Physics Test 1

Lesson 8: Velocity

Two branches in physics examine the motion of objects:• Kinematics: describes the motion of objects, without looking at the cause of the motion

(kinematics is the first unit of Physics 20).• Dynamics: relates the motion of objects to the forces which cause them (dynamics is the second

unit of Physics 20).

As we work through these two units on kinematics and dynamics (and through the rest of physics) we will discuss two kinds of measurements (quantities):

• scalar: scalars have magnitude (a number value), but no direction.Examples: time, mass, distance. Mass is a great example, since it has a number value (like 58 kg), but we don't give it a direction (like “East”).

• vector: have magnitude and directionExamples: velocity, force, displacement. Force has a magnitude (like 37 N) and a direction (like "pushed to the left").

Displacement & TimeIn kinematics we need to be able to have a way to describe the motion of the objects we will be studying, whether it's a car or an atom.

• The most basic information you must have to describe the motion of an object is its displacement, and the time it took to move that far.

• The displacement of an object is always measured from some reference point (which is usually “zero”, at a location at the start of the motion of the object).

• Although we use the words “distance” and “displacement” interchangeably in everyday language, they mean very different things in physics.

• The distance between two objects is scalar, since it doesn't matter which direction you measure it from. e.g. “We are standing 2.3m apart.”

• The displacement of an object is a vector, since you have to state the direction the object has traveled. e.g. “The car moved 2.56km east.”

The most simple formula for calculating the displacement of an object is…

Δd = df - di

• The Δ symbol is the greek letter “delta” and means “a change in…”• The subscript “f” and “i” stand for final and initial.• So, in this formula, we calculate the displacement of an object by taking the final position

minus the initial position.

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“Quantity” is the root word for "quantitative" measurements. This means you're supposed to get a number answer. A “qualitative” measurement describes qualities of the data, like “the apple is red.”

Example 1: A truck is passing a mark on the road that says 300m, and then passes another one 10s later that says 450m. Determine the distance the truck moved.

Δd = df - di = 450 - 300 = 150m

Note: If the example had asked for the displacement, we would have to include a direction (like “east”) in our answer.

Example 2: You start walking home from school. After walking 1.3 km North, you get a phone call on your cell from your mom asking if you can meet her at the mall. You will have to turn around and walk 2.5 km South. Determine your distance and displacement to get to the mall.

Let's start by looking at a quick sketch of the situation, as shown at right.● From the school you first walked 1.3 km [N].● You then turned around and walked 2.5 km [S].

If we want the distance you walked, we need to look at all the walking you did, without considering direction.

d = 1.3 + 2.5 = 3.8 km

When we look at your displacement, we need to consider the direction that you walked. Even though you walked North at first, turning around and walking South canceled out all of your initial movement. When we measure displacement we are only where you started and where you finished, not all the stuff in between. We will consider moving North to be positive, and South to be negative.

d=1.3 km−2.5=−1.2km [South]

Notice that the displacement is smaller and negative when compared to your distance. That's because even though you actually moved your body around the city 3.8 km, all you really accomplished by the end was moving 1.2 km South of where you started.

Velocity vs SpeedOne note for you before we begin to talk about how displacement relates to velocity. Avoid using the word speed when describing any velocity.

• Speed is a scalar quantity (it doesn't have direction).• We usually want the velocity of an object, since velocity is a vector. That means we also know

the direction the object is traveling. The more information the better.

Example 3:What speed did you drive today along Yellowhead? => I drove at 72km/h.What was your velocity along Yellowhead? -> I drove East at 72km/h.


Notice that the word “determine” has been bolded in the question. This is a “directing word” telling you what to do in the question. Get used to seeing these, since they are used on the Physics 30 diploma.

Illustration 1: Walking around after school.

1.3 km [N]

2.5 km [S]

Average VelocityThis leads to the first major formula for the calculation of average velocity.

v=dt

v = velocity (m/s)d = displacement (m)

t = time (s)• It is called average velocity because it looks at your overall velocity for the entire trip, not at

any one particular velocity you might have been traveling at during the trip.• What you have to measure is the total displacement divided by the total time.

• If you drive 275 km to Calgary in 3.00 hours, I calculate your velocity based on this information. The velocity I calculate, 91.7km/h, will be your average velocity. ○ I'm looking at the entire trip. It would certainly be rare if you had driven at exactly

91.7km/h every single moment during your drive to Calgary. ○ Sometimes you would have been going faster, sometimes slower, but overall your

average velocity was 91.7km/h. • You basically have to look at the start and finish only, how far did you move from

where you started (displacement) in a certain amount of total time.• These questions can also involve moving at different velocities for different periods of time; we

need to be really careful with these questions...

Example 4: I try to run the 100m race to break the world's record. Unfortunately, it takes me 16.83s to complete the run. Determine my average velocity.

This is an easy calculation... nothing fancy.

v=dt=

10016.83

=5.941771=5.94m / s

This is my average velocity. It does not show that I have to speed up at the start of the race, or that maybe I was slowing down near the end.

Example 5: A car drives along the highway at 115 km/h for 2.50 h. Once in the city, the car drives at 60.0 km/h for the next 0.500 h. Determine the average velocity of the car.

The average velocity is based on he total displacement of the car for the entire time it was moving, so we first need to figure out the total displacement and the total time.

First part of the drive... Second part of the drive...

v=dt

d=vt=1152.50d=287.5km

v=dt

d=vt=60 0.50d=30km


When you write the formula, you can skip putting in the arrows you see on the data sheet. They're just there to remind you that those measurements are vectors.

Warning!You MUST resist the temptation to just add the two velocities and divide by 2!

So, in total, the car moved 317.5 km in 3.00 h. Its average velocity is...

v=dt

v=317.53.00

v=105.83̄=106km /h

In everyday conversation we usually talk about velocity in kilometres per hour (km/h).

• To convert m/s to km/h multiply by 3.6 (this is an exact value and has an infinite number of sig digs.)

• The answer from Example 5 would be 21.4 km/h.• If you ever do a calculation like this, use the original number on your

calculator, not the rounded off answer.• Also, you can convert km/h to m/s by dividing by 3.6. This is more

important because you must make sure all your numbers are in standard units before starting a calculation.

Note that in the above example, the displacement and the velocity were positive numbers.• Positive and negative tell you which direction you are going with respect to the reference point.

(Remember, these are vectors.)• A positive velocity means you are moving forward , to the right, or up, while negative means

you are going backwards, to the left, or down.• This is why it is so important to pay attention to the numbers you are using in your calculations.

Example 6: A train is moving backwards at a velocity of 13.5 km/h for 6.40 minutes. Determine the train's displacement.

First we need to make sure we are dealing with standard units, and that the numbers have the correct sign.

v = -13.50 km/h = -3.75 m/st = 6.40 minutes = 384 s

v=dt

d=vt=−3.75 (384)

d=−1440=−1.44e3m

Example 7: Look back at Example 2. Determine your speed and velocity if the walk took you one hour and ten minutes.

First thing you should do is change the time into seconds. 1 hour = 3600 s10 minutes = 600 s1h 10min = 4200s

I can give myself a bunch of sig digs here. Since I wrote out the time in the question in words, I basically have as many sig digs for the time measurement as I want.


Warning!This conversion from km/h to m/s by dividing by 3.6 only works for velcoity! Do NOT use it for any other conversions!

To change minutes into seconds, we need to use the ratio of 60s/min. Do not change 6.40 minutes to 640s, because that is wrong.

To figure out the speed, we need to use the distance (in metres!) you traveled in 4200 s. That way we are using only scalar measurements.

v=dt

v=3.8e34200

v=0.9047619=0.90 m /s

To figure out the velocity we need to use the displacement you traveled. Now we are using only vector measurements.

v=dt

v=−1.2e34200

v=−0.28571429=−0.29m / s

Uniform MotionIn many of the questions we will be doing we have to assume that the object is moving at exactly the same velocity the whole time.

• Although this is not very realistic, it makes doing the questions a lot easier.• For now we will assume that the object is not accelerating at all.• If the velocity of an object is always the same, we say it has a constant velocity. We can also

call this uniform motion.

You still use the same formula as for average velocity.• Uniform motion is the easiest kind of motion to describe and measure, since it is always the

same. If the object is accelerating in any way we have to use different formulas (coming up in Lesson 10).

• In the examples you've done so far, and in most questions you'll do for now, you assume that it is uniform motion unless you are told otherwise.

Instantaneous VelocityIn real life we often have to deal with an object traveling without uniform velocity. Things are always speeding up and slowing down.

• This is the situation if you ask your friend how fast she is driving when you're in a car.• She'll glance down at the speedometer and tell you how fast she is going, but that is only

how fast she was going at that instant of time!• A split second later, she might be going a bit faster or a bit slower. Most people don't

drive their cars at a totally constant velocity.• That's why we call the measurement she gave you an instantaneous velocity.

Instantaneous velocity is the velocity of an object at one moment of time. • It is often easier to measure instantaneous velocity if you are looking at a graph of the motion

of an object (this also comes up in a later lesson).


http://studyphysics.ca/2007/20/01_kinematics/10_acceleration.pdf

Homework

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Lesson 9: d-t & v-t GraphsGraphing the motion of objects gives us a way to interpret the motion that would otherwise be difficult.

● Graphs will also allow you to show a large amount of information in a compact way.

Essentially you need to be able to sketch and interpret two main kinds of graphs in kinematics:1. Displacement – Time Graphs

● Sometimes called d-t graphs, or position – time graphs.2. Velocity – Time Graphs

● Sometimes called v-t graphs.

Displacement - Time (d-t) GraphsThis type of graph is based on the most basic things we need to know about the motion of an object (position and time).

● Typically you will be given a table of values that show the displacement of the object over a particular period of time.

● If the graph shows complex motion (such as Illustration 1 below), you do not just draw a single best fit line. Instead, you need to look at each section of motion and determine what kind of line best fits the data.

● Don't worry too much about sketching these complex situations... it is much more likely that you will draw an object moving in one way only.

● For the example graph shown below, imagine that you are running in a marathon, and we have decided to graph your movement.

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Illustration 1: d-t graph of a person running a marathon.

Now lets look at a description of the person's movement in each of the major sections.

Zero to 90s

Look at how you are running in those first 90 seconds. ● Every 30 seconds you have moved about another

150m away from the starting point… you must be moving at a constant positive velocity!

● A constant positive velocity is shown on a d-t graph as a straight line that slopes upwards. It is a linear relationship.

● In fact, if you found the slope of the line in this section, it will be the velocity that you were running at.

slope=riserun

=dt=v

90s to 150s

Yikes! You ran too fast at the start and now you’re out of breath!

● During this time period, your position on the graph has stayed the same…450m.

● This just means that you are standing in the same spot, exactly 450m away from where you started.

● A flat horizontal line means you are stopped.

150s to 240s

You must have started running forward again, since a positively sloped line means a positive velocity. ● Notice that this section of line is a little steeper than the first section. You are now running

about 200m every 30s. ● A steeper line (which has a bigger slope) means that you are moving at a faster constant

velocity.

240s to 300s

In this section the line slopes down, which means it has a negative slope. ● Since slope is equal to velocity, this must mean that you are running backwards.● A negative slope means a constant negative velocity.

● You must have forgotten to pass a check point, so you ran back to it.

300s to 360s

Again, we have a horizontal line. You must be stopped.

360s to 510s

You know that you have only one chance to still win the race… run as fast as you can! ● During this time period, the line curves upwards. ● The line becomes steeper and steeper as it continues. This means that the slope of the line is

getting bigger and bigger.


The slope of a d-t graph always equals the velocity of the object at that time.

The slope of the line is the velocity, and the slope of a flat line is zero. So, the velocity is zero.

● Since slope is related to velocity, your velocity must be increasing. You are accelerating!

● A curved line on a d-t graph means acceleration.

Here’s how you can remember if it was positive or negative acceleration on a d-t graph.

● If you see any part of the happy clown's face on a graph, it is positive acceleration.● If you see any part of the sad clown's face, it is negative acceleration.

Velocity-Time (v-t) GraphsYou need to remember that the rules you learned above for d-t graphs do not apply to v-t graphs.

● A common mistake by Physics 20 students is when they assume that all types of graphs work the exact same way.

● The graphs can be related to each other, but that doesn’t mean you look at them the same way.● The following v-t graph is based on the same data as we used for the d-t graph, but we will need

to look at what’s different.


A curved line on a d-t graph means acceleration is happening.

Illustration 2: Happy Clown

d

t

Illustration 3: Unhappy Clown

d

t

Zero to 90s

Remember that in the first 90 seconds you were running at a positive constant velocity.

● On this graph we see a horizontal line that reads “5m/s” for those same first 90 seconds.

● On a v-t graph a flat line means constant velocity.

90s to 150s

This is the section of time when you stopped because you were out of breath. ● Notice that "stopped" is shown by a horizontal line at exactly 0m/s. ● It's a flat line which means constant velocity. It just so happens that your constant velocity is

0 m/s.

150 to 240 seconds

You are running forward again. ● To show a faster velocity than earlier, we have a flat line that is higher than the previous one.

240 to 300 seconds

This is when you are running back to the check point. ● You are running at -3.3m/s.


Illustration 4: v-t graph of the same person running.

A flat line on a v-t graph means constant velocity.

● A negative velocity is shown as a negative slope.

300 to 360 seconds

Again, we have a horizontal line at zero. You must be stopped.

360 to 510 seconds

This is the section in which we already figured out you must be accelerating; you run faster and faster.

On a d-t graph the line curves upwards, but not on a v-t graph. ● On a v-t graph the line is straight and has a positive slope. ● A straight sloped line on a v-t graph means acceleration. ● The slope of the line is equal to the acceleration; a

positive slope is a positive acceleration, and a negative slope is a negative acceleration.

slope=riserun

= vt

=a

There is one other trick you need to know about v-t graphs. ● If you multiply velocity by time, what do you get? According to our formula...

v=dt

manipulated to d = v t

..displacement!● So, if I have a v-t graph and I calculate the area under the line

(which means I’m calculating velocity multiplied by time), I will know the object's displacement.

Homework

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A slope on a v-t graph means acceleration. The slope of the line is equal to the acceleration.

The area under the line of a v-t graph is the displacement of the object.

Lesson 10: Acceleration

Acceleration is a vector which measures the change in the velocity of an object.● Don’t forget that velocity is a vector, so it has magnitude and direction. ● This means acceleration could be any of the following three…

1. a change in speed, the magnitude of the velocity (from 34 km/h to 67 km/h) 2. a change in direction (from East to North-East) 3. a change in both speed and direction (from 34 km/h East to 12 km/h West)

Acceleration is a measure of the rate at which velocity changes.● Since velocity is a measure of the rate of change in displacement and had the equation…

v=dt

● the equation for acceleration should be similar, since it is a rate of change in velocity...

a= vt

a = acceleration (m/s/s or just m/s2)Δv = change in velocity (m/s)

Δt = time interval (s)

● For day to day use this formula is usually expanded to show that the change in velocity means final minus initial.

a=v f −v it

Notice the units for acceleration.● Since we take velocity and divide by time, we would get (without any simplification) m/s/s for

units. ● What this really means is that for every second that passes the velocity of the object will change

by this many metres per second. ● To make it a bit easier to read we use a little math to get m/s2 for the units.

Example 1: A car moving at 50km/h accelerates to 60km/h in 7.0 seconds. Determine its acceleration.

We first need to change the velocities from km/h into m/s:50 km/h = 13.88888889 m/s 60km/h = 16.66666666 m/s

Now we use the formula.

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Warning!Keep the actual 13.88888889 that you see on your calculator for your calculations… never round off until the end of a problem!!! You do need to keep in mind that the original number only had 2 sig digs.

a=v f −v it

a=16.666−13.888

7.0sa=0.3968254=0.40m / s2

If you got a different answer than this, check the following…● Did you use the rounded off values of 17m/s and 14 m/s, or did you use the actual values

you saw on your calculator? ● When you got to the end, did you use the rules for rounding off according to sig digs?

Negative AccelerationIn physics, acceleration is not always an increase in velocity. It can also be a decrease in velocity.

● Even though you might have heard people use the word deceleration to describe an object slowing down, this isn’t really proper physics.

● Deceleration can only mean one thing, slowing down. We will see that we have to expand our way of thinking about acceleration to include more than just this one idea.

● Instead, we call a decrease in velocity of an object traveling forward a negative acceleration. ● This actually helps us when doing calculations and also allows us to give more meaning to

acceleration.

Example 2: You are walking down the street when you see an enormous, 112kg pickle rolling towards you at 12 m/s. You are, of course, surprised by a pickle of this size, let alone the fact that it is rolling down the street. You jump in front of it and begin pushing on it until you finally bring it to a stop 17.5s later. At this point you are arrested for interfering in the “World’s Largest Pickle Rolling Championships”. Determine the acceleration of the pickle.

In this example you have an object that is initially moving and ends up at rest. If you think of the Δv = vf - vi part of the acceleration formula, you’ll notice that a final velocity of zero (you stopped it), minus another number gives you a negative Δv. Since time is never negative, your acceleration will be negative.

a=v f−v it

a=0−1217.5s

a=−0.685714=−0.69m/ s2

In this case the negative sign means you were “taking away” velocity from an object that was moving in the positive direction to begin with. The object was slowing down.


Example 3: Once you have been arrested, the officials start rolling the pickle back towards the starting line so that Haans van der Winkle, the current champion from the Netherlands, can have a second try. After pushing for 8.8s, they get the pickle rolling backwards (towards the starting line) at 4.31m/s. Determine the acceleration of the pickle.

In this example, they cause an object to speed up, but it’s moving in the opposite direction. Since velocity is a vector, we can just put a minus sign in front of its speed and say that it is moving in the negative direction.

a=v f−v it

a=−4.31−0

8.8sa=−0.4897727=−0.49m / s2

In this case, does the negative acceleration mean the pickle was slowing down? Not at all! It was speeding up in the negative direction. A negative acceleration can mean…

1. An object moving in the positive direction (positive velocity) is slowing down.

or

2. An object moving in the negative direction (negative velocity) is speeding up.

So how do you keep track of all these types of acceleration. “Speeding up backwards means… I don’t know!” Try this. It always helped me.

● We usually visualize speeding up as positive, and slowing down as negative. I put that in the top row.

● A positive velocity means it’s going in the positive direction (like forwards), and a negative direction is backwards.

● Now play a little “Battleship.” positive X positive = positive positive X negative = negative negative X negative = positive


● So, reading off the chart, an object moving backwards (-v) that is going faster and faster (speeding up) means it has a negative acceleration.

● Go ahead and quickly sketch this out (from memory) on a piece of scrap paper at the start of an exam to help remind yourself!

Change in DirectionAcceleration as a change in direction can be felt if you are in a car going through a turn.

● Even if it’s speed is the same (e.g. 50 km/h), the direction is changing.

● You feel this acceleration as the you feel your body being “pushed” towards the outside of the curve.

● Your direction is changing at every instant!

This is why it is important to distinguish between “speed” and “velocity” and why we define acceleration as a change in velocity.

● Acceleration, like velocity, has direction. ● If we change either speed or direction or both, velocity changes and we accelerate.

Homework

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We will see later that this is not exactly what is happening (even though it feels that way). We will eventually even have a way to do simple calculations in these situations.

Lesson 11: The Other Formulas

There are several other formulas that are very useful when the acceleration is uniform.● Do not use these equations if the acceleration is changing! The acceleration must be constant.● These other formulas are based on combinations of the basic velocity and acceleration formulas,

as well as interpreting graphs.● Although the formulas are shown here in a particular numbered order, you do not need to

identify them this way.

Formula 1The first formula is based on knowing information about displacement, final and initial velocities, and time.

d= v f v i2 t● At first this may seem to be an odd acceleration formula, since acceleration does not appear in

the formula as a variable.● Notice that there are two velocities, vf and vi, so we know that there must be

acceleration.● Only use this formula when you know for certain that the object has been going through

a constant acceleration, even though “a” doesn't appear in the formula.

Example 1: Determine how far a vehicle moved if it started at 12 m/s and accelerated up to 47 m/s in a time of 34s.

d=( v f+v i2 ) td=( 47+12

2 )34

d=1003=1.0e3m

Formula 2We will do problems where we have information about displacement, initial velocity, time, and acceleration. The formula for these situations is...

d = vit + ½ at2

● Be careful with this formula. Only the time is squared in the last term, not acceleration and time.

● As a bonus, a lot of the time vi will be zero, which cancels out the first term and leaves you with…

d = ½ at2


In your text book, this formula is written slightly differently as d =1/2 (v

i + v

f) t. It is still the exact

same formula.

Example 2: Occasionally the US Air Force calls me in to test fly their “birds”. A few weeks back I was flying along in my F-22 at 97m/s when I decide to kick in the afterburners for 12.3s. If the afterburners can generate enough thrust to accelerate the F-22 at 26m/s2, determine how far I travelled during that time.

d = vit + ½ at2

= (97) (12.3) + (0.5) (26) (12.3)2

d = 3159.87 = 3.2e3 m

Example 3: I am in a F-22 that is on the runway. From rest, I accelerate the plane at 3.9m/s2 for 14.5s to reach take off velocity. Determine how long the runway must be.

This is an example of a question where the initial velocity is zero (since I’m starting from rest), so…

d = vit + ½ at2

d = ½ at2

= (0.5) (3.9) (14.5)2

d = 409.9875 = 4.1e2 m

Formula 3There is a formula related to formula 2 that can be used when we know the final velocity instead of the initial.

d = vft - ½ at2

● Notice that the differences are final instead of initial velocity, and the minus sign instead of addition.

● Otherwise, this formula is used the same way as formula 2.

Example 4: A car drives 83m while accelerating at 2.4m/s2 for 4.9s. Determine the final velocity of the car.

We're going to have to manipulate the formula to solve for vf. Keep in mind that you may have to manipulate any of the formulas we are looking at.

d=vf t−1/2at 2

v f=d+0.5a t 2

t

vf=83+(0.5)(2.4)(4.9)

2

4.9vf=22.81878=23m /s


Formula 4Another very useful formula is the following…

vf2 = vi

2 + 2ad

● Very handy when you have a question with both velocities, acceleration, and displacement.● Don’t forget to do the square root at the very end if you are solving for a velocity, as the

following example shows…

Example 5: Determine the final velocity of a car that starts at 22 m/s and accelerates at 3.78 m/s2 for a distance of 45 m.

v f2=v i

2+2ad

v f2=222

+2 (3.78)(45)

v f2=824.2

v f=28.70888=29m / s

Many people forget that in the last step you need to square root in order to get the velocity instead of the velocity squared!

How to Choose the Right Formula!So, how do you figure out which formula to use for a particular problem?

● As you look back through the formulas, you'll see that of the five basic things we measure about the motion of an object (vf , vi , a , t , and d) , each formula only has four.

● To choose the correct formula, figure out the one thing that you are not given and not asked for in the question. Choose the one formula that does not have that variable.

● The following table may help.

Formula a vf vi d t

a=v f−v it * * * X *

d= v f v i2 t X * * * *d = vit + ½ at2

* X * * *d = vft - ½ at2

* * X * *vf

2 = vi2 + 2ad * * * * X


For example, let's say I had a question where I am given acceleration, displacement, and time, and asked to find initial velocity.

● The only thing I wasn't given, and I wasn't asked for, is final velocity. ● The only formula that does not have final velocity is d = vit + ½ at2. This is the formula I should

use.

Remember that for all of these formulas, you may be required to manipulate the formula to find the answer you are looking for.

● Always follow the rule of finding the formula that has all the knowns and unknown that you have.

● Write down the original formula as it appears on the data sheet.● Then manipulate it for your unknown, and solve.

Example 6: A pain can is knocked off the top of a building. If it falls for 3.5 s, determine the height of the building.

d=v i t+12at2

d=0+0.5(−9.81)(3.52)

d=60.08625=60m

Homework

p.48 #1,2p.50 #1,2p.51 #1,2p.52 #1,2p.53 all odd


Can you see why I am using this formula?

Lesson 12: Gravity

AristotleFrom the time of Aristotle (384-322 BC) until the late 1500’s, gravity was believed to act differently on different objects.

● This was based on Aristotle's observations of doing things like dropping a metal bar and a feather at the same time. Which one hits the ground first?

● Obviously, common sense will tell you that the bar will hit first, while the feather slowly flutters to the ground.

● In Aristotle’s opinion, this was because the bar was being pulled harder (and faster) by gravity because of its physical property of having more mass.

● Because everyone could see this when they dropped different objects, it wasn’t questioned for almost 2000 years.

GalileoGalileo Galilei was the first major scientist to refute (prove wrong) Aristotle’s theories.

● In his famous (at least to Physicists!) experiment, Galileo supposedly went to the top of the leaning tower of Pisa and dropped a wooden ball and a lead ball at the same time.

● Both were the same size, but different masses.● Down below an assistant watched for them to hit the ground.

● They both hit the ground at the same time, even though Aristotle would say that the heavier metal ball should hit first.

Galileo had shown that the different rates at which some objects fall is due to air resistance, a type of friction.

● Get rid of friction (air resistance) and all objects will fall at the same rate.● Galileo said that the acceleration of any object (in the absence of air resistance) is the same.● To this day we follow the model that Galileo created.

ag = g = 9.81m/s2

ag = g = acceleration due to gravity

Since gravity is just an acceleration like any other, it can be used in any of the formulas that we have used so far.

● Just be careful about using the correct sign (positive or negative) for the variables in the problem. I would strongly suggest you always stick with up being positive, and down being negative.


Warning!In any of the questions we will do, we must assume that the entire problem happens near the surface of the Earth. We will learn in later chapters that the acceleration due to gravity does decrease as you move further away from the centre of the Earth.

Upwards+ velocity+ displacement

Downwards- velocity- displacement- gravity

http://en.wikipedia.org/wiki/Aristotle

http://en.wikipedia.org/wiki/Galileo

Examples of Calculations with Gravity

Example 1: A ball is thrown up into the air at an initial velocity of 26.3m/s. Determine its velocity after 1.80s have passed.

In the question the velocity upwards is positive, and we'll keep it that way. That just means that we have to make sure that we use gravity as a negative number, since gravity always acts down.

a=v f−v it

v f=v i+atv f =26.3+(−9.81)(1.80)

v f=8.642=8.64m / s

This value is still positive, but smaller. That means it's still going up, it's just not moving as fast as it was originally.

Example 2: I throw a ball down from the top of a cliff so that it leaves my hand moving at 12m/s. Determine how fast is it going 3.47 seconds later.

Be careful! At the very start of the question I said that the ball is being thrown down. This means that any initial velocity it had must be negative.

a=v f−v it

v f=v i+atv f=−12+(−9.81)(3.47)

v f =−46.0407=−46m /s

In this example, the final velocity is bigger than the initial. That's because as it falls it will be going faster and faster. The negative sign simply means that the velocity is down.

Example 3: A rabbit jumps up into the air. As he leaves the ground he is traveling at 16.0 m/s. Determine how fast is he is going after 2.8s.

a=v f−v it

v f=v i+atv f =16.0+(−9.81)(2.8)

v f=−11.468=−11m /s

Why did I get a negative answer?● The rabbit reached its maximum height, where it stopped (instantaneously for the

briefest moment of time), and then started to fall down.● Falling down means a negative velocity.


The RulesThere’s a few rules that you have to keep track of. Let’s look at the way an object thrown up into the air moves.

As the ball is going up…

1. It starts at the bottom at the maximum speed.2. As it rises, it slows down because gravity is a negative acceleration.3. It reaches its maximum height, where for a moment its instantaneous velocity is zero. This is

exactly half ways through the flight time.

As the ball is coming down…

1. The ball begins to speed up, but downwards. Because gravity is negative, the velocity of the object increases in the negative direction.

2. When it reaches the same height that it started from (like the ground, or the person's hand), it will be going at the same speed down as it was originally moving up at. The only difference is that this velocity is negative because it is pointing down.

3. It takes just as much time to come down as it did to go up.

Applying these rules might seem complicated, but since they stay the same all the time you can get used to the problems by just practicing them over and over again.

Example 4: I throw my ball up (again) at a velocity of 12 m/s.a) Determine how much time does it take to reach its maximum height.

It reaches its maximum height when its velocity is zero. We’ll use that as the final velocity. Also, if we define up as positive, we need to remember to define down (like gravity) as negative.

a=v f −v it

t=v f−via

t=0−12−9.81

t=1.223242=1.2 s

b) Determine how high it goes.It’s best to try to avoid using the number you calculated in part (a), since if you made a mistake, this answer will be wrong also. If you are working on a problem where you must use your previous answer (there's no other formula), then you just have to.

v f2=vi

2+2ad

d=v f

2−v i

2

2a

d=0−122

[2(−9.81)]

d=7.33945=7.3m


c) Determine how fast it is going when it reaches my hand again.Ignoring air resistance, it will be going as fast coming down as it was going up. This means its final velocity as it reaches my hand is -12m/s (negative because it is coming down).

Gee'sYou might have heard people flying fighter jets or rockets in movies say how many “gee’s” they were feeling.

● All this means is that they are comparing the acceleration they are feeling to regular gravity.

So, right now just sitting in a chair, you are experiencing 1 gee… regular gravity.● This just means that you are experiencing one times the acceleration of gravity. One times

9.81m/s2 is equal to 9.81m/s2. ● During lift-off the astronauts in the space shuttle experience about 4 gee’s.

● That works out to about 4 x 9.81 = 39m/s2.

Example 5: Gravity on the moon is 1.67m/s2 . Determine how many gee's this is.This is just a quick easy way to compare the acceleration due to gravity on the moon to the value you feel from day to day here on Earth.

1.67m /s2

9.81m /s2 =0.170 gee ' s

Example 6: A space probe sent to one of Jupiter's moons, Callisto, is taking pictures with a digital camera that records one picture every tenth of a second (measured to a precision of three sig digs). While doing this, a small screw falls off a part of the probe right in front of the camera and can be seen to start falling past the camera as it takes the pictures. Over a series of six pictures, the screw can be seen to fall 15.7 cm starting from rest. Determine how many gee's there are on this moon.

To figure out the gee's, we'll first need to figure out the actual acceleration due to gravity on this moon. Then we can convert it into gee's. We also need to remember to use 0.500 s as the time. Six pictures were taken in total, but the first one is t = 0 just as the screw starts to fall from rest. The next five pictures (0.500 s ) record its fall of 0.157 m.

d=v i t+1/ 2at 2

a=2d

t 2

a=2(0.157)

0.5002

a=1.256=1.26m /s2

changed into gee's 1.26m /s2

9.81m /s2 =0.1280326=0.128 gee ' s

Homework

p.63 All even from #4 – 14


http://en.wikipedia.org/wiki/Jupiter's_moons

Lesson 13: Vectors in One Dimension

Up to this point we have been focusing on the number crunching sort of questions you can do in physics.

● In this chapter the focus will start to be shifted toward more complicated problems that might not always be solved by just “plugging numbers into a formula.”

● For this reason, we will start to use vector diagrams as a way to organize our information and to help us solve our problems.

● As we start to use these diagrams, keep in mind that we are drawing diagrams that truly represent the motion of the object.

As we learned back in Lesson 8, just about anything you measure in Physics can be divided into two categories: scalars and vectors.

Scalars: Any measurement that is given as a single number, and nothing else. It has magnitude, but no direction.Vectors: A measurement that is given as a number and a direction. It has magnitude and direction.

We often use arrows to represent vectors. In fact, for the rest of the course you should see them as being interchangeable; an arrow in a diagram is a vector.

● When you have several of these vectors drawn together, you have a vector diagram. ● Although vector diagrams are drawn for different reasons in different kinds of problems, the

rules that govern how they are drawn are always the same.

Vector Drawing Rules1. The vector is drawn pointing in the direction of the vector. This is probably the key feature

of what makes a vector a vector... direction. If an object is moving East, you better make sure that the arrow points East. Always remember that when a direction is written down with the magnitude of a measurement, the direction should appear in square brackets.

2. The length of a vector is proportional to the magnitude of the measurement. This just means that the bigger your measurement, the bigger your vector. If I wanted to show you a vector for a car moving at 10km/h [East], and another one moving at 20km/h [East], the second vector would be twice as big.

3. A vector can be picked up and moved around in a vector diagram, as long as when you place it in its new location it is still the same size and pointing in the same direction.

If you are extremely careful, you can even use the rules to draw all your vector diagrams to scale and solve them by measuring stuff without using physics formulas.

● I would caution you that calculations will usually give you more accurate answers than a diagram.

Adding Vectors In One DimensionOne dimensional vector diagrams are the easiest ones to solve.

● The only thing you have to really watch out for is how you touch the vectors to each other in the diagram.


● To add vectors , the vectors much touch with the head of one (the pointy tip) touching the tail (nothing there) of the next vector.

Example 1: Sketch a diagram that shows how you wold add the following two vectors, A and B.

Right now A and B. are not touching head-to-tail, so we will have to rearrange them. Keep in mind that I can pick up and move around the vectors as long as I make sure that when they are plunked back down they are still pointing the same direction and the same size.

Drawing 2 shows A + B. Notice how the vectors individually are still exactly the same, but they have been rearranged so that A is first and then touches B head-to-tail. If I wanted to draw B + A instead, it would look like Drawing 3.

You'll notice that it doesn't matter in what order I add the two vectors. A + B = B + A.

The idea that I get the same result when I add the vectors shown above is a very important one in Physics.

● It would be like if I told you that I was planning to walk 4 km East, and then another 6 km East. It would give me the same result (10 km East) as I would get by first walking 6 km and then 4 km.

● Since the end result is 10 km East, I would say my resultant is 10 km [E].● A resultant is the sum total of two or more vectors added. It shows you what you would get as

an end result of the other vectors put together.

Example 2: Sketch the resultant of the addition of the two vectors in Example 1.Since we already figured out that A + B = B + A we can draw the resultant for either combination and get the same thing. We'll just go ahead and draw it for A + B.

The resultant is just another vector drawn as an arrow. The difference is that the resultant touches tail-to-tail and head-to-head. This is because the resultant has to start at the beginning (the tail of A) and finish at the end (the head of B). That's the only way for the resultant to show that it has the same overall result as the two original vectors.


Drawing 2: A + B

Drawing 3: B + A

In the text book they call vectors that point in the same direction (like these two) collinear.

Drawing 1: The original vectors A and B

A

B

Drawing 4: Resultant of A + B.

There is an “old-school” way to remember what's happening with these vectors. It involves going back to my younger years playing the video game Pac Man.

● Let's assume that the original vectors A and B are showing where Pac Man was moving. Starting at the tail of A, Pac Man moves to the right. Then he continues along B a little more to the right.

● Overall, Pac Man has moved to the right a whole bunch... that's what the resultant shows us. Pac Man can follow the two individual vectors A and B to get from the beginning to the end, or he can just follow the one resultant and end up in the same place from the same beginning.

Subtracting Vectors in One DimensionThis is where things get a bit more interesting.

● What we need to remember here is that in Physics a negative sign simply means “in the opposite direction.”

● We can take A - B and simply change it into A + -B.● The negative sign on the B just means that we will need to take the original vector B and

point it in exactly the opposite direction (180° from where it's pointing right now.● Then we will simply add them just like we did in the previous diagrams (touching head-

to-tail of course!) to get our resultant.● The reason you may have to do subtraction of vectors is because some physics formulas require

you to subtract vectors. For example, Δv = vf – vi.

Example 3: Sketch a vector diagram of A – B.The most important thing to remember is that A - B is equal to A + -B. So, all we need to do is take the original vector for B and spin it around so it points in the opposite direction.

Now we just add them head-to-tail and draw in our resultant.

Using the Pac Man analogy, You can see that if Pac Man moved to the right a lot (A), and then moved back the left a bit (B), then it would be the same if he'd moved to the right just a bit (resultant).

Be careful since subtraction is not commutative (that just means that A - B ≠ B - A).

Homework

p73 #1p74 #2p75 #8


Drawing 5: Vectors A and -B.

A

-B

Drawing 6: The resultant of A + -B.

Lesson 14: Vectors in Two Dimensions

Two dimensional problems are a little tougher, because we are no longer just lining up collinear vectors and doing quick math.

● Instead, we need to pay attention to how the vectors form a more complex (but not very complex) diagram. The majority of these diagrams will involve right angle triangles.

● If they are right angle triangles, just use your regular trig (SOH CAH TOA) and pythagoras (c2 = a2 + b2).

● You'll want to be thinking about physics as you set up your diagram (so that you get everything pointing head-to-tail and stuff) and then switch over to doing math just like any trig problem.

Components of VectorsOne of the most important ideas in vectors is components.

● Just like a component stereo system is made of several individual parts working together, components of vectors are the individual parts that add up to the overall resultant.

Example 1: A car drives 10km [E] and then 7 km [N]. Determine its displacement.First, draw a proper diagram:

The red and the blue vectors are the components of the resultant. ● The red and blue components show you how walking East and the North will result in

you moving more or less in a North-East direction.Notice how this diagram even shows the vectors being added in the correct order according to the question.

● 10 km [E] is shown leading up to 7.0 km [N]. Start at the tail of the red arrow and follow the path it takes you along. You eventually end up at the head of the blue vector.

● If you added them with 7.0 km [N] and then 10 km [E] you would still get the same final answer, just with a different angle because of a different reference point.

● The resultant is drawn in head-to-head and tail-to-tail, just like a resultant is always supposed to be.

This is certainly a right angle triangle, so just use c2 = a2 + b2 to find out the magnitude (size) of the resultant.

c2=a2

+b2

c2=102

+7.02

c=12.206556=12kmSince this problem is about vectors we would also have to give a direction in our answer.


Warning!To truly be called components, the two vectors must be perpendicular to each other. Usually this is tied back in to math by referring to them as the x-component and y-component.

Illustration 1: Car moving.

10 km

7.0 kmresultant

● This will involve measuring an angle, but it will also involve a way to make sure everyone knows what that angle means.

● We will leave that part of the problem until after we have finished the next part...

Vector DirectionsThe problem with angles is that on their own they don't mean very much.

● If I say to you “20o ”, you need to ask “20o pointing where?”● Over the years there have basically been two systems that are

accepted for giving meaning to the angles used in Physics:1. Cartesian Method2. Navigator Method

● You are expected to understand and be able to use both.

Cartesian Method

Although you might not always refer to it using the name “Cartesian”, you have almost certainly used this system in math at some time. It is basically the “x-axis” system you learned in Science 10.

● It divides space into four quadrants using an x and y axis.● The positive x axis is considered to be the beginning and

is assigned an angle of 0o.● As you move counter-clockwise, the angles get bigger. At

each axis point you have added another 90o.The Cartesian method is fine for math class, but it does have one serious drawback; the positive x axis being set as 0o is an arbitrary (for no good reason) decision.

● For this reason I will not use this method in the notes as often. That does not mean that you can ignore it.

Navigator Method

The navigator method gets rid of the arbitrary nature of the Cartesian method by using compass points.

● We will use one of the four compass points as our reference, and then measure how many degrees towards one of the other compass points we have moved.

● We do have to be a little careful, because when using the navigator method there are two different styles of writing down the answer, and every answer can be given using two different angles!

● The following examples will show you how this can work.


Did You Know?

The Cartesian Method is named after the mathematician and philosopher René Descartes, who created the system around 1637.

Illustration 2: Cartesian Method

0O

90O

180O

270O

Illustration 3: Navigator method example.

http://en.wikipedia.org/wiki/Cartesian_plane

Example 2: Determine the direction of the vector shown in Illustration 3.If you look at where the angle is placed in this diagram, you'll probably agree that we are measuring an angle that is 30° away from the North.

● It's like we are using North as a reference line in this situation. The vector is 30° away from North.

Looking at the diagram, we're basically going clockwise towards East a little bit. ● So, we know that we are using North as a direction, but we're leaning over towards East

somewhat.In the physics style of giving a direction, we would write [N30°E], which is read as "North 30° East". The reference line (North) is given first, and then the number of degrees away from it (30°) going towards another of the reference lines (East).

We always put directions in [square brackets] to show that they are the direction, and not some weird formula!

Since all the way from North to East is a full 90°, we know that this could also be drawn showing that the vector is 60° (we get it from 90° - 30°) counterclockwise from the East.

● This would mean that we could also measure this vector as [E60°N] . ● Between these two ways of measuring, [N30°E] is considered more, ummm, polite, since the

angle is smaller. Either one is still correct.

Your text book has chosen to use a slightly modified (and I don't like it much) style of using the Navigator method.

● In this system the order is changed to (1) the number of degrees, (2) towards a reference, (3) from a reference.

● For example, the text book would say that the direction in Illustration 3 is [30° E of N]. Notice how it says you are 30 degrees towards the East from North. In fact, we would read it as “30 degrees East of North.”

● I will avoid using this style in the notes, but you must be aware of it and be able to use it, especially when doing questions in the text book.

Example 3: At the beginning of this lesson in Example 1, you figured out the distance a car moved, but didn't give its direction. To give an answer of his displacement (a vector measurement) you need to determine the direction of the vector.

As a hint, you should probably use tan to figure out this angle. Using either sin or cos will involve using the resultant you just calculated. If you got the resultant wrong, you'd get your angle wrong also.Traditionally you measure from the start of the resultant, at its tail. This means we will measure the angle at the bottom left of the diagram.

tanθ=oppadj

=7.010

θ=34.99202=35o

This means that the final answer for the question you were asked at the beginning of this lesson is that the displacement of the person is 12 km [E35oN].


Warning!Make sure your calculator is in degree mode (not radian mode), or else all your calculations will be wrong. Also remember to take the inverse of tan at the end, since we want the angle, not the tan of the angle.

Homework

p78 #1,2


Lesson 15: Solving Vector Problems in 2 Dimensions

We can now start to solve problems involving vectors in 2 dimensions.● We will use all the ideas we've been building up as we've been studying vectors to be able to

solve these questions.● The majority of questions you will work on will involve two non-collinear (not in a straight

line) vectors that will become part of a right-angle triangle. If there are more that two vectors, you will probably be able to use a trick or two that will allow you to get a triangle out of them.

Since the triangles will be right-angled, you will be able to use a bunch of your basic math skills.● Just use your regular trig (SOH CAH TOA) and Pythagoras (c2 = a2 + b2).● Usually you'll want to be thinking about physics as you set up your diagram (so that you get

everything pointing head-to-tail and stuff) and then switch over to doing it like any math trig problem.

Example 1: On a hot summer day a person goes for a walk to see if they can find a 7 Eleven to buy a Slurpee at. He first walks 3.5 km [N], then 4.2 km [E] , and finally 1.4 km [S] before getting to the 7 Eleven. Oh, thank heaven! Determine the displacement of the person.

A quick sketch will help us organize things so we can get to work.

This shows the vectors being added head to tail, and the resultant that the person actually moved, start to finish. Now consider this; at the end when the person walked 1.4 km [S] he was basically undoing some of the original 3.5 km [N] that he had originally walked. We can add these vectors in any order we want, so let's just rearrange things so that we add the first and the last vectors together (since they are collinear), then take care of the other one.

Assuming North is positive and South is negative...+3.5 km + -1.4 km = +2.1 km

This basically shows that by walking North and then South, the person overall moved 2.1 km [N]. Now we can use just this one vector moving North along with the vector pointing East and draw one simple triangle diagram to get our answer.


Illustration 1: Adding three vectors to get the resultant.

3.5

km

4.2 km 1.4 km

Resultant

Illustration 2: Adding collinear vectors.

3.5

km

2.1

km

1.4 km

c2=a2

+b2

c2=2.12+4.22

c=4.695743c=4.7km

tanθ=oppadj

tanθ=4.22.1

θ=63.43495O

θ=63O

The person walked a displacement of 4.7 km [N63OE].

You can also break a resultant apart to get components if you need to.● On its own this might not be very useful (like Example 2 shows).● Sometimes it can allow you to do questions that are more complex (as Example 3 shows).

Example 2: A wagon is being pulled by a rope that makes a 25O angle with the ground. The person is pulling with a force of 103 N along the rope. Determine the horizontal and vertical components of the vector.

It would probably be helpful to draw a quick sketch to help organize your thoughts. Then you can start using trig to find the components.

We want to know the components of the 103 N vector, so we basically need to draw a triangle and then solve for the unknown sides. The triangle will be drawn using the force vector from above along with horizontal and vertical components.

Now solve for the two unknown sides of the triangle.Horizontal

cosθ=adjhyp

adj=cosθ(hyp)

adj=cos25O(103)

adj=93.3497021=93N

Vertical

sinθ=opphyp

opp=sinθ(hyp)

opp=sin25O(103)

opp=43.529681=44N


Illustration 4: Pulling a wagon with a rope.

wagon

103 N25O

Illustration 5: Triangle with unknown sides.

wagon

103 N25O

Illustration 3: Adding two vectors to get a resultant.

4.2 km

2.1

km

Resultant

θ

Example 3: A plane flies 34 km [N30OW] and after a brief stopover flies 58 km [N40OE]. Determine the plane's displacement.

Draw a very careful diagram for questions like this one. Make sure that you carefully draw in the angles and properly show the two vectors being added head-to-tail.

We can't add the vectors like this, so we need to use a way to simplify things. The easiest thing to do is take each of the vectors individually and break it into its components.

Once you have two x and two y components, you can add them x to x and y to y.

Then you will have one set of x and y components which can be added to give you your resultant.

x−component

sin=opphyp

opp=sinhyp

opp=sin30O 34

opp=17km

y−component

cosθ=adjhyp

adj=cosθ(hyp)

adj=cos30O(34)

adj=29.444864km

x−component

sinθ=opphyp

opp=sinθ(hyp)

opp=sin40O(58)

opp=37.2816814km

y−component

cosθ=adjhyp

adj=cosθ(hyp)

adj=cos40O(58)

adj=44.4305777km

Now here's the nifty part. Both of the y components are pointing up, so we'll call both of them positive. When we add them we get...

29.444864 km + 44.4305777 km = 73.8754417 kmThis is positive so it is also pointing up.

The x components are a little different. The first one points to the left so we will call it negative. The second x component pointing to the right we will call positive. This gives us...

-17 km + 37.2816814 km = 20.2816814 km This is positive so it is pointing to the right.


Illustration 6: Two vectors added head-to-tail.

30O

40O

34km

58km

Illustration 7: First vector broken into components.

30O34km

x

y

Illustration 8: Second vector broken into components.

40O58km

x

y

Draw a new diagram made up of just these two new vectors and find the resultant.

c2=a2

+b2

c2=73.87544172+20.28168142

c=76.6089256=77km

tanθ=oppadj

tanθ=20.281681473.8754417

θ=15.351686O=15O

The plane's displacement is 77 km [N15OE].

Homework

p82 #2p84 #1,3p88 #2


Illustration 9: Final resultant.

73 k

m

20 km

Res

ulta

ntθ

Lesson 16: Relative Motion

Relative motion is just a way of saying that sometimes different people will say different things about the motion of the same object.

● This is not because one of them is wrong, but because they are using different frames of reference.

● The best way to see how this is possible is to look at some examples.

● In all of the following examples, ignore air resistance.

1-D Relative Motion Example 1: Let’s say I am standing on the back of a pickup truck (that is motionless), and I am throwing apples forwards. I know that I can throw an apple at exactly 15m/s every time.

● If a person were standing on the sidewalk, how fast would she say the apples are moving?● Since she will see them exactly the same way as me (we're both in the same reference

frame), she will say 15m/s.● Now the truck starts to move forwards at 20m/s. I am still throwing apples forwards, exactly the

same as I was throwing them before, at 15m/s.● If I am really not paying attention to what’s going on around me (like the fact that I am

standing in the back of a moving truck), how fast would I say the apples are moving?● Still 15m/s! Relative to me, I can only make an apple move away from me at 15m/s, so

that’s how fast I measure the apple moving away from me.● How fast does my friend on the sidewalk say the apple is moving?

● Well, even before I throw it, she’ll say that the apple is moving at 20m/s (the speed of everything on the truck).

● When I have thrown the apple forward, adding more velocity to it, she will say it is going at (20m/s + 15m/s) 35m/s!

● Now I turn around and start throwing the apples from the rear of the truck, backwards!● I will still say that my apples are moving at 15m/s, because from my way of looking at

it, that’s how fast the apple is moving. The only thing I might say that is different is that it is -15m/s, since even I should be able to notice they are going in the opposite direction now.

● My friend on the sidewalk will say that the apple is moving at (20m/s + -15m/s) 5m/s!

In each of the above examples, we are really talking about two different people having two different frames of reference while measuring the relative velocity of one object.

Frame of reference: When you are standing on the ground, that is your frame of reference. Anything that you see, watch, or measure will be compared to the reference point of the ground. If I am standing in the back of a moving truck, the truck is now my frame of reference and everything will be measured compared to it.

Relative velocity: In the above examples, each person was measuring the velocity of the apples relative to (compared to) the frame of reference that they were standing in. Relative to a person standing on the sidewalk, the apple may be moving at 10m/s, while for a person in the frame of reference of the truck, the apple is moving at 15m/s relative to him.


A frame of reference can be thought of as any spot your doing your measurement from as long as it is not accelerating. This is called an inertial frame of reference.

Example 2: Sitting at your desk, how fast are you moving?● Relative to the ground: Zero. You’re not moving relative to the frame of reference of the

ground.● Relative to the sun: 2.97e4 m/s! That’s a pretty big difference, but since the Earth is orbiting the

sun at this speed, an observer standing on the sun (ouch!) would say that you are moving at 2.97e4 m/s.

● Both of these answers are correct in their own frame of reference.

Example 3: You might have even noticed relative velocity while sitting at a red light…● Have you ever been sitting at a red light with a bus stopped next to you?● You’re kind of daydreaming, looking out the window at the side of the bus, when all of a

sudden it feels like your car is rolling backwards!● Then you realize that it was just the bus moving forwards.

● Your brain knows that the bus was just sitting there on the road… it became part of the frame of reference of the ground.

● When your brain saw the bus moving forwards, it had already “decided” that the bus won’t move. The only option remaining is that you must be moving backwards.

2-D Relative Motion One other thing that we will need to keep in mind as we begin to solve these problems is that the components are mutually independent.

● This might sound like an oxymoron, since mutually means together, but independent means apart.

● What it means is that the two components are obviously working together (mutually), touching head-to-tail and stuff, but they are still measuring separate things (independent).

● If one of the components changes it will not affect the other. For example, if x got bigger, y would still be the same. Only the resultant would change.

A perfect example of this is a boat trying to cross a fast moving river.

● Let's say you point your boat directly East across a river that is flowing North.

● As your boat moves itself through the water towards the East, the river is constantly pushing it to the North.

● In the end someone watching from the shore would say that you were moving in a direction roughly North-East .

● This happens because the two components mutually work together to give us our resultant.


Illustration 1: A boat traveling East across a river flowing North.

Did You Know?

Frames of reference and relative motion is actually the reason that people get car sick. Your brain is getting two different sets of information about your body's motion that might not exactly agree with each other; information from your eyes, and information from your inner ear. Some people are more sensitive to these differences, which causes them to feel car sick as they watch the road "whiz" by. If you are prone to getting car sickness, try to look forward at a point far in the distance and stay focused on that.

Still, the two components are independent of each other. ● Let's say you went out and bought a new motor for the boat that could move the boat twice as

fast. Do you expect that would change the speed of the water flowing in the river? Of course not. Just because a boat is moving on a river, it does not have an effect on the speed of the river itself.

● What if the motor on the boat stayed the same, but a dam on the river broke and the water started moving faster. This wouldn't change the maximum speed of the motor on the boat.

● In both cases changing one of the components has no effect on the other.

Example 4: You are in a boat that can move in still water at 7.0 m/s. You point your boat directly East across a river to get to the other side that is 200 m away. The river is flowing at 4.0 m/s [N].

a) Determine your velocity measured by someone on the shore.

Calling the vectors "boat" , "river", and "how you end up moving" is not really accurate enough for the kinds of questions we will be looking at in detail. One way to label the vectors with subscripts is shown here, but you may also use any (reasonable) system of your own that you feel comfortable with.

bvw = velocity of the boat in still water

wvs = velocity of the water with respect to the shore

bvs = velocity of the boat with respect to the shore

The question is asking for the black vector called bvs , which is the resultant of two components.

c2 = a2 + b2 = 7.0 2 + 4.0 2

c = 8.0622577 = 8.1 m/s = bvs

tanθ=

oppadj

=vsw

vwb

=4.07.0

θ=29.74488=30O

The boat is moving at 8.1 m/s [E30°N].


Illustration 2: Subscripts used to name vectors.

b) Determine how much time it takes for the boat to cross the river.You must choose vectors that point in exactly the same direction to solve questions like this!Do not be tempted to use your answer from part (a) unless you have good reason to! If you think about it, this boat is going across a river that is 200 m wide measured straight across. Which vector is pointing straight across the river? The red one showing the velocity of the boat as though it was moving in still water. Even though it gets pushed off course by the water, the components are mutually independent! The boat is still moving East across that river at 7.0 m/s. The displacement the boat has to travel (200 m [E]) and the velocity bvw (7.0 m/s [E] ) are the only ones that point in exactly the same direction.

v=dt

t=dv

=2007.0

t=28.57142857=29 s

The boat takes 29 s to cross the river.

c) Determine how far downstream from directly across the river the boat will hit shore.

Which velocity is pushing the boat downstream? That has to be wvs . And from the last question we know how long the boat will be on the water to be able to be pushed downstream.

v=dt

d=vt=4.0(28.57142857)

d=114.2857143=1.1e2m

The boat will hit the shore 1.1e2 m [N] of where it was originally pointed at.

If you look back at the answer for (a) in the previous example, you should probably be able to see how giving you the velocity of 8.1 m/s [E30°N] measured by someone on the shore could be broken into components.

● This way you can figure out the velocity of the water(wvs) and the velocity of the boat in still water (bvw).

● You can either put components together to get a resultant, or break a resultant apart to get components depending on your needs.

The other popular type of question to do with components (that is similar to a boat crossing a river) is a plane flying with a wind blowing it off course.

● You will find examples of these types of questions in the worksheet for this chapter.

Homework

p95 #1,2 p97 #2 p99 #1, 3 p100 #1 p101 #1, 6, 8, 9, 11


Lesson 17: Projectiles Launched Horizontally

The study of projectile motion brings together a lot of what you have learned in the past few seconsti.● You need to know about gravity, velocity, acceleration, and vector components to be able to

fully understand (and figure out) these questions.● Don’t worry, though. Even with all that stuff to keep track of, learning how to do these

questions and understand what is happening a little bit at a time makes it all manageable.

The Wile E. Coyote EffectI’m sure you have seen the cartoon where the Coyote is chasing after the Road Runner and runs off of the cliff. He hangs in mid air for a second, looks down, and then starts to fall.

● The question is how true is this, and how many people believe it is true?

A few years ago some researchers in the U.S. went to elementary schools, junior and senior high schools, and universities and asked them to look at the following:

"Ignoring air resistance, which of the following correctly showswhat an object would do if it rolled off a cliff?"

The kinds of answers they got was almost exactly the same at all ages.● About 60% said number 1 was correct. The object will stop in midair, and then start to fall

straight down. Because this is what the Coyote always did in cartoons (and because some people actually referred to it in their explanation of why they chose this answer), the researchers called it the Wile E. Coyote Effect.

● About 25% said number 2 was correct. The object will move forward at first, but will eventually just fall straight down.

● Only about 15% answered number 3. The object will continue to move forwards the entire time it is falling.

The correct answer is actually number 3, and if you think about it using the physics you've studied it makes sense.

● Let's say a coyote does run off a cliff. As he leaves the cliff he has a horizontal velocity.● From studying forces and acceleration you already know that the only way to change that


Illustration 1: A ball rolling off a cliff.

horizontal velocity (cause an horizontal acceleration) is to exert an horizontal force on the coyote.

● If we are ignoring air resistance (which is a very good idea since it will be practically zero), then there is no horizontal force to cause an horizontal acceleration.

● Since there is no horizontal acceleration, the coyote will travel horizontally at the same speed the whole time!

That doesn't tell us anything about what is happening vertically, which is completely separate from what the object is doing horizontally.

● As soon as the coyote leaves the cliff he will experience a vertical force due to gravity.● This force will cause him to start to accelerate in the vertical direction.● As he falls he will be going faster and faster in the vertical direction.

Looking at this problem as what is happening horizontally and vertically, you should get the idea that this is just like the components that we were just working on a couple of lessons back.

● The horizontal and vertical components of the motion of an object going off a cliff are separate from each other, and can not affect each other.

● In a lot of books you will see the horizontal component called x and the vertical component called y.

If we look at the horizontal and vertical components of an objects velocity as it rolls off a cliff, we would get something that looks like this.

● The x-component is there from the start, and stays the same the entire time.● The y-component doesn’t even exist at the beginning, but grows bigger as the object falls.● The shape of the path that it follows is actually a parabola . If you’ve studied those in math

class, great! Don't worry about it if you haven't, just as long as you recognize the shape and know the name.


Illustration 2: Components as a ball rolls off a cliff.

http://en.wikipedia.org/wiki/Parabola

To understand how to actually figure out questions involving these situations, it’s probably best to look at an example.

● Keep in mind the characteristics of the object as it falls while you go through the example.● When you are doing a part of a question that has to do with vertical movement “THINK

VERTICAL” and only use vertical ideas (like gravity).● When you are doing a part of a question that has to do with horizontal movement “THINK

HORIZONTAL” and only use horizontal ideas (no gravity/acceleration).

Example 1: I throw a ball off the edge of a 15.0m tall cliff. I threw it horizontally at 8.0m/s.

a) Determine how much time it takes to fall. b) Determine how far from the base of the cliff it hits the ground. c) Determine how fast it is moving vertically when it hits the ground. d) Determine what its total velocity is when it hits the ground.

a) THINK VERTICALWe’re talking about something falling, and that is vertical motion, so we will only use vertical ideas and numbers. It actually would take the exact same amount of time for the object to hit the ground if I just dropped it straight down from the edge of the cliff, so let’s just calculate the time to fall that way. Remember to think vertically, so make sure your displacement falling down is negative, and your acceleration down is negative.

d=v i t+1/2at2(sincev i iszero...)

d=1 /2at 2

t=√2da

=√2(−15.0)

−9.81t=1.7487435=1.75s

b) THINK HORIZONTALWell, we know it was in the air for 1.75s (from the previous question), and it was moving at a constant speed of 8.0m/s in the x-direction the whole time, so…

v=dt

d=vt=8.0(1.7487435)d=13.98994834=14m

It will move 14m horizontally, so it hits the ground 14m away from the base of the cliff.

c) THINK VERTICALIt has been accelerating down the whole time. We know that gravity is causing this acceleration, and that it wasn’t moving vertically at the start, so we can figure out how fast it is going (vertically) when it hits the ground.

vf2=vi

2+2ad

vf=√vi2+2ad=√0+2 (−9.81)(−15.0)

v f=17.15517415=17.2m /s


d) Combination!It’s total velocity is found by adding the horizontal and final vertical components of the velocity to find the resultant.

c2=a2

+b2

c2=8.02

+17.155174152

c=18.92881401=19m /s

tanθ=oppadj

tanθ=17.15517415

8.0θ=64.998873=65O

The object is moving at 19m/s at an angle of 65° below the horizontal when it hits the ground.

Although there will always be slight differences in actual problems, this is the standard sort of questions that you will be asked for these types of questions.

Homework

p107 #1-3


Illustration 3: Resultant velocity.

Video Killed the Radio Star!Still having trouble with horizontal projectile motion? Then click here.

http://youtu.be/a5kOoXNsP3c

Lesson 18: Projectile Motion at an AngleTo do questions involving objects launched from the ground upwards at an angle (like kicking a football up into the air and watching it as it arcs in the air and comes back down), you need to add a few more steps to the way you did the questions for objects launched horizontally.

● There are actually two ways to do these types of problems, one based on the vertical velocity of the object, the other based on the vertical displacement.

● The only big difference in these methods is how we are going to calculate the time that the object spends in the air.

● Choose whichever method you are most comfortable with, and whichever one suits the particular question you are doing. We will look at each and then it’s up to you to figure out which way you will approach a problem.

Imagine for a moment that you are watching an object as it rises into the air after you kick it upwards at an angle. Look at Illustration 1 below as you read through this description.

● When it left your foot, it was going at the fastest that it can possibly move during its flight.● The instant it leaves your foot, gravity is pulling down on it, causing it to have less and less

vertical velocity.● Remember that there will be no change in the horizontal component of its velocity.● When it reaches the highest point in its flight, it isn’t moving up, and it isn’t moving down, for

an instant of time… its vertical velocity is ZERO!● By the time it reaches the ground again, it will still be moving with its original horizontal

velocity and will have just as much vertical velocity as when it left your foot. It will have the exact same velocity as it left your foot with!

We can use this information about its vertical movement to do some calculations. We know...● that there is gravity (-9.81m/s2) causing the acceleration on the object vertically.● the initial vertical velocity of the object.● the final vertical velocity of the object. We can even use this two ways, since we can say that

the final vertical velocity happens at the halfway point (zero m/s), or when it gets back to the ground (same as it left the ground at).

This gives us enough information to calculate the maximum height of the flight, and the time it spends in the air. After that, we can calculate just about anything…


Illustration 1: Object launched at an angle.

Example 1: You kick a soccer ball at an angle of 40° above the ground with a velocity of 20m/s. Determine...

a) How high will it go?b) How much time does it spend in the air?c) How far away from you will it hit the ground (aka range)?d) What is the ball’s velocity when it hits the ground?

Before we can calculate anything else, we first need to break the original velocity into components.

● We do this so we have a vertical component to do the first couple calculations with. The horizontal component will be used later.

Y−Component

sinθ=opphyp

opp=sinθ(hyp)

opp=sin40O(20)

opp=12.85575219=13m/s

X−Component

cosθ=adjhyp

adj=cosθ(hyp)

adj=cos40O(20)

adj=15.3208889=15m /s

a) THINK VERTICAL!At its maximum height, halfways through its flight, the object won't be going up or down, so we'll say that its final velocity at that point is zero.

vf2=v i

2+2ad

d=(v f

2−V i

2)

2a

d=02−12.855752192

2(−9.81)

d=8.42356597=8.4m

The ball will reach a maximum height of 8.4 m.

b) THINK VERTICAL!The following three methods show slightly different ways of approaching the problem.

● In the first, we will assume that when the ball strikes the ground it is traveling at the same velocity down as it originally started at up.

● In the second, we will only calculate the time it gets to the halfway point, at exactly the highest point it reaches. At that point the vertical velocity is zero. This just means you have to double your final answer to get the whole time it


Illustration 2: Components of original trajectory.

20 m/s

x

40O

y

was in the air.● In the third method we will assume that the ball strikes the ground at the same

height that it left the ground from. This means that it has a vertical displacement of zero.

● Note: I used the unrounded value of 12.85575219m/s in each of these calculations, but just wrote 13m/s to save space.

MethodOne

a=v f−v i

t

t=v f−v i

a

t=−13−13−9.81

t=2.6s

Method Two

a=v f−v i

t

t=v f−v i

a

t=0−13−9.81

t=1.3s

Multiplied by two=2.6s

MethodThreed=v it1/2at2

0=v it1 /2at2

−v it=1 /2at2

−v i=1 /2at

t=−2v i

a

t=−213

−9.81=2.6s

The unrounded answer for each method shown here is 2.620948459 s.

c) THINK HORIZONTALIt is moving at a constant velocity horizontally during the whole time we just figured out, so let’s use the horizontal component of the velocity to figure out the displacement horizontally.

v=dt

d=vtd=15.3208889(2.620948459)

d=40.15526015=40m

The horizontal displacement (range) is 40m.

d) The ball’s velocity when it hits the ground is exactly the same as when it was originally launched… 20 m/s at 40° up from the horizontal. The only difference is that now it's spiking into the ground.

Homework

p109 #2p111 #1p112 #6


Lesson 19: Galileo & Newton

Before we really start looking at dynamics (the study of forces) in depth, it would be wise to learn a little something about the two people that contributed the most to this branch of physics:

● Galileo Galilei ● Sir Isaac Newton

Galileo GalileiBorn in Pisa, Italy on February 15, 1564, died in Arcetri, Italy, January 8, 1642.

● First studied medicine in 1581 at the insistence of his parents.● Today he is remembered mostly for his work in astronomy,

mathematics, and physics.● He carefully constructed experiments to reduce error and

ensure true observations, unlike many other "scientists" at that time.

Galileo showed that falling bodies do not have velocities proportional to their weights.● According to Aristotle, heavier objects fall faster because gravity pulls on them more.

● Drop a book and a piece of paper at the same time and you’ll see what Aristotle was talking about!

● Galileo contradicted Aristotle, saying that air resistance is to blame, not gravity.● To prove this he (supposedly) climbed to the top of the Leaning Tower of Pisa and

dropped two objects, one made of metal, the other of wood.● According to Aristotle's theory, the heavier metal weight should have hit the ground

first.● In fact, both hit at the same time… gravity without air resistance) acts the same on

everything.● You can show this for the book and piece of paper you dropped by placing the paper on

top of the book, and then dropping them. You got rid of the air resistance acting on the paper, so they both fall together.

In 1609, based on information from Holland, he built a telescope.● What he saw made the Catholic Church…umm...

“grumpy.”● This was because the majority of what he said either

contradicted Aristotle, or in some way went against the vision of humanity at the centre of the universe.

● Although common knowledge today, the four ideas that follow were radical in their time.

1. He found “mountains” on the Moon.“But God put it there, it should be perfect,” said the Church.

2. Discovered four of Jupiter’s moons.“But everything is supposed to orbit us on Earth,


Illustration 1: Galileo Galilei

Illustration 2: "Mountains" (craters) on the moon.

http://en.wikipedia.org/wiki/Galileo_Galilei

http://en.wikipedia.org/wiki/Image:Moons_of_solar_system_v7.jpg

http://en.wikipedia.org/wiki/Moon

http://en.wikipedia.org/wiki/Aristotle

http://en.wikipedia.org/wiki/Isaac_Newton

because God put us at the centre,” said the Church.

3. He looked at the sun (not a good idea) and saw sun spots, areas where the sun appears blotchy.

“But God created the sun, so it’s perfect,” said the Church.

4. He supported the Copernican theory that the Sun is at the centre, not the Earth.“Alright, enough’s enough, Galileo! You’re in trouble!” said the Church.

In 1633 the Inquisition (sort of like a Catholic Church courtroom trial) forced him to abjure (renounce) his theories

● As punishment he was placed under house arrest for the rest of his life.

● It is claimed (but often disputed) that as Galileo stood up from his recanting, he uttered "Eppur si muove" which is Latin for "And yet it moves".

● This is a reference to his belief that the Earth was not the centre of the universe, that it actually moves in an orbit around the sun.

● In 1979 Pope John Paul II called for Galileo’s conviction to be annulled.● In 1992, after looking at the legal issues involved, Galileo’s conviction was reversed.

Sir Isaac NewtonBorn December 25, 1642, died March 20, 1727

● He was born in Woolsthorpe, England in 1642, the same year that Galileo died.

● He was educated and taught at Cambridge University, specifically Trinity College.

● He was President of the Royal Society (a “think tank” for scientists) from 1703 till his death.

● He was knighted in 1705, not for his scientific accomplishments, but for putting the Queen's face on a coin while he was head of the British Mint!

Cambridge had closed shortly after he received his degree in 1665, due to the Plague sweeping the country.


Illustration 3: A view of sun spots, with a comparison to the size of the Earth.

For more information......you can read the text of Galileo's adjuration and the annulment of Pope John Paul II by visiting the Multimedia page on studyphysics.ca. You can also click here to read a copy of his book Dialogue Concerning the Two Chief World Systems.

Illustration 4: Sir Isaac Newton

http://en.wikipedia.org/wiki/Royal_Society

http://en.wikipedia.org/wiki/University_of_Cambridge

http://en.wikipedia.org/wiki/Woolsthorpe-by-Colsterworth

http://en.wikipedia.org/wiki/Sun_spot

http://www.law.umkc.edu/faculty/projects/ftrials/galileo/dialogue.html

● Luckily, Newton had already moved to Woolsthorpe (away from big towns) to live with his mother.

● It was over the next year that he did the best work of his lifetime:● He worked out the Three Laws of Motion.● He was “occasioned by the falling of an apple”.

● No, he was never hit on the head by the apple, he just noticed it and wondered why things fall.

● Because of this event he applied his three laws to describe the force of gravity● Newton revealed very little of his work until he published the book “Philosophiae Naturalis

Principia Mathematica” (Mathematical Principles of Natural Science) known more commonly as simply The Principia.

Newton was not known as a friendly person.● He had few friends, and constantly got into arguments● He also had two nervous breakdowns during his life, one caused by the death of his mother after

which he spent six years in isolation.● Although egotistical, he always acknowledged the work of those before him, especially Galileo.

● “If I have seen further than other men, it is because I have stood on the shoulders of giants.” The ‘giant’ is Galileo


Lesson 20: Galileo's Force Theories

Over time questions started to be asked about why objects moved.● Just describing the motion (kinematics) wasn't good enough.● The study of dynamics, an analysis of the forces acting on an object, was born.● By putting these two ideas together, you get a complete explanation of the motion of objects, a

branch of physics known as mechanics.

kinematics + dynamics = mechanics

Greeks (like Aristotle) had noticed 2300 years ago that to keep something moving you had to keep pushing on it (a force). If you stopped pushing, it stopped moving.

● From this they came up with the idea that the natural state of any object is to be at rest (not moving).

● They believed a constant force results in a constant velocity.● From our everyday experience this seems to be true… but is it?

Galileo’s WorkGalileo imagined a world where there was no friction.

● He believed friction complicated his study of dynamics, and by ignoring it he could simplify his theories.

● He came up with two “thought experiments” (they could only be imagined, not done in real life). Remember… NO FRICTION!

Experiment One

Ignoring everything else, an object rolling down a slope will speed up. The only reason its velocity will increase (positive acceleration) is because it is going down a slope while gravity is pulling down on it.

Ignoring everything else, an object rolling up a slope will slow down. The only reason its velocity will decrease (acceleration) is because it is going up a slope while gravity is pulling down on it.

If the ball is moving on a level surface it has no reason to speed up or slow down! With no acceleration, it will move at a constant velocity forever.


Illustration 1: Experiment One

This first thought experiment is Galileo's way of challenging Aristotle's idea that an object will naturally be at rest.

● Galileo shows that an object can naturally be moving... forever!● The only special rule that Galileo wants us to follow is that there is no friction.● The idea is that if we can understand this simple situation better, we can add friction back in

later.

Experiment Two

In the first drawing, a ball rolls down a slope on one side, then rolls just as high up a similar slope on the other side.

Next, the ball now has to roll up a slope that is not as steep, but rolls to the same height. Notice the distance it must go along the slope is greater.

Finally, since it has no slope to go up in the last drawing, it should keep moving forever along the level surface at a constant velocity. Galileo said that this is the natural motion of the object.

Here Galileo gives an even more detailed example of why an object in motion can be a perfectly normal thing.

● Since this was a challenge against Aristotle, it was not particularly popular.● Galileo didn't take this research much further, partly because of the limitations he had doing the

math.● We need someone else to come along and invent Calculus to be able to figure this stuff

out in detail.● This is exactly what Sir Isaac Newton did, and we'll be looking at his work later lessons.

For now, here’s one more way to look at Galileo’s ideas without all the slopes…● What do you have to do to keep the box moving at a constant velocity of 2.0 m/s in each of

these situations?


Illustration 2: Experiment Two


Illustration 3: Pushing objects on various surfaces.

Lesson 21: Free Body Diagrams

Free Body DiagramsWe will need a way to organize forces that are acting on a particular object. The easiest way to do this is by using a free body diagram.

● A free body diagram is just a simple sketch of the object showing all the forces that are acting on it.

● To draw a proper free body diagram, you must follow these steps:1. Draw a quick sketch of the object. Often a simple box will do.2. Place a dot in the centre of the object. We basically treat this as the spot that all the

forces are thought to act upon.3. For every force acting on that object (we don't care about forces acting on any other

objects), draw a vector that shows the size and direction of the force. Each vector must start from the dot and point outwards.

4. Label each vector based on the type of force it is. Do not include numbers or calculations!

Example 1: Sketch a free body diagram for a book being held up by a person.

We draw a quick sketch of the book, and then put a dot in the centre.

Next, we identify that there are two forces acting on this book; the force of gravity pulling it down, and the applied force of a person pushing it up. Since the book is just being held up (not accelerating up or down), we can assume that the two forces are equal, so we will draw the vectors the same size.

Free body diagrams must show a wide variety of forces acting on an object.● These are the common forces acting on objects that you need to memorize:

Fg = force due to gravityFa = applied forceFf = force due to frictionFT = force of tensionFN = normal forceFNET = net force

Each of the forces that have not been mentioned yet will be discussed in detail as they are introduced in the following examples.

Example 2: Sketch a free body diagram of a laptop sitting on a table. ● We know that there will be a force due to gravity (Fg) pulling the laptop down, but if that

was the only force it should be dropping down towards the ground. ● There must be a force acting against the force of gravity that is holding the laptop up.

This is happening because the table top is strong enough to hold up the laptop. ● We call this upwards force the normal force (FN ). A normal force is exerted

upwards by a surface (like a table or a floor) and is perpendicular to the surface.


Drawing 1: Free body diagram of forces on a book.

F

a

Fg

Warning!It is critical that you label the forces on your diagrams using exactly the symbols shown here. The use of capitals and lower case letters does matter!

● As long as the object is not breaking through the table top or flying up into the air, the force due to gravity and the normal force must be equal. That way they just cancel out and the object stays where it is.

Example 3: You are trying to push a heavy box across the floor. It's causing you some trouble because the floor is not very smooth. Sketch a free body diagram of the box.

● We still have force due to gravity and normal force, since it is still an object on a surface.

● You are trying to push it sideways, so that will be an applied force that you are exerting.

● It's tough to push because of friction between the box and the floor, so we'll also need to draw a force due to friction.

● Force due to friction (Ff) always opposes the motion of an object. It is parallel to the surface the object is on.

● Notice the normal force and force due to gravity are still equal. The applied force and force due to friction are also equal to each other. This would mean that the object is moving at a constant velocity (more on this in later sections).


Drawing 2: Free body diagram for example 2.

FN

Fg


FN

Fg

FaFf

Example 4: A sled is sliding down a sloped hill. Sketch the free body diagram of the sled.We will have a situation similar to the one above, except that it is on a slope and there is no applied force (you don't usually push yourself down a hill on a sled).

Notice that... ● Fg still points straight down. ● FN points upwards at an angle, since it has to be

perpendicular to the surface of the slope. ● Ff points back up the hill, since friction will try to

slow down the sled.

Homework

p.131 #1,2



θ

FN

Fg

Ff

Lesson 22: Net Force

The net force is the vector sum of all the forces acting on an object.

● If the forces are parallel we can just add them together as positive and negative forces.

● If the forces are at an angle we have to add them as components of vectors.

Example 1: A car is stuck in a snow drift. Niels and Katrien attach two ropes to the vehicle and try to pull it out by pulling in the same direction. Niels pulls with a force of 75N while Katrien pulls with a force of 68N. There is a force due to friction of 40N acting on the car. Sketch a free body diagram of the situation and determine the net force acting on the car.

Sketch a top-down diagram, remembering to show the friction acting against the two people pulling.

● Notice how Niels and Katrien and both pulling in the same direction, parallel to each other. Since we've shown it to the right, we will call these two forces positive.

● Friction is acting against them, in exactly the opposite direction. We will call this force negative.

To determine the net force, we will need to write out a formula.

● We won't find this on the data sheet, since every net force question can be based on different forces.

● Net force will be equal to all the forces from the free body diagram added together.

FNET = Fan + Fak + Ff

= 75 + 68 + -40Fnet = 103 N

The net force is 103 N acting to the right.

Example 2: As Niels and Katrien pull the car, they notice a patch of ice on the road directly in front of them. To keep on pulling without slipping on the ice, they must begin to pull at an angle as they walk around the ice. Niels still pulls with a 75N at [E15ON] and Katrien pulls with 68N at [E20OS]. Friction is still 40N. Sketch a new free body diagram and determine the new net force.

Be careful with the new sketch, since you'll need to use it to figure out the components of the vectors to be able to add everything. We need to figure out each of the applied forces' components. Note: In all of the calculations that follow I will not be using sig digs. Rounding off in the middle could screw up the final answer. Only the final answer will be rounded for sig digs.


When someone talks about his gross pay and net pay, what do they mean? Gross pay is how much you are paid before any deductions. Net pay is how much you actually get on your paycheck after all the deductions.It’s the same sort of thing when we examine net force. After you have added and subtracted all the forces you are left with the net force acting on the object.

Drawing 1: Free body of Niels & Katrien pulling the car.

Fan

Ff Fak

Warning!Do not subtract any forces. We always add to find the net force. The direction of a vector determines its sign, and that's the only way minuses should appear in the formula.

Drawing 2: Free body of Niels & Katrien pulling at an angle.

Fan

Ff

Fak

15O

20O

Niels' Components

cos=adjhyp

adj=coshyp

adj=cos15O75

x=adj=72.444N

sin=opphyp

opp=sinhyp

opp=sin15O75

y=opp=19.411N

Katrien's Components

cos=adjhyp

adj=coshyp

adj=cos20O 68

x=adj=63.899N

sin=opphyp

opp=sin hyp

opp=sin20O68

y=opp=23.257N

Now we figure out what all of our x-components added together give us It's sort of like doing a special “x-component only” net force. Don't forget that friction is a x-component.

x-componentsFx = Fan + Fak + Ff

= 72.444 + 63.899 + -40Fx = 96.343 N

Now figure out the total of the y-components. Be careful that you show Katrien's y-component as negative, since it is pointing down. Friction does not have a y-component since it was totally pointing in the x direction.

y-componentsFy = Fan + Fak

= 19.411 + -23.257Fy = -3.846 N

Finally, add your x and y-components as a vector diagram to get the resultant.

c2=a2

b2

c2=96.3432

3.8462

c=96.420N=96N

tan=oppadj

tan=3.846

96.343=2.3O


Drawing 3: Niels' vector broken into components.

75N

15O

x

y

Drawing 4: Katrien's vector broken into components.

68N

20O

x

y

Drawing 5: Resultant based on components.

Resultant

96.343N

3.846 N

The net force is 96N [E2.3OS].

Example 3: A 20kg sign is supposed to hang from a pair of wires attached to the wall and a support beam as shown in the diagram. The wires that will be used can withstand a force of tension up to 300N each. Determine the tension in wire one (FT1) and wire two (FT2), and explain any concerns you may have.

● Start off by drawing a free body diagram of the situation. Be sure to include the force due to gravity.

● In the diagram I've moved the 40O angle into a different spot, but this is ok. Since I am still measuring the angle from a straight horizontal line in both cases, they are congruent angles.

● The important thing is that for the sign to hang without moving or falling, the net force acting on it must be zero. When several forces acting on each other cancel each other out, resulting in zero net force, we say that the forces are in a state of equilibrium.

● FT1 is pulling the sign up and to the right, while FT2 is pulling it to the left, and Fg is pulling it down.

● FT1 will have to be broken into components (FT1x & FT1y). ● FT1y is the only force holding the sign up, so it must be equal in

magnitude to Fg pulling it down. That way the net force acting vertically will be zero.

Fg=mgFg=20.0−9.81

Fg=−196.2N

FNET=FgFT1y0=−196.2FT1y

FT1y=196.2N

● We can use this value of FT1y to calculate FT1 based on the triangle shown in Drawing 8.FT1

sin=opphyp

hyp=oppsin

=196.2sin40

hyp=305.23N


Drawing 6: Sign hanging from two support cables.

Mr.C's Awesome Physics Shop

40O

FT1

FT2

Drawing 8: FT1

broken into components.

40O

FT1

FT1x

FT1y

Drawing 7: Free body diagram of hanging sign.

40O

FT1

FT2

Fg

● We can also see from the free body diagram that the force of tension in the second wire is exactly opposite to the x-component of the tension in the first wire. The net force is also zero horizontally. If we calculate FT1x (look back at drawing 8), it must be equal but opposite to FT2 .

FT1x

tan=oppadj

adj=opptan

=196.2tan40

adj=233.82N

FNET = FT2 + FT1x 0 = FT2 + 233.82 NFT2 = -233.82 N

● So the final values for the tensions in the two wires are 3.1e2 N [up] in wire one, and 2.3e2 N [left] in wire two. Since the tension in wire one is greater than the 300 N the wire can withstand, it will have to be replaced with a stronger wire.

Homework

p.132 #1p.133 #2p.135 #2p.136 #3, 7


Lesson 23: Newton's First Law (Inertia)Newton’s Laws of Motion, as written in his book the Principia, are actually very difficult to read.

● At the time, Newton didn’t care too much about the “readability” of his book. He just wanted to get the stuff put down on paper.

● As a result, the way we state his laws today, and the formulas we use, are in some ways different from the way he originally wrote them. They still mean the same thing.

He actually wrote the laws in a specific order for a specific reason.● As we go through the laws, you should realize that he builds one on top of the other.

The First Law (The Law of Inertia)“Every body continues in a state of rest or uniform velocity in a straight line,

unless an external force acts on it.”

“Every body…”Means any physical object in the universe that has mass. It can be here on the Earth, on the moon, floating in space, wherever.

“…continues in a state of rest or uniform velocity in a straight line…”If it is sitting still, it will stay that way. If it is moving, it will keep on moving forever at that velocity in a straight line.

“…unless an external force acts on it.”Unless something else pushes it.

It is important that you understand the idea of “external forces.”● Let's say your car is stuck in a snow drift, so you ask your friend sitting in the passenger seat to

push you out.● He agrees and starts pushing as hard as he can on the dashboard…. the car doesn’t move!● Your friend is an internal force. He would have to get out of the car and push from out there to

be an external force to cause your car to move.

An object resisting a change in its “state of motion” (stopped or moving in a straight line) is something that Newton called inertia.

● That’s why this law is sometimes called the Law of Inertia.● Basically, the idea of inertia is that however an object is moving right now, it will keep on

moving that way.● In day to day experience you don't necessarily see this because of the effects of friction.

● If you roll a ball across the floor it will slow down and eventually stop.● This is not because it is violating inertia, but because there is an external force acting on

it... friction.

We will basically say that as long as the net force on the object is zero, we will not see a change in its velocity.

If FNET = 0, then Δv = 0


Example 1: Describe the motion of a hockey puck that is shot down the ice.A hockey puck will keep moving in the same direction at (almost) the same speed unless someone stops it or changes its direction. This would be done by applying a external force. The effect of friction on the puck will be quite small.

Example 2: Describe the motion of a book sitting on a desk.A book sitting on a desk won’t start to move all on its own. A force needs to be applied to it. Since it is in a state of rest, it will stay that way.

Example 3: Use Newton's First Law to explain why people are injured in car accidents when they do not wear seat belts. Assume the person was in a head on collision.

During the collision the car is rapidly brought to rest by a force acting against it. In the car the person still has inertia, so the person will continue to move forward at the same speed as the car was originally traveling at until a force acts against him. This force will be supplied by the steering wheel, dashboard, or windshield as they hit it. It is this force that causes injuries to his body.

Newton’s First Law goes against what Aristotle said, but is basically what Galileo had said a few years earlier.

● Aristotle had said that if you stop pushing an object, it will come to rest.● He believed that “at rest” was the natural state for any object.

● Galileo told us to ignore friction and basically came up with Newton’s First Law.● It is called Newton’s First Law because he was the one that formally published it and

had the mathematical proofs to back it up.

Homework

p142 #1, 3, 4


Lesson 24: Newton's Second Law (Motion)

To really appreciate Newton’s Laws, it sometimes helps to see how they build on each other.● The First Law describes what will happen if there is no force.● The Second Law describes what will happen if there is a force.

The Second Law (The Law of Motion)“When an external, unbalanced force acts on an object,

the object will accelerate in the same direction as the force. The acceleration varies directly as the force, and inversely as the mass.”

“When an external, unbalanced force…”We are still talking about these external forces, but now we’ve added in the idea of “unbalanced”.

● Unbalanced just means that there isn’t anyone (or anything) pushing against the force hard enough to cancel it out.○ There is a net force acting on the object.

“…accelerates in the same direction as the force…”Some people think this means that the object will move in the same direction as the force… not necessarily!

● The object might be moving to the right, while a force is pushing left.● That means the object will slow down.● It’s acceleration is in the direction of the force (to the left), but it is still moving to the

right.

“The acceleration varies directly as the force…”This just means that if the force increases, the acceleration will increase. If the force decreases, the acceleration decreases.

● This makes sense… push something harder and it will accelerate more! They depend directly on each other.

“…and inversely as the mass.”This means that if the mass is bigger, the acceleration is less. If the mass is less, the acceleration is more.

● This makes sense also… if something has less mass, it is easier to make it move faster! They depend inversely on each other.

Mathematically this would be written as…aF

a1m

● The little swimming fish symbol ( α ) is the Greek letter “alpha” and means “varies as” in math.● It does not mean they are equal to each other, but it does show they are related to each other in

some way.


● We are showing that acceleration is directly related to force● If the net force exerted is increased, the acceleration will also increase.

● Acceleration is inversely related to mass.● If the mass increases, the acceleration will decrease.● If the mass is bigger, the object will have more inertia, so it will resists changes to its

motion more.

When you combine these two relations, you get one of the most basic and important formulas ever discovered in physics.

F = ma

F = force (Newtons, N)m = mass (kg)

a = acceleration (m/s2)

The unit of force is called the Newton. ● It's equivalent to kg m/s2.● It was named in honor of the outstanding work that Newton did in physics.● By definition a one kilogram mass will be accelerated at 1 m/s2 if a 1 Newton force is applied to

it.

Example 1: Determine the force acting on a 5.46kg object if it is accelerated at 17m/s2.F=ma

F=(5.46)(17)

F=92.82=93N

Example 2: Determine the acceleration of a 1000 kg car if a 2.5e3 N force acts on it?F=ma

a=Fm

=2.5e31000

=2.5m /s2

Applications of Newton's Second LawIt is best to start the more complex diagrams by analyzing what forces are involved, come up with a net force formula, and draw a free body diagram.

● In these examples the forces will all be acting vertically or horizontally, not both.

Example 3: A 3.70e3 kg elevator is being raised by a cable that exerts a 4.00e4 N force upwards. Determine the acceleration of the elevator.

● There are only two forces acting on the elevator, the force of tension in the cable pulling it up and the force of gravity acting down.

● It is the net force that causes the acceleration of the elevator, so we're basically solving for that. We can substitute in FNET = ma

● Remember to use negative for down and positive for up.


Drawing 1: Free body diagram of elevator.

Fg

FT

FNET=Fg+F T

ma=mg+FT

a=mg+F T

m

a=3.70e3(−9.81)+4.00e4

3.70e3

a=−36297+4.00e4

3.70e3

a=37033.70e3

a=1.0008108=1.00m/ s2

The acceleration of the elevator is 1.00 m/s2 [upwards].

Example 4: Determine the acceleration of the elevator from Example 3 if the tension in the cable is only 3.30e4 N.

FNET=Fg+FT

ma=mg+FT

a=mg+FT

m

a=mg+FT

m

a=3.70e3(−9.81)+3.30e4

3.70e3

a=−36297+3.30e4

3.70e3

a=−32973.70e3

a=−0.8910810811=−0.891m /s2

The acceleration of the elevator is now 0.891 m/s2 [down].

Example 5: A 0.0500 g piece of paper is dropped. While it falls there is a frictional force of 4.71e-4N. Determine the acceleration of the paper.

● The frictional force must be pointing upwards since it will resist the motion of the paper falling down.

● Don't forget to turn the mass into kilograms.


Drawing 2: Free body of falling sheet of paper.

Fg

Ff

You don't need to do all this fancy canceling (as shown on the fourth

line). You can just start substituting in your numbers.

FNET=Fg+Ff

ma=mg+Ff

a=5.00e-5(−9.81)+4.71e-4

5.00e-5

a=4.905e-4+4.71e-4

5.00e-5

a=1.95e-55.00e-5

a=−0.390m /s2

The acceleration of the paper is 0.390 m/s2 [down].

Example 6: Two boxes are attached by a thin wire as shown in Drawing 3. A person pulls on the wire attached to the yellow box. There is a force of friction of 213 N acting on the boxes. Determine the force applied to the boxes if they accelerate to the right at 2.45 m/s2.

● Since the boxes are attached by a wire, they must accelerate at the same rate. We can also treat them as one big mass instead of two small masses.

● We will assume that the wires have no mass.● Fg and FN cancel each other vertically, so we can ignore them.● To the right is positive, and to the left is negative.

FNET=Fa+Ff

ma=Fa+Ff

F a=ma−Ff

F a=30.0(2.45)−(−213)

Fa=286.5=287N

The applied force is 287 N [right].

Example 7: “Atwood's Pulley” is a classic physics problem involving two unequal masses hanging from a wire over a pulley. Since one of the masses is heavier, the whole thing will move so that the heavier mass moves down and the lighter mass moves up. If mass one is 12.00 kg and mass two is 7.50 kg, determine the acceleration of each mass.

● Since we are dealing with two masses that are attached, we will do the same thing as Example 6 and add the masses.

● We will need to be careful, since the net force is pulling the blue box down and the yellow box up, but is made up of two forces due to gravity pulling down on different sides.


Drawing 3: Two boxes attached by wire.

25.0 kg5.0 kg

Drawing 4: Free body diagram of single mass.

30.0 kg FaFf

FN

Fg

Drawing 5: Atwood's Pulley.

12.00kg

7.50kg

● To simplify this, we will take our single mass and pretend that one force due to gravity is pulling it to the left, while the other force due to gravity is pulling it to the right.

Mass OneMass OneFg1=mg

Fg1=12.00−9.81

Fg1=−117.72NPulls to the left with 117.72N.

Mass TwoMass TwoFg2=mg

Fg2=7.509.81

Fg2=73.575NPulls to the right with 73.575N.

FNET=Fg1Fg2FNET=−117.7273.575

FNET=−44.145FNET=−44.15N

FNET=ma

a=FNET

m=

−44.1519.50

a=−2.26m/ s2

Homework

p149 #1,2p150 #1p152 #2p158 #8,9


Drawing 6: Rotate the two forces due to gravity horizontally, and show them acting on the combined mass.

12.00kg

7.50kg

FFg1g1 FFg2g2

19.50kgFFg2g2FFg1g1

The minus sign, based on Drawing 6, just shows that mass one will drop as mass two rises. Both will have an acceleration of 2.26 m/s2.

When doing questions that involve a pulley, just remember that a pulley kind of changes the direction a force is acting in. That's why we can change the directions some of the forces are pointing in these questions, even though we typically can not change the direction of vectors.

Lesson 25: Newton's Third Law (Action-Reaction)

Newton came up with one more law when he started thinking about the interaction of objects.● He had already talked about what happens when there is no force (1st Law). ● He then talked about what happens when there is a force (2nd Law). ● But what happens when you have objects interacting, affecting each other?

The 3rd Law (The Law of Action-Reaction)

“For every action force there is an equal and opposite reaction force.”

Anytime something applies a force, there will be an equal and opposite force back in the opposite direction.

● Push on the handle of a lawnmower to make it go forward, and it will push back against you in the opposite direction with just as much force.

● This is the pressure that you feel of the handle against your hand.

There is one ultra important thing to remember when you are looking at action-reaction pairs.● The two forces that you are looking at are each acting on different objects! ● If you are examining what you think are action-reaction forces, but the forces are both acting

on the one object, it is not an action-reaction pair. ● In the above example, you exert a certain force on the lawnmower. The lawnmower exerts an

equal force on you. Two objects, two forces.

Some examples of action-reaction forces depend on the objects being in direct contact, meaning that the two objects involved are actually touching each other to exert forces on each other. These are called "contact forces."

1. Action: the tires on a car push on the road…Reaction: the road pushes on the tires.

2. Action: while swimming, you push the water backwards... Reaction: the water pushes you forward.

Action-reaction pairs can also happen without friction, or even with the objects not touching each other, known as "action at a distance" forces …

1. Action: a rocket pushes out exhaust…Reaction: the exhaust pushes the rocket forward. One of the original arguments that flight in the vacuum of space was impossible was that there would be nothing to push against. This action-reaction explains how a rocket can fly in space where there is no air to push against.

2. Action: the earth pulls down on an apple…Reaction: the apple pulls up on the earth. How can this example be true?!?


Example 1: If the apple has a mass of 0.150 kg determine the acceleration of the Earth.● There is an action-reaction pair of forces given by Fa = - FE

● We know that the apple will accelerate towards the earth at 9.81 m/s2, but does the earth accelerate towards the apple at the same rate?

● If this were true you would expect the earth to be constantly jumping up towards falling objects.

● Carefully remember Newton’s Second Law (F = ma). In this example the forces are equal, but the mass of the earth is considerably larger than the apple!

● The earth has more inertia than the apple, so the same force will only accelerate it a little bit.

The Force of the Earth on the Apple

Fb=ma=mgFb=0.150(−9.81)

Fb=−1.4715NThis is the force of the Earth acting on the apple, but because of Newton’s Third Law, it is also the force of the apple on the Earth, just in opposite directions.

Fa = -1.4715 NFE = 1.4715 N

The Acceleration of the Earth because of the Apple

FE=ma

a=FE

m=1.47155.97e24

a=2.46482e-25=2.46e-25m /s2

This is such a small acceleration of the Earth towards the apple that it can’t even be measured. We can see that although the forces are equal, the accelerations do not have to be.

Example 2: When a rifle fires a bullet, the force the rifle exerts on the bullet is exactly the same (but in the opposite direction) as the force the bullet exerts on the rifle… so the rifle “kicks back”. The bullet has a mass of 15 g and the rifle is 6.0 kg. The bullet leaves the 75 cm long rifle barrel moving at 70 m/s.

a) Determine the acceleration of the bullet.We will assume that the bullet is traveling forwards, while the rifle is moving backward.

vf2=vi

2+2ad

a=vf2−v i

2

2d

a=702

−02(0.75)

a=3266.6̄=3.3e3m /s2

The positive acceleration shows that the bullet is speeding up while traveling forwards.


Did You Know?

Sir Isaac Newton was uncomfortable with his own theories about gravity being an “action at a distance” force. He believed so strongly that there must be some material that connects objects that have a gravitational pull on each other, that he was one of the first scientists to seriously suggest there was a mysterious substance called the aether (sometimes spelled ether) that connected all objects in the universe.

http://en.wikipedia.org/wiki/Luminiferous_aether

b) Determine the force of the rifle on the bullet.F b=mbab=0.015(3266. 6̄)=49N

c) Determine the acceleration of the rifle.Fb=49NF r=−49NFr=mr ar

ar=Fr

mr

=−496.0

ar=−8.16̄=−8.2m /s2

The negative acceleration shows the rifle is speeding up while going backwards.

d) Explain why the bullet accelerates more than the rifle if the forces are the same.Although have the same amount of force acting on them, they each have a different mass (and therefore a different inertia).

Example 3: You just got a couple of identical 20.0 kg shopping carts to do some shopping. The reason you have two is that they are unfortunately stuck together. You push them into the store to get help pulling them apart. You are running towards the store accelerating the shopping carts at 1.2 m/s2. Determine the force the second cart exerts on the first cart.

● You are pushing the first cart forward, and the first cart is pushing the second cart forward. That means that the second cart is pushing back on the first cart.

● We know the mass of the second cart and its acceleration, which takes a certain amount of force. As much as it takes that much force to push it forwards, it will push back just as hard on the first cart.

Force pushing Cart Two forwards...F2=m2a2

F2=20.01.2=24N

Force of Cart Two pushing back on Cart One...F12 =−24N


Illustration 1: Two shopping carts being pushed.

The notation 2F

1 just means “the force of 2

acting on 1.”

Example 4: Once you get in the store you see an employee pushing two boxes out of the stock room (see Illustration 2) to the right. The employee is pushing with a force that causes the two boxes to accelerate at 1.12 m/s2. You know that the friction acting on the box of dog treats is 14.8 N.

Determine the force the dog treat box is applying against the dog food box.

● Draw a free body diagram of the box of dog treats. ● We can ignore the vertical forces since they balance out

against each other.● Remember that it is the net force that is resulting in the

actual 1.12m/s2 acceleration of the box. ● If we can figure out the applied force moving the treats

forward (the dog food box pushing the treats), we will know how hard it is pushing back against the dog food.

FNET=FaFfma=FaF f

2.361.12=Fa−14.82.6432=Fa−14.8Fa=2.643214.8

Fa=17.4432=17.4N

● If it takes an applied force of 17.4 N to push the treats forward, it must be pushing back against the dog food just as hard with a force of -17.4 N.

Example 5: If I push on a lawn mower, it pushes back on me with an equal, but opposite force. Explain why we don’t both just stay still.

● The answer is that these forces are acting on different bodies (and there are other forces to consider).

● It doesn’t matter to the lawn mower that there is a force on me… all that matters to the lawn mower is that there is a force on it, so it starts to move forward.

● Another action-reaction pair you need to consider is that I am pushing backwards on the ground, and it pushes forwards on me.

Homework p168 #2,5,7


Illustration 2: Pushing two boxes to the right.

Dog Food4.50 kg

Dog Treats2.36 kg

Illustration 3: Free body diagram of the box of dog treats.

FN

Fg

Fa

Ff

Illustration 4: The two sets of action-reaction pairs in this situation.

Lesson 26: Friction

Friction is a force that always exists between any two surfaces in contact with each other.● There is no such thing as a perfectly frictionless environment.● Even in deep space, bits of micrometeorites will hit a moving object, causing some friction

(although it is incredibly small).

There are two kinds of friction, based on how the two surfaces are moving relative to each other: 1. Static friction

The friction that exists between two surfaces that are not moving relative to each other. 2. Kinetic friction

The friction that exists between two surfaces that are moving relative to each other.

In any situation, the static friction is greater than the kinetic friction.● Have you ever tried to push a really big object? Did you notice that you were pushing harder,

and harder, and HARDER, until suddenly it was like glue that was holding it to the floor snapped? Then, it felt easier to push the object than it did just to get it started.

● When it was still, you were trying to overcome the static friction (bigger force).● When it finally started to move, you were now pushing against the kinetic friction

(smaller force).

Nobody is exactly sure why friction acts the way it does…● Some physicists’ theories on friction involve the idea of the minute (tiny) imperfections in the

surfaces grinding against each other.● Imagine two pieces of sandpaper rubbing past each other… they’d have a difficult time!● Now remember that any surface, no matter how smooth it might appear to the naked

eye, has tiny bumps.● These bumps on any surface will grind past other bumps on the other surface and cause

friction.● There is also the hypothesis that there are small electrostatic attractions between atoms of the

two surfaces, pulling on each other.● Think of the electrons in one of the surfaces being attracted to the protons in the other

surface.● As you hold one object against another, billions of these attractions between the

electrons and protons of the two objects cause them to stick to each other somewhat.● This pulling on each other could also be a source of friction.


Did You Know?

One of the problems that NASA would need to solve before sending astronauts on a long journey (like Mars) is protection from the microdust and micrometeorites in space. One of the most serious problems is that as the spacecraft travels through space at high speeds, the front will be damaged the most. Most plans have some kind of ablative shield that would cover the front of the craft. Ablative is just a what you call any material that you expect to wear away because of some form of damage, while it protects whatever is underneath.

Video Killed the Radio Star!Watch a video of me explaining the difference between static and kinetic friction by clicking here.

http://en.wikipedia.org/wiki/Micrometeorites

http://en.wikipedia.org/wiki/Ablative_heat_shield

http://youtu.be/VMleQwil_KQ

Friction always acts in the direction opposite to the motion of the object.● Just look at the direction the object is traveling. The direction of the force due to friction will be

exactly 180° opposite.● Friction is also proportional to the normal force, which is how we'll be able to calculate it.

F f FN

The actual formula for friction is…F f= FN

Ff = force due to friction (Newtons)FN = normal force (Newtons)

μ = Greek letter “mu”, coefficient of friction between

two surfaces (no units)μs is static, μk is kinetic

Obviously, some surfaces have less friction than others…● A rubber hockey puck against ice has less friction than a car tire on an asphalt road.● That's why there are also two measurements of friction (static & kinetic) for any combination

of surfaces.● The static friction that you calculate is a measurement of the maximum it can be. It can

be any value up to or equal to that maximum amount.F f staticsFN

● The kinetic friction is the value of the friction.F f kinetic= kFN

● When we measure the coefficient of friction (μ), the smaller the number, the less the friction between the two surfaces.

● By gathering empirical evidence of different combinations of surfaces physicists have been able to come up with values to use for coefficients of friction.

● You are not expected to memorize this table…


Did You Know?

Some people think that my tarantula can climb the walls of her tank because of some sort of "stickiness" on her feet. Actually, she's using friction more than anything else. A tarantula's feet are covered with thousands of microscopic hairs. When she touches her feet to the glass, these hairs jam into the micro-cracks in the surface of the glass and hook on. This is why you'll often see her tap one of her feet against the glass a few times before it takes hold.

"Empirical" evidence means that you actually have to perform the experiment each time to get results. There is no shortcut, regular pattern, or formula that you can use to get the results.

Surfaces μs μk

steel on steel 0.74 0.57

aluminum on steel 0.61 0.47

copper on steel 0.53 0.36

rubber on concrete 1.0 0.8

wood on wood 0.25 – 0.5 0.2

glass on glass 0.94 0.4

waxed wood on wet snow 0.14 0.1

waxed wood on dry snow - 0.04

metal on metal (lubricated) 0.15 0.06

ice on ice 0.1 0.03

teflon on teflon 0.04 0.04

synovial joints in humans 0.01 0.003* depends on type of wood

Example 1: A 12kg piece of wood is placed on top of another piece of wood. There is 35N of maximum static friction measured between them. Determine the coefficient of static friction between the two pieces of wood.

First calculate FN ...FN=Fg=mg

FN=12(9.81)=117.72N

Then use this answer to calculate Ff ...Ff=μsFN

μs=Ff

FN

=35

117.72μs=0.2973157=0.30

Example 2: I have a steel box (mass of 10 kg) sitting on a steel workbench. I try to push the box out of the way…

a) Sketch a free body diagram of the box.

With no definite information about the amount of force being applied, we'll just draw all the vectors equally for now


As long as the surface is completely horizontal, we can say F

N = F

g. We will

calculate it as a positive value, since FN

points up.

Illustration 1: Free body diagram of box.

FN

Fg

FaF

f

b) I push against the box with a force of 25 N. Determine if anything will happen.Let’s calculate the maximum force due to static friction. First we figure out the normal force...

FN=Fg=mgFN=109.81

FN=98.1=98N

Use that to calculate the maximum static friction. We can get the steel-on-steel value of μs from the table on the previous page...

F f=μsFN

Ff=0.74(98.1)=72.594F f⩽73N

So, does this mean that when I push with Fa = 25 N, the friction will push back with 73 N?

● No. That wouldn't make sense, since that would mean that if you gently pushed the box, it would actually start to accelerate back towards you!

● The force due to static friction can go up to a maximum of 73 N, but can also be less.

● It will be equal to whatever the Fa is, up to the maximum calculated here. In this case the friction only has to be as high as 25N to “beat” my applied force.

FNET=FaFf

FNET=25−25FNET=0N

With zero net force acting on it, the box will continue to do what it was already doing (Newton's First Law). The box will just sit there motionless.

c) Determine what will happen if I push with a force of 73 N.This exactly equals the maximum static frictional force between these two surfaces.

FNET=FaFfFNET=73−73

FNET=0NAs above, with no net force acting on it, the box will not start to move.

d) If I push with a force of 100 N , determine if anything will happen.This applied force is greater than the static friction, so it will start to move… but remember that we will now be using kinetic friction!

● Calculate the kinetic friction, which is the value you must use. No more of this “maximum” stuff.


Ff=μ FN

Ff=0.57(98.1)

Ff=55.917N

FNET=Fa+F f

FNET=100+−55.917FNET=44.083N

FNET=ma

a=FNET

m

a=44.083

10a=4.4083=4.4m/ s2

Using Friction in Questions Involving AnglesAngles will complicate questions because you need to take into account components and pay attention to vertical and horizontal net forces.

Example 3: A person is pushing a 49 kg box so that the applied force of 950 N is at an angle as shown in Illustration 2. The coefficient of kinetic friction is 0.37. Determine the acceleration of the box.

This is a more realistic question, in that you probably do not often pull or push something exactly parallel to the ground. Pushing at an angle as shown in the diagram is more reasonable. We will need to calculate the components of Fa.

First draw the free body diagram.

Now break Fa into components. X−Component

cos=adjhyp

adj=coshyp

adj=cos20O950

adj=892.707N

Y−Component

sin=opphyp

opp=sinhyp

opp=sin20O 950

opp=324.919NThe x-component is what is actually pushing the box forward. Now we need to figure out how much friction is pushing back to be able to calculate the net force horizontally moving the box.

● To be able to calculate the friction we will need to know the normal force. In the vertical direction the normal force is equal in magnitude (but opposite in direction) to the force due to gravity plus the y-component calculated above. That way the net force in the vertical will be zero.


Illustration 2: Pushing a box at an angle.

Fa

20O

Illustration 3: Free body diagram.

20O

FaF

g

FN

Ff

Illustration 4: Components of applied force.

950 N

x

y20O

FNET=FgyFN

FN=FNET−Fg−yFN=0−49−9.81−−324.919

FN=805.609N

Now we calculate the friction...Ff=FN

F f=0.37805.609

Ff=298.075N

...and the net force in the horizontal direction...FNET=xF f

FNET=892.707−298.075FNET=594.632N

...and finally the acceleration.FNET=ma

a=FNET

m=

594.63249

a=12.13535=12m /s2

Homework

p185 #1,2p190 #3-6


Illustration 5: Net force in vertical direction.

Fg

FN

y

Lesson 27a: Torque (AP Only)

Torque Basics

Torque is a concept that is very closely related to the ideas we've been developing about force. • Just like force is a push or a pull on an object, torque is a

twist.• Torque is the tendency of a force to twist a rigid object.

◦ We measure the “twist” from a point where the object can pivot, often called the fulcrum point.

As an example of what torque means and why it is important, think of the example of a door.• Usually the door knob is placed far away from

the fulcrum (where the door is attached to the wall with hinges) as shown in Illustration 1.◦ If you pull on the knob to open the door, you

have a pretty easy time because the force you are using causes a lot of torque (twisting) around the fulcrum (hinges) to open the door.

• Imagine that the door knob is now placed near the centre of the door, closer to the hinge, as shown in Illustration 2.◦ If you pull on the door knob with the same

force as before, it will be more difficult to open the door.

◦ This is because the torque (twisting force) has decreased.

The formula to measure the amount of torque is a short one, but applying it depends on the question being asked.

=F ℓτ = torque (N∙m)

F = force (N)ℓ = lever arm distance (m)

• At the end of a calculation of torque, it is common to give the torque a positive or negative value, depending on the direction it is turning. ◦ Clockwise (CW) torque = negative◦ Counterclockwise (CCW) torque = positive

• In the illustrations shown above, the force was kept constant but the lever arm distance was changed. This is why we did not get as much torque in the second situation.

12/3/2009 © studyphysics.ca Page 1 of 4 / C&J 9.1 – 9.2

We will only discuss “rigid” objects in our study of torque. A rigid object will not bend or flex as a force is applied to it.

Illustration 1: The torque is quite large when the force is applied far away from the fulcrum.

fulcrum (hinges) door knob

Fa

torque

Illustration 2: If the force is applied closer to the fulcrum, the torque is smaller.


Fa

torque

Did You Know?

The symbol τ is the Greek letter tau.

Example 1: You are trying to loosen a rusty bolt by using a wrench. You apply 250 N of force to try to turn it counterclockwise. Determine the torque if you hold the wrench (a) 3.00 cm and (b) 12.0 cm from the bolt. Explain which method is best.

a) =F ℓ

=2500.0300=7.50 N⋅m

b) =F ℓ

=2500.120=30.0 N⋅m

The second method is better, since it maximizes the effect of the force being exerted so that the torque is larger.

Torque if the Force is Applied at an Angle

In some situations a force may be applied at an angle so that the torque is less than it could be.• In these situations we must draw a line of action that points in the same direction as the force,

and measure the lever arm distance from the fulcrum to this line.

Example 2: A person is pulling on a door with a force of 210 N to open it. The door knob is 72.0 cm from the hinges. Determine the difference in torque if the person pulls straight back on the door knob, as opposed to pulling at a 65o angle from the door.

In the first situation the person pulls straight back, so the force is perpendicular to the door and the lever arm length is 0.720 m.

=F ℓ=2100.720=−151 N⋅m

The torque is negative since it is turning clockwise.

When the person pulls against the knob at an angle, they are not using the force as well as they were previously to create torque. The distance from the door knob to the hinge is no longer the lever arm distance. We extend the line of action out from the force, and then draw the new lever arm distance perpendicular to this line and pointing to the fulcrum. This makes a right angle triangle, so we can use trig to figure out the length of our new lever arm...

sin=opphyp

opp=sin hypopp=sin 65o

0.720opp=0.653 m


Illustration 3: Pulling straight back on the door knob maximizes the torque.


Fa = 210 N

ℓ = 0.720 m

Illustration 4: Pulling on the handle at a 65 degree angle decreases the lever arm distance, so the torque also decreases.


Fa = 210 N

d = 0.720 m

ℓ = 0.653 m

65o

line of action

Since the question said we are turning it counterclockwise, the torque on the bolt is positive in both answers.

Now we can calculate the torque (negative for clockwise turning) for this situation... =F ℓ

=2100.653=−137 N⋅m

The difference in the torque between the two situations is about 14 N∙m.

Torque for Objects in Equilibrium

We need to look at lots of situations where all the forces acting on an object cancel out (FNET = 0) so that the object is not moving.

• This implies that the net torque acting on the object must also be zero, since otherwise the object would start to rotate.

• We can use this in situations involving rigid objects in equilibrium to predict things about the forces they will experience.

Example 3: A 900N person is walking across a bridge that is 15.0 m long. The bridge itself weighs 3000 N. The person has walked 2.00 m away from one end of the bridge. Sketch a diagram of this situation and determine the force the support at the opposite end of the bridge need to exert to hold the bridge up at that moment of time.

We will treat the end of the bridge that the person is walking away from as the fulcrum point, and assume that the bridge is trying to rotate clockwise because of the torque being applied. All distances will be measured from the fulcrum. If we can figure out how much the torque is on the bridge, we can figure out the force acting down on the support at the opposite end. The force exerted upwards by the support must be equal to this to be able to keep the bridge in equilibrium.

We draw in force vectors to show where the weight of the bridge and the person are. We draw the weight of the bridge exactly in the middle, 7.50 m from the fulcrum (assuming the weight is evenly spread across the two supports). The weight of the person is 2.00 m away from the fulcrum.

Now we calculate the torque due to these two forces, and the total torque acting on the bridge... Bridge Person Total Torque

bridge=F ℓbridge=3000 7.50bridge=−22 500 N⋅m

person=F ℓperson=900 2.00 person=−1800 N⋅m

total=bridge person

total=−22500−1800total=−24300 N⋅m


Illustration 5: Person standing on a bridge.

2.00 m

7.50 m

Fg of bridgeFg of person

Finally, we need to see how much force this torque causes on a point 15.0 m away from the fulcrum, which is where the far support is located.

=F ℓ

F=l

F=2430015.0

F=1620 N=1.62e3 N

There is a combined force of 1.62e3 N on the support at the far end. This means that the support must be able to push up with this much force.

Torque and Levers

You can use the concepts of torque to solve problems involving levers.• A lever is usually just a rigid rod that has a fulcrum somewhere along it.• Levers come in three types, depending on where the fulcrum is placed. In some cases this

results in an increase in force at some point along the lever, in other cases it is done for convenience.

Example 4: A person is using a lever arranged somewhat like a see-saw to help lift a tree stump out of the ground. The lever is 1.60 m long, and the fulcrum is placed 0.400 m from the tree stump. Assuming that the lever has negligible weight (you don't have to count it), and that all forces are perpendicular to the lever, determine the force at the tree stump if the other end is pushed down with a force of 500 N.

A quick sketch of the situation would look something like this...

The force that is being applied on the left side will cause a torque that is equal to the torque on the right side.

left=F left ℓleft=5001.20left=600 N⋅m

right=F right ℓ

F right=right

l

F right=600

0.400F right=1500 N=1.50e3 N

So, by having different distances from the fulcrum we can have the advantage of increasing the force exerted on one part of the lever.


We use a positive value for torque here, since the clockwise or counterclockwise torque does not necessarily relate to the up or down direction of the force.

0.400 m

1.20 m

Fleft

= 500 N

Fright

Lesson 27: Gravity on Inclined Planes

You need to be especially careful when you are doing problems involving gravity pulling something down a slope.

● The physics involved is considerably more complex than it might first seem, mostly because everything is tilted.

● Let’s look at a standard question of gravity on an inclined plane (slope) to see how we would figure it out.

Example 1:Determine the acceleration of a 15 kg box down a 30° slope if the coefficient of friction is 0.13 on this surface.

The first thing we should do is sketch a free body diagram of the situation.

There are a few special things you have to notice about this diagram:Force due to Gravity (Fg) is pointing straight down. Even though we are on a slope nothing will ever change that gravity points straight down. The only reason for this box to move down the slope will be a component of gravity's force.Normal Force (FN) Remember that a normal force is always perpendicular to the surface that you are on. Since this surface is slanted at a bit of an angle, the normal force will also point at a bit of an angle. In these questions Fg ≠ FN

Force due to Friction (Ff) will always be opposite to the direction that something is moving. In this situation the object is moving down the slope, so friction points back up the slope.There is no applied force (Fa), since this would mean that there was something other than gravity actually trying to shove it down the slope. In some questions you might have an applied force also, but not in this question.

We can calculate the force due to gravity...Fg=mg=159.81=147.15N

We need to break Fg up into components that point down parallel to the slope (F//) and perpendicular to the slope (F⊥).


Illustration 1: Box on an inclined plane.

If you are wondering how I figured out that the 30°angle is at the top of the red triangle, take a look at this...

● I created the yellow triangle by just extending the line of Fg down a bit.● Since this makes a nice right angle triangle, I know that the angle at the top of the

yellow triangle must be 60° (since the angles have to add up to 180°).● Since F⊥ is perpendicular to the slope, the angle in the top of the red triangle must be

30° so that they will add up to 90°. ● Since we know Fg and an angle on the triangle, we can use basic trig to calculate the

other two sides.Determine F//, the force pulling the box down along the slope...

sin=opphyp

sin=Fl lFg

Fl l=sinFgFl l=sin30O

147.15

Fll=73.575N


Illustration 2: Fg broken into components.

Illustration 3: Angle of slope to angle in components.

Determine F⊥...

cos=adjhyp

cos=F⊥Fg

F⊥=cosFg

F⊥=cos30O147.15

F⊥=127.436N

Look back at Illustration 3 and you should notice that the force perpendicular F⊥ is equal in magnitude to the normal force FN (although they point in opposite directions).

Determine the force due to friction Ff using the value you just got for normal force.Ff=μ FN

Ff=0.13(127.436)

Ff=16.5667N

Now you know the force that is taking it down the slope, and the friction that is slowing it down. Determine the net force FNET...

FNET=F l l+Ff

FNET=73.575+−16.5667FNET=57.008N

Now, finally we can determine the acceleration of the box...FNET=ma

a=FNET

m

a=57.008

15a=3.8006=3.8m/s2

Our final answer is 3.8 m/s2 , and yes, we round it off.

Notice that in each step I had you sketch or determine something.● I bet you can see how each of those could be a part of a multistep question like (a), (b), (c), etc.

○ In fact, I could have about 7 or 8 parts to this question.○ If you ever do a question like that, then yes, you must round off your final answer for each

step to the correct number of sig digs.○ You should still keep the unrounded number written at least somewhere, since you should

be using unrounded numbers for the questions that follow.

Re-read through this example a few times. It's long and confusing at some parts, but try to look at each individual tiny calculation. Taken in little bits, each part isn't as hard.Homework p192 #1,3,9,14,17


I've made Ff negative because it is working against the F//. One of them must be negative if the other is positive.

Lesson 28: Mass, Weight, & Fields

Newton's work was all related to one goal he had for himself... to explain gravity.● Newton realized that to explain the force due to gravity, he would first have to come up with a

set of rules to explain forces in general. That's the stuff we've been working on in the last few lessons.

● Newton stated that a gravitational force exists between any two masses. ● According to his Third Law, this means that when you fall the Earth is pulling you

down, while you are pulling the Earth with just as much force up.● We don't see the Earth move because it has so much more mass than you that the Earth's

inertia (tendency to keep doing what it is already doing) is enormous.

Weight vs MassNormally when a person wants to know his or her mass, they will just stand on a scale.

● Since this depends on the force due to gravity pulling you down, you are actually measuring your weight, a force acting on your body.

Weight = Fg = mg

● Your weight is measured in Newtons (just like any other force) and is different in different locations on Earth (since “g” varies from place to place).

Mass is considered to a be constant anywhere in the universe.● Floating in space, you could hold a car in your hand… it’s easy because it has no weight.● Throw it at someone and it hits them with its inertia…

it hurts!● This is because it still has mass, so it will tend

to keep doing what it is already doing.● The amount of material that makes up the car is the

same in space as it is on Earth, so they have the same mass.

● The mass of an object is like asking “how many atoms are in that object?”… this number will always be the same, no matter where you are in the universe.

● Mass is always measured in kilograms.

The best way to determine the mass of an object is to apply a known force to it and measure its acceleration.

● This is known as the inertial mass, since it depends on the inertia of the object.● Changes in the local acceleration due to gravity would not change this measurement.

Example 1: I have a 5.00kg rock.a) Determine how much it weighs on the Earth and on the Moon.b) Determine its mass on the Earth and on the Moon.

a) Weight is measured in Newtons! On Earth...Fg= mg = 5.00kg (9.81m/s2) = 49.1N

7/27/2011

Some textbooks make a distinction between gravitational mass and inertial mass. They are both still a measurement of mass in kilograms. The only difference is really just how it is measured. Gravitational mass is measured by comparing a known mass to an unknown mass. Inertial mass is measured by seeing how much the mass accelerates when a force is applied to it.

On the Moon...Fg= mg = 5.00kg (1.67m/s2) = 8.35N

b) The mass of the object on the Earth and the moon is 5.00kg! The object has the same matter making it up even if I take it to a different place.

Example 4: An object is accelerated at 3.24m/s2 by a 68.0 N force. Determine its inertial mass.F=ma

m=Fa

=68.03.24

=20.98765=21.0kg

Gravitational FieldsNewton realized that the gravity that keeps you on the Earth is the same gravity that keeps the moon in its orbit around the Earth.

● To explain this action-at-a-distance force, physicists often use the idea of fields.

● A field is a area around an object that has an effect on nearby objects.

● In the case of gravitational fields, the field always points in towards the centre of the mass.

● All masses have a gravitational field, but only the gravitational fields of large objects (like planets) are easily noticeable.

To measure and show the gravitational field around an object, we would place a known test mass nearby.

● The test mass is any mass we choose, as long as it is small enough that it does not have a significant gravitational field of its own (theoretically its mass

should be 1∞kg , which is so close to zero that it

doesn't even really matter).● The test mass will always move towards the centre

of the object, so we draw vectors pointing in towards the centre.

● By measuring the force of gravity pulling the test mass towards the object, we have a measurement of the gravitational field near the object.

g=Fg

m

You'll notice that we are actually measuring the acceleration due to gravity at that location.● We could certainly measure it in m/s2, or we can choose to use units that have more to do with

the experiment we just did, N/kg . ● On many data sheets you'll see that the acceleration due to gravity is also listed as a

gravitational field strength. ●

7/27/2011

Did You Know?

The idea of “fields” is also used in Social Studies classes. The difference is that they call it a “Sphere of Influence.” For example, the former USSR had a sphere of influence that extended outwards to many nations that were nearby.

Illustration 1: Although weak, there is a gravitational field around a cereal box.

RiceKrispies

● You'll probably see the value listed near Earth's surface as 9.81 m/s2 and 9.81 N/kg. The two ideas are used pretty interchangeably in most questions.

It is also reasonable to say that the effect of gravity is greatest when closer to the object. ● As you move further and further away from the centre, the force exerted by gravity becomes

weaker (although it never truly disappears). ● Illustration 1 shows this by the way the vectors are further apart from each other when more

distant from the object. Closer in the vectors are closer to each other. ● This means the gravitational field is stronger when gravitational field vectors are drawn

closer.● As a relationship, this is shown by...

g=1r2

● This inverse square relationship became the basis of one of Newton's greatest formulas, the Law of Universal Gravitation.

Homework

p202 #1,2,4,5,7,9

7/27/2011

Lesson 29: Newton's Law of Universal Gravitation

Let's say we start with the classic “apple on the head” version of Newton's work. ● Newton started with the idea that since the Earth is pulling on the apple, the apple must also be

pulling on the Earth (Newton’s 3rd Law).● Ask a person on the street where gravity comes from in this situation, and they'd probably say

“the Earth.” They would never consider the apple as a “source” of gravity.● But if the apple is pulling on the Earth, that must mean that an object doesn’t have to be huge to

have a gravitational pull on other objects. There is nothing special about the Earth compared to the apple... both are sources of gravity.

● That would mean that one apple should be able to have a gravitational pull on another apple… that means any mass pulls on any other mass.

● The reason we don’t see the effect of, for example, you being pulled towards your computer monitor, is that the masses are so small that the force is also very small.

● Still, the force is there, and Newton wanted to come up with a way of calculating it.

Using a lot of calculus and some pretty tough physics he came up with two key concepts:1. The force due to gravity between two objects is proportional to the two masses.

If one or both of the masses is big enough (like the Earth) then the force becomes noticeable. As a relationship this can be written as.

Fgm1m2

m1 and m2 are both masses (kg)

2. The force due to gravity is inversely proportional to the square of the distance between the two masses.This distance is always measured from the centres of the two objects, so we usually consider it to be a radius in the formulas. As the distance increases, the force drops off exponentially.

Fg1r2

r = distance between the centres (m)

Example 1: The force due to gravity between two objects is measured to be 160 N. Determine the magnitude of the force if ...

a) one of the masses is doubled.b) the radius is doubled.

a) We will use the first relationship from above...Fgm1m2

2Fg2m1m2

The old rule says “do it to one side, do it to the other.” Since we doubled the right side, we also double the left. If the force is doubled, it is now 320 N.


b) We use the second relationship this time...

Fg1r2

Fg

4

12r2

Notice how we doubled the radius, which is squared and on the bottom. So, on the other side we are dividing the force by four. That gives us an answer of 40 N

We can take the two relationships and put them together to make a formula.

Fg=Gm1m2

r2

Fg = force due to gravity between the two objects (N)G = the Gravitational Constant

m1 and m2 = the two masses (kg)r = the distance between the two objects’ centres (m)

Newton turned his attention to trying to find the value for the Gravitational Constant, “G”.● Newton looked for a way of calculating the value for G from the formula above. If we solve

that formula for G we get...

G=Fgr

2

m1m2

● Let’s look at how we will substitute numbers into this formula.● Newton realized that the only thing he could measure Fg for would be an object on

Earth’s surface. An example would be you. We could calculate the force due to gravity on your body easily using Fg = mg.

● We need to know the distance from the centre of the Earth to your centre, which we do have: 6.38e6 m. And yup, they even had a pretty good estimate of this in Newton's time!

● We need to know your mass, which would be m1… that’s no problem.● The last thing we need, m2, is the mass of the Earth. Oh, oh. That one is a problem.

● In Newton’s time no one had any idea how heavy the Earth really was.● If we knew G, then we could calculate the mass of the Earth, but that’s what we

are trying to calculate here!● Newton continued to look for some way to calculate G indirectly, but never found a

way.


Warning!The gravitational constant “G” is not the same as the acceleration due to gravity “g.”

Did You Know?

This is the first formula that you'll see from a family of formulas called the "inverse square formulas". They all look pretty much the same, and lead physicists to believe that there are many common connections and relationships throughout all of physics.

http://en.wikipedia.org/wiki/Inverse_square_law

Henry CavendishAbout a 100 years later, a man named Henry Cavendish finally figured out a way to measure the value for the Universal Gravitational Constant “G.”

● He attached a really heavy pair of metal spheres to the ends of a long metal rod, and then hung the rod from a wire.

● He then brought another pair of really heavy metal spheres near the spheres on the rod.

● Cavendish knew that because they had mass they should pull on each other, but very weakly.

● To measure this weak pull, he carefully measured how much the wire was twisting (torque) whenever he brought the other masses near by.

● This is why the device he used is called a torsion balance.

After a lot of very careful, very tedious tries, he found that G was 6.67e-11 Nm2/kg2.● Cavendish realized that because he knew the value for G, he could now calculate the mass of

the Earth. That’s why he titled the paper that he published “Weighing the Earth.”

Example 2: Using values that you know, determine the mass of the Earth.● No cheating! Don't look up the value for the mass of the Earth on your data sheet; you're

supposed to calculate it. ● We know that the force exerted on my body by the Earth is my weight, which we can calculate

using Fg = mg , where little “m” is my mass (we think of it as being like a little test mass).● I also know that the force could be found using Newton's big Universal Gravitation Formula,

where one mass is a little “m” (my “test” mass), and the other mass is a big “Me” (the mass of the Earth).

● Let's make those two formulas equal to each other (they're both force due to gravity after all) and see what happens.

Fg=Fg

mtestg=GmtestMe

r2

g=GMe

r2

Me=gr2

G

Me=9.816.37e6

2

6.67e-11Me=5.99e24kg

● This is pretty close to the accepted value on your data sheet.


Illustration 1: Cavendish's Torsion Balance.

cancel the little test mass

http://en.wikipedia.org/wiki/Henry_Cavendish

Example 3: You have a 200g pickle next to a 345g sandwich on your plate. Their centres are 4.5cm apart. Determine the force of gravity of the pickle pulling on the sandwich.

Keep in mind that according to Newton's Third Law, as much as the pickle pulls on the sandwich, the sandwich pulls on the pickle just as hard. The value that we are going to calculate could be used for either situation.Also, don't forget to convert the masses to kilograms, and the distance to metres.

Fg=Gm1m2

r2

F g=6.67e-11(0.200)(0.345)

0.0452

Fg=2.272741e-9=2.3e-9N

Example 4: Io is one of Jupiter's moons. It is known to have 400 active volcanoes. One of the reasons it is so active is because of the gravitational pull of Jupiter on Io. On one particular occasion the Sun, Io, and Jupiter lined up in a straight line so that both the Sun and Jupiter were pulling Io in opposite directions. Using the information given on the chart, determine the net force acting on Io at this time.

Celestial Body

Distance from Io (m)

Mass (kg)

Io n/a 8.93e22

Jupiter 4.22e8 1.90e27

Sun 7.98e11 1.99e30

The arrangement of the three bodies would look something like this...

While Jupiter is pulling Io to the right (positive direction), the Sun is pulling Io to the left (negative direction). Calculate each force separately and then add them.

FJ=Gm1m2

r2

FJ=6.67e-118.93e221.90e27

4.22e82

FJ=6.35486e22N

FS=Gm1m2

r2

FS=6.67e-118.93e221.99e30

7.98e112

FS=1.861335e19N


Illustration 2: Jupiter's moon Io.

Illustration 3: Arrangement of bodies.

SunIo Jupiter

FJ

FS

FNET=FJFS

FNET=6.35486e22−1.861335e19FNET=6.35299e22=6.35e22N

Notice how the effect of the Sun is unimportant in this situation. This is because it is so far away from Io compared to Jupiter. It would be the same as asking you which is a more important force acting on your body, the pull of the Earth or the pull of the Sun? Although the Sun is a much bigger mass, it is also very far away, so the effect of its pull on your body isn't noticeable.

Homework

p207 #2p208 #1p209 #2


Lesson 30: Gravitational Field Strength

In the last lesson we were able to combine our two formulas for force due to gravity to get a new formula.

Fg=Fg

mtest g=GmtestMe

r 2

g=GMe

r 2

● The great news is that the mass (shown above as the mass of the Earth “Me”) can actually be any mass. It could be the mass of the moon, Mars, an asteroid, whatever!

● In fact, we should replace mass of the Earth in the formula with just mass, since it can be any mass.

g=Gmr 2

g = gravitational field strength (m/s2)G = Universal Gravitational Constant

m = mass of object producing the field (kg)r = distance from centre of mass (m)

● This formula lets you calculate the the gravitational field strength (the acceleration due to gravity) caused by that mass ata specific distance from its centre.

Example 1: The planet Mars has a mass of 6.42e23 kg and a radius (from its centre to the surface) of 3.38e6 m.

a) Determine how much a 60.0 kg person would weigh on Mars. b) Compare it to his weight on Earth. c) Determine how heavy the 60.0 kg person would “feel” as an apparent mass in kilograms on Mars.

a) To determine someone's weight, we first need to know the acceleration due to gravity on Mars.

g=Gm

r 2

g=6.67e-116.42e23

3.38e62

g=3.75m/s2

So the person's weight will be...Fg=mg

Fg=60.03.75Fg=225N


b) On Earth the person has a weight of…Fg=mg

Fg=60.09.81Fg=589N

Probably the easiest way to compare the person's weight on Mars and the Earth is to find the ratio of the two.

Fgearth

FgMars

=589225

=2.62

Since this is a ratio, it has no units (the Newtons canceled each other out). It simply means that you weigh 2.62 times more on the Earth as compared to Mars.

c) The reason I asked for the person's apparent mass is because I want to know how heavy the person thinks he feels in kilograms. In reality, the true mass of a person never changes. I am asking how heavy he feels based on the facts that he feels lighter on Mars, and that they are used to the effects of gravity on Earth. We will take the person's weight on Mars and the gravity of Earth to find out the apparent mass.

Fg=mg

m=Fg

g

m=2259.81

m=22.9kg

Keep in mind that the person's mass is still really 60.0 kg. He just feels like he is only 22.9 kg because he is on Mars.

The Elevator QuestionThe concept of gravity becomes a bit more complicated when you examine a complex system like an elevator going up and down.

● It might sound strange to call an elevator complex, but it really does make a challenging problem.● How do you think you would solve a question that asks you about your weight as an elevator

accelerates up or down?

Going up...

When the elevator is accelerating up, what would happen to your weight?● Have you ever noticed that when an elevator first starts to move up, you feel as though you are

being pushed down a little?● This is because you can feel the elevator’s acceleration. The elevator is pushing you up, so

(according to Newton's Third Law) you push down against the floor. ● It basically makes you feel a bit heavier for a moment.

● A scale would show this as an increase in your weight (temporarily). If the scale shows


kilograms, it would show your apparent mass as being bigger than your true mass.

Going down...

What would happen to your weight if the elevator started to accelerate down?● You would feel the elevator drop out underneath you.

● If it really dropped out underneath you, it would feel just like being on the “Space Shot ” at Galaxyland as it is falling … you’d feel weightless!

● You feel this way because as the elevator accelerates down (away from you) it is not pushing up against you as hard as it was before. Since the floor is not pushing as hard up against you, you are not pushing as hard down against the floor (Newton's Third Law again).

● A scale would show your weight as being less. If the scale shows kilograms, it would show your apparent mass as being smaller than your true mass.

Let’s look at how we would actually figure out some numbers for this type of question by looking at an example.

● Keep the following in the back of your mind. A regular scale that you buy in a regular store is made to measure things in kilograms, and it is built for Earth’s regular gravity of 9.81m/s2. You’ll see why this is important later.

Example 2: You are standing on a scale in an elevator. You have a mass of 75kg. Determine what a scale would show as your apparent mass (in kilograms) if…a) the elevator starts to accelerate upwards at 3.0m/s2.

We have three parts in the formula that we will have to use:FNET = the overall force acting on the person causing acceleration upwards.Fg = the force due to gravity pulling the person down.FN = the force of the scale on the floor pushing up against the person.

We should mostly be concerned with what the normal force is, since however hard the scale has to push the person upwards will show up as a reading on the scale. We know the force due to gravity, since that's just the person's weight. The net force is just overall what is happening to the person.

FNET=FgFN

ma=mgFN

ma−mg=FN

ma−g =FN

75[3.0−−9.81 ]=FN

75[12.81]=FN

FN=9.6e2N

The question wanted to know the apparent mass, not the weight of the person.● I can change this into a reading in kilograms by remembering that the scale we're using

has no idea what is going on... it was originally calibrated to be sitting in someone's bathroom where gravity is a nice constant 9.81m/s2.

● This is not the true mass of the person, since mass never really changes.


We start off with a standard net force formula. We do a quick substitution, since F

NET = ma and F

g = mg. Next step

is to move “mg” to the other side. Since both formulas on the left side have “m” we can factor it out. Now start putting in the numbers, watching out for the directions of the accelerations. This lets us calculate the normal force. As hard as the scale is pushing the person up, the person must be pushing down just as hard which tells us the weight the scale will show in Newtons.

FN=mg

m=FN

g

m=9.6e29.81

m=98kg

So a regular scale shows an apparent mass of 98 kg!

b) the elevator starts to accelerate downwards at 4.0m/s2.We’ll handle this part of the question the same way.

FNET=FgFN

ma=mgFN

ma−mg=FN

ma−g =FN

75[−4.0−−9.81 ]=FN75[5.81 ]=FN

FN=4.4e2NSince the scale is pushing up against the person with 4.4e2 N of force (the weight that shows on the scale in Newtons), the person's apparent mass will be...

FN=mg

m=FN

g

m=4.4e29.81

m=44kg

So a regular scale shows your apparent mass as 44 kg.


Physics Test 1

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