Home, Science, Science Concepts

Newton’s Three Laws of Motion (Part 2)

Hi guys,

This is the second part of my ‘Newton’s Three Laws of Motion’ Series. I will be writing about Newton’s Second Law of Motion. This post will be rather short due to the simplicity of Newton’s Second Law of Motion.

The acceleration of a body is directly proportional to, and in the same direction as, the net force acting on the body, and inversely proportional to its mass. Thus, F = ma, where F is the net force acting on the object, m is the mass of the object and a is the acceleration of the object.

Its pretty self-explanatory, the net/resultant force applied to an object is the product of its mass and its acceleration. This formula is the cornerstone to Classical Physics as it can be differentiated or integrated into many of the other formulas in Classical Physics.

What is ‘net force’? Net force is the force that is applied to an object after all other forces have been taken into consideration. Do remember that the SI Units for Force, Mass and Acceleration are the Newton, N, the Kilogram, kg and Metres per second per second or metres per second squared , m/s^2.

So how do you use this formula?

Lets imagine that a 5kg metal cube is being pushed with 10N of net force. What is the acceleration of the object?





Therefore we can deduce that the object is accelerating at 2m/s^2.


And that’s the end of Part 2 of Newton’s Three Laws of Motion.

It was pretty short but it shows us how simple yet powerful Newton’s Second Law is. It supports the whole of Classical Physics yet it can be simplified to just 4 characters: F=ma.


Thank you for reading this post. Once again, please post any errors or additional points in the comments to benefit the other readers and feel free to comment or send me a message via the contact form page.

See you soon!

Clyde Lhui 🙂




Home, Science, Science Concepts

Reference Frames

Hi guys,

In my previous post about Newton’s First Law of motion:

When viewed in an inertial reference frame, an object either is at rest or moves at a constant velocity, unless acted upon by an external force.

I did not explain reference frames which may have confused some of you. So i shall be explaining reference frames in this post.

Essentially, reference frames are frames which can be used to describe a location of an event(something that occurs at a very specific point and does not include things that occurs in an area).

To further explain what is an event, imagine you are in a science fair (preferably one with lots of amazing science experiments being carried out (but it doesn’t really matter)) to you the science fair may seem like an event, however in terms of reference frames, the science fair is not  an event (because it occurs over an area).

So you may be thinking right now: What in the world is an event since practically everything has a volume and takes up space? Well classically nothing can physically be an event, reference frames are just used to describe things based upon (non-existent) events or “events” (objects which still take up space) in classical physics.

However in quantum physics, events do physically exist in the form of elementary particles in the standard model of particle physics. Examples of such particles include: Electrons, quarks, muons and all three types of neutrinos (electron neutrino, muon neutrino and tau neutrino). These particles supposedly do not take up any space/volume however they do have mass. (This is where classical physics and quantum physics clashes. Since these particles posses a certain amount of mass but no volume, they posses an infinite density and in classical physics they would be black holes/singularities) i may do an in depth post about the standard model of particle physics in the future.

So how do these reference frames look like? They have a x, y and z axis which each represent a dimensions in our 3 dimensional world. However in the context of physics, there is a fourth dimension of time which is represented as t not within the reference frame.

Reference frames

Fig 1.1: two reference frames with the origins being O and O’

Notice that any point in space can be described using coordinates taken with reference to the origin of the reference frames hence the name “Reference frames”

NOTE: Coordinates can also hold a negative value so events behind the origin can still be described by using a reference frame.

In the above example, observer O is ‘stationary’ and observer O’ is ‘moving’. You may wonder why i use inverted commas to describe their motion. That is due to the fact that there is no such thing as an object that is absolutely moving or absolutely stationary (although there was once an ‘ether’ which was an absolute reference frame, however it was later dismissed through a ‘ether wind’ experiment)

Now I shall move on to Galilean and Lorentz Transformations.

So what are these? Galilean and Lorentz transformations are a set of formulas which can be used to calculate the velocity, location and time that is relative to one observer by using values obtained from the other observer and the first observer him/herself.

What is the difference between Galilean and Lorentz Transformations? Galilean Transformations reflect Galilean Relativity which is based upon common sense. While Lorentz Transformations reflect Einstein’s Theory of Special Relativity which is based upon the idea that the speed of light is constant in any reference frame.

Do remember the following points:

  1. x=Length
  2. y=Height
  3. z=Breadth
  4. t=Time
  5. c=Speed of Light (3.0 x 10^8)

Points 1-4 are the four dimensions of reference frames that you should take note of. Do note that there is also a velocity transformation which has the values u and v which are the velocities of the two observers.

Allow me to reuse Figure 1.1 as an example to describe the Galilean Transformations.

Reference frames

In Figure 1.1, there are 2 observers: Observer O and Observer O’

Observer O’ is moving forward at a velocity of v relative to Observer O who is moving at a velocity of u.

Galilean Transformations are as follows:






After transforming the values, you will get a different value, which is what observer O observes. An example will be if observer O was moving at a speed of 3m/s and observer O’ was moving at 5m/s, observer O will see observer O’ moving forward at a velocity of 2m/s relative to him. In this case what is the true velocity of observer O’? It is both 5m/s and 2m/s, this is due to the fact that these 2 velocities are observed in different reference frames, and are thus both correct.

Lorentz transformations are different from Galilean transformations as they take into account the change in time and space due to the velocity of the observers (aka Special Relativity). I/Jackson will explain this phenomenon in a later post.

The Lorentz Transformations are as follow:

x’=((x-vt))/√(1-v^2/c^2 )




The Lorentz Velocity Transformations are a little difficult to understand due to the definitions of u, v and u’ and will be explained in the Special relativity post.

As previously mentioned in the post, an ‘Ether’ Reference Frame was proposed.

This was to resolve the conflict between the Galilean Relativity and Maxwell’s Theory. Maxwell’s Theory is essentially the idea of which Special Relativity was based upon. They are based on 4 equations (The 4 Maxwell’s Equations) which are named after James Clerk Maxwell, the publisher of the early form of the equations which are the foundation of classical electrodynamics. Maxwell’s equations prove that the Speed of Light-c is a constant. and this is in direct conflict of Galilean Relativity which show that if you move a t a sufficiently high velocity, you will be able to see light moving.

Therefore to solve this conflict, the ‘Ether’ Reference Frame was proposed. This reference frame is an absolute reference frame, meaning objects which are stationary relative to this reference frame are truly at rest and objects which are in motion relative to this reference frame are truly in motion.

However this was dismissed due to the Michelson-Morley Experiment which tested for ‘Ether Wind’.


Fig 1.2: The Michelson-Morley experiment. On the left is a directed light source. The beam of light is directed at a beam splitter which splits the beam of light and directs them in 2 directions toward 2 mirrors of equal distance. The mirrors bounce the light back which then is sent to a light detector.

How does this experiment prove the ‘Ether’ Reference Frame wrong? This experiment was conducted on Planet Earth of course. During that period, it was already known that Earth rotates around an axis. thus the setup of the experiment would be rotating together with the Earth. This would mean that should the ‘Ether’ Reference Frame be present and correct, the light going toward the mirror on the right would take shorter/longer (depending on whether the experiment was set-up facing the east or the west) as compared to the light hitting the top mirror. However the detector detected that the light was hitting together at the same time, thus dismissing the idea of the ‘Ether’ Reference Frame. This also prompted the development of Einstein’s Theory of Special Relativity.

Reference Frames are key in the understanding of both Classical and Quantum Physics as they are useful in the understanding of many experiments and theories.

And with that i end this post on Reference Frames.

Should you spot any mistakes or have any questions, please leave a comment to benefit the other readers and of course myself! We will be posting the second part to Newton’s Laws of Motion and another post on Black Holes by my new co-writer Jackson! Lastly, we will be blogging about new upcoming projects in the series ‘The Power Surge’ (For reals this time). If you have any ideas for more projects, please feel free to comment/contact me in the contact form.

Thank you For Reading and I hope to see you soon!

Clyde Lhui 🙂


P.s: An Inertial Reference Frame is a reference frame that either is stationary or moving at a constant speed.


Special Thanks to Mr Damian Boh for his explanation of the topic to me which greatly aided me in the writing of the content of this blog post.

Home, Science, Science Concepts

Newton’s Three laws of motion (Part 1)

Hi guys,

I’m terribly sorry for being inactive for such a long time. I have been extremely busy lately therefore not having time to update the blog. So to make up for it i’m going to explain Newton’s three laws of motion. I will be posting these in 3 separate posts, explaining 1 law per post and this is part 1. I feel that this fundamental concept is extremely important in understanding other concepts as it is extremely simple, easy to understand and at the same time very significant.

Okay so on to the three laws of motion.

So what are the three laws of motion? They are Laws of physics which can predict the motion of an object pretty accurately (Although not as accurately as Quantum Physics which was developed by Einstein around 300 years after Newton formulated these laws.)

The three laws of motion are as follows:

  1. When viewed in an inertial reference frame, an object either is at rest or moves at a constant velocity, unless acted upon by an external force.
  2. The acceleration of a body is directly proportional to, and in the same direction as, the net force acting on the body, and inversely proportional to its mass. Thus, F = ma, where F is the net force acting on the object, m is the mass of the object and a is the acceleration of the object.
  3. When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction to that of the first body.

Okay that might have been a little too complicated. but on to the explanations.


Let’s analyze Newton’s first Law of Motion

“When viewed in an inertial reference frame, an object either is at rest or moves at a constant velocity, unless acted upon by an external force.”

So what does this mean?

Newton’s first law of motion is also known as the law of inertia and inertia is the tendency of an object to remain in its state of motion be it not moving at all or moving at 10000000 km/h. Heavier bodies possess more inertia while lighter objects possess less inertia so heavier objects are more difficult (requires more force) to speed up but more difficult to stop and lighter objects though easier to speed up, are easier to stop.

Imagine a bus and a feather in front of you. Which is easier to move? The feather. Imagine the bus is now moving at 40 km/h and the feather is also moving at 40 km/h. Which one is easier to stop now? The feather.

So what does any of that have to do with anything?

Well simply put Newton’s first law of motion states that an object will stay stationary or continue moving in a constant velocity in a straight line if no forces (resultant) are acting upon it. This idea is expressed in terms of mass with inertia.

So why do things eventually stop?

Well that is because there are other forces acting upon the object.

Friction, Gravity, Normal Forces and practically any type of force can oppose other forces, cancelling them out and thus causing the object to stop.

Try pushing both you hands against each other and use as much force as possible to push your hands against each other. Despite applying so much force, your hand remains fairly stable and in roughly the same place. Why does this happen? The forces cancel each other out, therefore your hand remains stationary.

As you may have noticed, I have been using the word “resultant” very often. Why is this so?

Resultant force is the final value of the force acting upon an object after taking into account all other forces. Lets say we have a book that is falling. In this case there are 2 forces acting on the object: Gravity and Air resistance. Assuming the gravity acting upon this book is 10N and the air resistance acting on the book is 5N the resultant force will be 5N downward with gravity.

This resultant force value can be calculated with a simple formula which I will explain in my second post.

And with that I end this post. Thanks for your continuous support of my blog and I wish you a good day!


Best Wishes,

Clyde Lhui 🙂


P.s: Once again im very sorry for being so inactive. I had to spend quite some time catching up with school work but I promise that I will work on this whenever I can!

Home, Science, Science Concepts

Terminal Velocity

Hi guys,

As I mentioned certain scientific terms in my previous post, i would like to go in depth on those concepts,beginning with terminal velocity, it being the most fundamental concept in my post.

So what is terminal velocity?

Terminal velocity is the velocity of an object when the drag force (dependent on the fluid the object is travelling through) acting upon it is equal to the downward force of gravity acting upon it. Simply put, when the air resistance of a falling object cancels out the gravitational force which is pulling it downwards and accelerating it.

So how do these forces affect the motion of the object? The forces cancelling each other out make the object remain at a constant rate of motion.

You may ask why does the object still move when the forces cancel each other out. This is due to the fact that in the beginning the force of gravity still manages to overcome the drag force, allowing the object to gain speed (accelerate) initially. But as the object increases in velocity, the drag force increases, this effect can also be seen in the case of friction (Drag and friction are pretty much the same thing). Lets assume that a boy is dragging a heavy box, full of files, across a distance of 100 meters, now we will imagine this scenario in two different ways, firstly in the case whereby the boy is walking slowly and in the second whereby the boy is running. So in the first case the boy walks, when he reaches the end, he feels the bottom of the box, where the box and the floor meet, it still feels the same as before, now in the second case, he runs, he once again feels the bottom of the box, this time it feels warmer than before. So what can we infer from this scenario? Before I reveal the answer, i would like to state a few properties of friction:

  1. Friction opposes motion
  2. Friction causes wear and tear
  3. Friction produces heat when kinetic energy is converted into thermal energy

So what can we infer? In the second scenario, there was more heat, therefore we can assume that there was more frictional force produced in the second case.

Now lets go back to what i mentioned previously, air resistance increases (Drag Force) as the object’s velocity increases. As seen in the example above, we can tell that this statement is true.

That’s essentially the definition of terminal velocity. Before we move on, lets do a recap:

  1. Terminal velocity is the velocity an object is at when the gravitational force acting upon it is equal to the drag force acting upon it in the opposite direction therefore cancelling out all forces therefore having a resultant force of 0
  2. The drag force acting upon the object increases as the object accelerates due to the downward force of gravity.

Ok so lets move on to the math behind terminal velocity and some examples of it.

The formula for terminal velocity is as follows: V_t= \sqrt{\frac{2mg}{\rho A C_d }}

Vt=Terminal Velocity

m=Mass of falling object

g= Acceleration of the object due to gravity

ρ= Density of fluid which the object is travelling through

A= Projected area of the object

Cd= Drag Coefficient

I went through everything in my previous post but lets do a recap on these terms:

mass= Amount of matter in an object (SI Unit : kg)

Acceleration = Rate of increase of speed (SI Unit: m/s^2)

Gravitational force= how much gravitational force does an object exert on another object (SI Unit: N/kg)

Density= Mass per unit volume of matter (SI Unit: kg/m^3)

Projected Area= Area of a falling object which is in contact of the air flowing through it. (SI Unit: m^2)- Same as Area

Drag coefficient= A value which depends on the shape of an object and can only be calculated by using the drag force of the object and other factors or by doing actual testing. This value has no units.

That’s all the terms. So now I shall be doing some examples.

So assuming I drop a metal cube which has a mass of 3 kg and has a projected area of 1 m^2 on Earth 90 degrees downward, through air at a temperature of 25 degrees Celsius, what would the Terminal velocity of the cube be?

All we have to do is input all the values into the formula. The acceleration due to gravity on earth is 9.81 m/s^2. The density of air at 25 degrees Celsius is 1.1839 kg/m^3 and the drag coefficient of a cube is 1.05 facing downward. The result is : 6.881101581m/s.

So there’s Terminal Velocity for you!

I would like to thank Mr Tan Ping Hock and Mr Yao Zhi Wei Adrian, My current and previous physics teachers respectively for clearing my doubts about certain concepts within this topic of terminal velocity!

Thanks for reading!

Clyde Lhui






Home, Science, Science investigations

I’m falling down down down down…

Hi everyone!

Recently I have been playing this game called League of Legends (LoL). I’m pretty sure that loads of guys out there play LoL. So I was checking out some info about the game. So in this game, there are 3 maps or Fields of Justice. The first one is Summoner’s Rift, the second being Twisted Treeline and the third is called Howling Abyss. But wait, I’m not here to tell you about LoL. If so why would I label it under ‘Science’? Ok here comes the interesting part. In the description of Howling Abyss LoL said that at screams could be heard in the wind and those were rumoured to be those of warriors who got pushed into the bottomless abyss. So I was thinking “Since its a bottomless pit, the warriors who fell in the abyss wont hit the ground and die so how long will it take for them to die and if they screamed when they died, how long would the screams take to reach the top?”

Ok so the first part’s really simple: How long would they take to die? Assuming there are no flying creatures that eat them up, they did not cry, sweat, pee or breathe through their mouth and they were regular humans, they would take 3 days to die of dehydration.

So that’s part 1 answered, now for part 2. Here’s where it starts to get difficult. How long would their final scream take to reach the top? firstly we have to find out how far down they would have travelled. So assuming they were falling like a skydiver (horizontally) we can find the projected area or the area which is in contact of the air below them. The average male Body Surface Area ( BSA )is 1.9 m^2 therefore to calculate the projected area, we can take the BSA divided by 2 as only one side of their bodies are in contact with the air. therefore the projected area would be 0.95 m^2. Next we have to find their mass. Assuming they were very fit and they had the most ideal body weight according to the Body Mass Index (BMI) and they were average American males in the United States of America who have an average height of 1.763 m, they would have a mass of around 150 pounds or 68.0389 kg. Now we have all the data we need to calculate the terminal velocity of the warriors or the speed they will be at when they stop accelerating. The formula for terminal velocity is as follows: V_t= \sqrt{\frac{2mg}{\rho A C_d }}

Vt= Terminal velocity

m=Mass of falling object

g= Acceleration due to gravity

p(rho)= Density of fluid which the object is travelling through (in this case, air)

A= Projected area

Cd= Drag coefficient

So what do all these mean? well we have been through practically everything except for the drag coefficient. So what is drag coefficient? Drag coefficient is a value which cannot be calculated. It is the value which puts the force generated by drag into account. The drag coefficient of a man is 1.0-1.3. Lets assume that our warriors have a drag coefficient of 1.2, not too skinny yet not too fat. The acceleration due to gravity on Earth is 9.8m/s^2 and the density of air at room temperature (25 degrees Celsius) is 1.1839 kg/m^3 so with everything in SI units we can calculate the terminal velocity of the warriors. The final answer is 54.44492062 m/s assuming all our calculations are right and in SI units. this can be rounded off to 54.4m/s (3 s.f.). The next part that we have to solve is the acceleration period. so from 0m/s to 54.4 m/s how long would it take? Once again assuming that the acceleration due to gravity is around 9.8 m/s^2 as on earth, all we have to do is divide 9.8m/s^2 from 54.4 m/s. The result we arrive at is 5.551020408 s. to find the distance travelled during this time, all we have to do is imagine that all this data is being plotted on a distance time graph and find the area during the acceleration period.

1/2*5.551020408 s*54.4 m/s =150.9877551 m

This can once again be rounded off to 151 m ( 3 s.f.)

Now lets move on to the part where we deal with the duration of the freefall. Assuming that they took EXACTLY 3 days to die of dehydration, they would take 24 h*3= 72 h to die. Now we minus the 5.551020408 s from the 72 h. 72 h = 72 h*60 min = 4320 min =4320 min* 60 s= 259200 s

259200 s – 5.551020408 s = 259194.449 s (my calculator only has this many decimals)

Therefore we can assume that our warriors remained at terminal velocity for 259194.449 s.

So to find the distance they travelled at terminal velocity, all we have to do is:

259194.449 s* 54.4m/s= 14100178.02 m

So our warriors travelled a total distance of:

14100178.02 m + 151 m = 14100329.02 m

Finally to calculate the time taken for the sound to reach the top, we have to divide the distance travelled by the speed of sound which is Mach 1 or 346.13 m/s at room temperature (25 degrees Celsius).

14100329.02 m / 346.13 m/s = 40737.09018 s

That is how long the sound would take to reach the place where our warriors fell from. Converting it to hours : 11.3159 h

So there’s the answer!

If you have anything else, like a science question of some sort or you found a mistake in my calculations and want me to rectify it, please post it in the comments section or you could send me a message via the contact me form!

Thanks for reading!

Clyde Lhui 🙂