1. Newton’s laws of motion

Statement of the three laws.

Newton's First Law (also known as the Law of Inertia) states

that an object at rest tends to stay at rest and that an object in uniform motion tends to stay in uniform motion unless acted upon by a net external force.

Net Force is when we vectorially add up the various forces on a body.

Vector nature of forces to be stressed.

Newton's Second Law states

that an applied force,

, on an object equals the rate of change of its momentum, , with respect to time.

F = ma as a special case of Newton’s second law.

when we use Newtons k = 1,

as 1 Newton = the force a 1kg mass gets as it accelerates by 1m/s2

Newtons 2nd law expt f=ma from wally

Vector nature of forces to be stressed.

To show that a ∝ F.

We need to keep the mass of the system conserved, so the total mass of the entire system that on the hanger and that on the trolley, remains the same.

Newton masses, apparatus capable of calculating acceleration,

Putting the spare masses on the trolley / glider and the remainder on the hook, over a free running pulley. Placing the minimum mass on the trolley calculate the acceleration of the trolley. Repeat until all masses have been shifted from the trolley to the hanger.

Tabulate all the data plot it in a graph

Newton's Third Law states

that for every action there is an equal and opposite reaction.


definitions and units.

Some very interesting problems for you to think about and solve.

Newtons 3rd law Verification.

Often students don't belive that there is a opposite force to all forces, "what if i push a wall, it doesn't move or push me back" ... ahh yes it does


2 x 1m stick, Laser pointer, retort stand, small mirror, upturned bin, odd timber, pin, blutak.


  1. At the height of the upturned bin attack a meter stick to the wall using blu-tak
  2. Between the stick and the bin lies the other bit of wood.
  3. The pin is placed between the wood and meter stick and a small mirror is attached to it.
  4. A Laser pointer held at an angle is shone at the mirror, its reflection should land on the other meter stick/
  5. Some one should push the wall
  6. Other member of the group should watch the reflection.

Result / Observations

When someone pushes the wall it is easy to notice the position of the laser reflection jumping, this is due to the wall and the attached meter stick moving, in turn leading to the mirror to move.



It can happen that the mirror may not return to its starting point.

taken from Physics on Stage 3

Vector nature of forces to be stressed.

Forces are vectors, so we can find a resultant force on an object, no matter how many forces are acting on it. If the resultant force is zero, the forces must be balanced.

Balanced forces cause no acceleration (This means that the object will remain stationary or carry on moving at a constant speed.)

Demonstration of the laws using air track or tickertape timer or powder track timer, etc.


• seat belts

• rocket travel.

Sports, all ball games.

Vector nature of forces to be stressed.

Appropriate calculations.

What force is achieved if a ball goes from rest to 80 m/s in 0.3 of a second, it having a mass of 124g.

Friction: a force opposing motion.

Importance of friction in everyday experience, e.g. walking, use of lubricants, etc.

to demonstrate the effect of friction

get 2 books, overlap the pages .... pull apart ?!?!

2 cat on a sloped roof,

which one slides down the roof first ?

the one with the lower mew ....

Air resistance is very similiar to friction

find out more here

A little quiz on Forces

Force and momentum

More on Momentum

Force and momentum

More on Momentum

And now the correct answer, with an explanation, to my last post which questioned,

"What is the reaction force of the weight of a book at rest on a table?"

From my count as of 7:40 pm Eastern Time here are the results:

A) The force of friction (5 people)

B) The force of the table upward on the book (120 people)

C) The force of the book on Earth (36 people)

D) The force of the book on the table (0 people)

So what is the correct answer? It's choice (C), "The force of the book on the Earth!"

The question states, "What is the reaction force of the weight of a book at rest on a table?" The weight of an object is the result of the gravitational pull from some other mass, and in this case from Earth. However, Newton's Law of Universal Gravitation states that every mass pulls on every other mass in the universe with an equal and opposite force. Just as the earth is pulling on the book, the book is pulling back on the earth with the exact same magnitude of force, but in the opposite direction. Using Newton's 2nd Law of motion for an object with constant mass, F = ma, one can see that that the acceleration of the book will be huge compared to the acceleration of the earth because their masses are a factor of 10^24 different! The acceleration on the book is 9.8 m/s/s downward, or what we typically call the acceleration due to gravity g. The acceleration of the earth toward the book is so small that it can't even be measured with any current (or likely future) technology.

Most of you said the answer is (B), "The force of the table upward on the book." This is a common misconception, as is obvious from the number of people that chose this incorrect response. Allow me to explain why this is incorrect.

The book is at rest on the table, and so obviously the table is applying an upward force on the book. Because the book is not accelerating up or down the force that the table applies to the book must be equal to the force that the book applies to the table. It just so happens that the magnitude of the force that the table exerts on the book is equal to the weight of the book, but that does NOT mean that it is the reaction force. The reason the size of the forces are the same is because the book's weight is the only force downward at the present time, and the table has enough strength to not fall apart, and therefore pushes upward with the same amount of force that is pushing downward on the table.

What you are thinking of are contact forces, and they also come in pairs. These contact forces are known to physicists as Normal Forces. Mathematically, normal means perpendicular to, and the normal force is always perpendicular to the surface.

Think of it this way: If the table were to suddenly disappear the book would fall to the ground due to its weight (weight is a force) and the earth would accelerate upward a tiny tiny bit from the equal amount of gravitational pull from the book. In this case there is still a force pair, and it is the same as the force pair when the book is at rest on the table.

Why does the book rest on the table? It rests on the table because the table is in the way and pushes upward on the book with a force that just so happens to equal the weight of the book. The force that is pushing upward (the normal force of the table on the book) is actually an electromagnetic force of repulsion between the electrons in the table and the electrons in the book (like charges repel). At the same time, the book is applying a force downward on the table (the normal force of the book on the table), which is also an electromagnetic force of repulsion between the same electrons.

The force pair for the book's weight (the pull from the earth on the book) is the gravitational force from the book pulling on the earth. The force pair for the contact force from the book on the table (the normal force from the book on the table) is the contact force from the table on the book (the normal force from the table on the book).

I found many of your responses quite amusing and others quite shameful, as some people were quite rude to people they thought has the wrong answer. Some of these people were bashing others for having the wrong answer, when in fact it was they that had the wrong answer. I suggest you quickly read the comments before many people delete their incorrect responses and pretend they never picked (B).

Some engineers were so convinced that they were right that they were saying they hate their nation for educating the youth incorrectly, even though it was the engineers that were wrong. Some high school physics teachers thought they were right, but were in fact wrong, and have been teaching their students incorrectly all along. Even a degree carrying physicist happened to choose the wrong answer, and then assert he was right because he has a degree in physics. Well, you were wrong.

I hope something can be learned from this, as even those people that think they know what they are talking about don't always know what they are talking about. Please, research things for yourself and question everything! You will never learn if you simply trust what is told to you. I highly recommend some of you pick up a good high school physics book and review it thoroughly before you bash others for disagreeing with you.

So, how do you know you can trust me? Well, because I'm awesome. But seriously, I promise you that the correct answer is (C).

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A bird is standing in a bird cage, which is itself resting on an electronic balance. The bird flaps its wings and starts to fly upwards while remaining in the cage. Neglecting any effects from the turbulence of the air created by the flying bird, will the mass reading of the balance

(A) decrease

(B) increase

(C) remain the same

(D) depends on the type of bird

(E) none of the above

The correct answer is (A) decrease.

In this problem the scale reading will only depend upon the weight of the cage and its contents. If the bird's weight is being supported by the cage or something attached to the cage then it will add to the weight of the cage. However, if the bird is flying then its weight is being supported by the air rather than the cage. The air is not physically attached to the cage and can move across its boundaries freely. Furthermore, the weight of the air is not supported by the cage. Therefore, the weight of the air and anything being supported by the air, i.e. the flying bird, will not affect the apparent weight of the cage.

So, if the bird is standing in the bird cage the scale will record the weight of the cage plus the bird's added weight. However, if the bird is flying around the bird cage then its weight is no longer being supported by the cage and will therefore not be registered on the scale. So, when the bird begins to fly the scale reading will decrease.

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