (60 6.37 106 m  )

1
m
[9.81 2 ]
3600
s
 1 h  1 day 
Tmoon  2.353110 6 s 


 3600 s  24 h 
 27.2 days
The true value for the moon’s orbit is 27.3 days. It should be remembered that
Newton was performing these calculations in the days without computers or calculators.
Newton was a great thinker but rather eccentric, and often mean and vindictive.
He did not bother to publish many of his works or ideas, often until much later or when
he was goaded to by friends like Edmund Halley (of Halley’s comet). Yet when other
scientists or mathematicians claimed credit for a discovery, he was merciless in
denouncing them, as with his lifelong attacks of Robert Hooke (of Hooke’s Law and
microscope fame) and Gottfried Liebniz (co-discoverer, with Newton, of calculus). In
his later years he became quite famous and was appointed Warden of the Royal Mint, a
lucrative and important post. One of his duties there was to chase down counterfeiters
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and Newton managed to convict many prominent ones who were then tried, executed by
hanging then drawn and quartered. He is reported to have died a virgin.