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.

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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 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

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(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.

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

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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.

(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.

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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 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?

(0.998 days, 3.08 km/s)

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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?

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Now roll a marble along the floor. Again, what happens and why?

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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.

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b) the bus turns left

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c) the bus slows to a stop.

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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.

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: Which has more inertia, a bus or a car moving at the same speed?

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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 Fnet = ma

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.

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 forcesm – 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 force- FA) 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 problemIf 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.

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.

FN = mg

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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.

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(Note: since the motion is along the horizontal, only the forces along the horizontal component are of interest. We can ignore the vertical forces)

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 apparent weight, is the normal force 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 magnitude of 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 magnitude of the acceleration of the two masses below.

b) Assuming they both start at rest, what is the final maximum speed 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-for-champions.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 0.53Nickel Mild Steel - 0.64Solids Rubber 1.0 - 4.0 -Teflon Steel 0.04 -Teflon Teflon 0.04 -Zinc Cast Iron 0.85 0.21

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Example 16: A 25.0 kg boy starts from rest down a slide at an incline of 25.00 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 ProblemGO 1: You will explain the effects of balanced and unbalanced forces on velocity

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

Extra Question: Calculate the coefficient of kinetic friction, (1 mark)

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Newton’s 3rd Law: Action and Reaction Quick lab 3-8

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.

“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.”

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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 Gravitationhttp://www.bing.com/videos/search?q=Bending+the+Fabric+of+Space&adlt=strict&view=detail&mid=CCEF161DAF039FC8BD89CCEF161DAF039FC8BD89&FORM=VRDGAR

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).https://www.youtube.com/watch?v=EE9TMwXnx-s

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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 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:

4.0 m 8.0 m

3 equal mass objects: 1.2 x 103kg

What is the net force on object B?

What is the net force on object C?

A B C

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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 space2.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 bottomPg229 #7-9

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 #1Read pg 213-214Do #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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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:

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.

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.

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