# ATAR Notes: Forum

## VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: #1procrastinator on September 09, 2011, 03:44:37 pm

Title: Newton's Laws and probably more questions
Post by: #1procrastinator on September 09, 2011, 03:44:37 pm
If a force of 100N was exerted on a box of mass 20kg, the box would push back on whatever's pushing it right?

1) Would this be -100N since it's in the negative direction? If so, then would a then have to be negative which would make no sense

I don't really get the whole F=ma thing, it made sense when I was just plugging in numbers but when I thought about it, I I realised I didn't understand it at all.

2) So F is the force exerted by an object m mass accelerating at a. When you have a problem where one object collides with another after being pushed (given the force, mass of both objects), how do you find the acceleration of the first object, and the force exerted on the second?

Object 1 = 20kg, object 2 = 30kg, force = 100N

Is it F/m = [a] to find the acceleration? This is what brought me to the first part, it's accelerating in the positive direction away from the force but the force is being applied to it I thought F=ma was to describe ITS force on others

3) do we always have to indicate F, a, v, etc. are vectors? Even if the final answer is just a magnitude?

Title: Re: Newton's Laws and probably more questions
Post by: Lasercookie on September 09, 2011, 05:33:01 pm
So you seem to be having trouble understanding Newton’s laws. I’ll try to address that first, and then later answer your questions directly. It’s inevitable that I will repeat something you are already familiar with, but I think you need some refreshing.

Just to cover all bases: Newton's first law describes how a rolling wheel will keep on rolling and rolling. It will never stop unless it is stopped by an outside influence (like friction, or a tree). A bit after it runs into the tree, it'll become a stationary wheel. Since it's at rest, it'll keep being at rest. Unless something (like a foot) applies a force upon it and changes it’s state.

Now merely apply the fancy wording:

First law: The velocity of a body remains constant unless the body is acted upon by an external force.

In the case of the rolling wheel, its velocity, which is non-zero (as it is in the state of motion), remains constant until it is acted upon by an external force (the force applied on the wheel by tree). After this event, its velocity will be reduced to zero and the wheel will be in a state of rest.  It will not leave this state of rest unless it is acted upon by an external force.

Let's take the situation you described: "When you have a problem where one object collides with another after being pushed (given the force, mass of both objects), how do you find the acceleration of the first object, and the force exerted on the second?"

And let's also make it a bit less abstract (though the values we'll use will be unrealistic):
You have a block of cheese and a knife. You cut the cheese with the knife. The cheese requires a force of 10N to cut.

So the knife was initially still, then it moved in order to cut the cheese. In other words it accelerated. But what caused this acceleration? The answer is a force. Ignoring the situation where the force is not big enough to do anything useful; If you apply a force to something, it will change what it was doing (Newton’s First Law). It’s velocity will change – again in other words it will accelerate $a=\frac{\Delta v}{\Delta t}$.

Of course, you can have more than one force acting on an object. For this reason we refer to all the forces that act on an object as the net force. In our simple situations, we usually only consider one force that has acted, so that single force will be our net force. This is an important distinction to make.

So, let’s summarise: we know that a net force causes acceleration. We also know that this effectively changes the inertia of an object. So net forces are proportional to inertia and acceleration. But we want hard numbers – what’s the measure of inertia?

Mass. The more mass an object has the more inertia it has. In other words, an object with greater mass will be harder to change its state of motion.

So there we have it: $F_{net}=mass*acceleration$.

This should fit quite intuitively with your idea of weight. If you have a light person and a heavy person and you pushed them both with an equal amount of force, the heavier person wouldn’t move as far. If you wanted to accelerate the heavy person with the same acceleration that the skinny person had, you would have to push harder (apply a greater force).

So, back to our cheese and knife. We need to apply a force of 10N and in our simple situation, we use the knife to apply the force. Since this is the only force we’re talking about, we apply a net force of 10N.

Newton’s Third Law, is probably the hardest to understand (but easiest to remember)

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

Push your knuckles together, as hard as you can. The amount they discolour will be the same, because they both experienced equal and opposite forces.

Note that it doesn’t matter which you consider to be an action and which you consider to be a reaction. Just don’t switch around half way while working out. You don’t talk about the action and then suddenly switch to talk about the reaction. I think this is the point where you were getting confused was where to apply Newton’s third law.

You also need to remember that the Newtonian pair must act on different objects. Let’s use the example of a person (let’s call him Howard) standing on the Earth. So, due to gravity, Howard is applying a force on the Earth (this will be force 1). The Earth is applying a force on Howard (this will be force 2). Force 1, which is Howard on Earth, cannot act on Howard. Force 2, cannot act on the Earth. It is Earth on Howard. If they acted on the same object, the forces would cancel each other out and we would live in a world where no force could ever be applied. (I don’t know if this paragraph is 100% correct – someone please confirm this).

So, we want to know acceleration of the knife. So we had:
$F_{net}=10 N$. This was the force of the knife on the cheese. The force of the cheese on the knife would be equal AND opposite. That would be -10N. We’re going to ignore that fact though, we’re only talking about the action here (though the complete interaction is comprised of the knife acting on the cheese and the cheese acting on the knife).

Let’s make up some values for the masses now:
$m_{knife}=0.500kg$
$m_{cheese}=1kg$
We'll now consider the force of knife on the cheese, and calculate it's acceleration.
$F_{net}=m_{knife}*a \rightarrow a = \frac{F_{net}}{m_{knife}}$
$a = \frac{10}{0.500} = 20 m/s^2$ (yeah, ridiculous behaviour for whoever was using the knife)

Why is it accelerating in a positive direction? Because we defined the force of the knife on the cheese to be positive.

Let's take a look at the reaction:
$F_{net}=m_{cheese}*a \rightarrow a = \frac{F_{net}}{m_{cheese}}$
$a = \frac{-10}{{1}}=-10 m/s^2$
So, the acceleration due to the reaction is negative because we defined the action to be +ve.
You could easily switch around the positive and negatives if you wish.
The force exerted on the cheese was 10N (0.500*20), while the force exerted on the knife was -10N (1*-10)

I know I didn’t address your entire question, but hopefully this cleared things up.

You don’t have to give directions if it only asks for the magnitude. You don’t always have to mark F and a as vectors with an arrow if you don’t want to – just don’t forget they are actually vectors. Technically you should be always providing directions, if asked for acceleration. VCAA exams will get you in the bad habit of leaving it out (very rarely will they want directions).
Title: Re: Newton's Laws and probably more questions
Post by: lukew on September 10, 2011, 09:18:22 am
If I may, I have a few questions I wouldn't mind some help with too.
ATM I'm working on flight which incorporates a bit of Newton's third law as well, Bernoulli'd equation, net forces, thrust, drag, weight, lift, momentum, impulse etc.

The following information applies to questions 13–16.
A jet plane is travelling at a speed of 200 m s-1 relative to the air around it. Each second the engines of the plane take in a total of 150 kg of air. The air is used to burn 2.0 kg of fuel each second. Finally, the air and burnt fuel mix is expelled from the rear of the engines at a speed of 450 m s-1 relative to the plane. Use the following relationships:

change in momentum (impulse), Δp = mΔv = FΔt (N s)
power, P = Fv (W)

13) What is the magnitude of the total force exerted on the plane by the intake of the air?
14) What is the magnitude of the force exerted on the plane by the air and burnt fuel being expelled from the engines?
15) What is the net thrust produced by the engines?
16) What is the power developed as a result of the net thrust?
Answer = 1.6 × 106 W

I calculated question 13 by subbing in p=mv=ft and therefore 150 x 200=f x 1.....therefore F=30,000N (30kN), but question 14 onwards, I just got random answers that didn't match up at all.

Sorry for how long this is but if anyone could offer help it would be greatly appreciated :)
Title: Re: Newton's Laws and probably more questions
Post by: Lasercookie on September 10, 2011, 10:12:16 am
ATM I'm working on flight which incorporates a bit of Newton's third law as well, Bernoulli'd equation, net forces, thrust, drag, weight, lift, momentum, impulse etc.
Flight was awesome.

A jet plane is travelling at a speed of 200 m s-1 relative to the air around it. Each second the engines of the plane take in a total of 150 kg of air. The air is used to burn 2.0 kg of fuel each second. Finally, the air and burnt fuel mix is expelled from the rear of the engines at a speed of 450 m s-1 relative to the plane. Use the following relationships:
Note the term relative.

I'll use a fancy approach and use Galileo's velocity addition formula (you might not know this formula explicitly, but you would have done questions similar to it (e.g. the ones where you had a plane flying into wind, how fast is it actually going)

$\vec{V_{Brelative_A}}=\vec{V_B} - \vec{V_A}$

14) What is the magnitude of the force exerted on the plane by the air and burnt fuel being expelled from the engines?
$p=mv=F \Delta t$

$m=150kg + 2kg = 152kg$
$t=1s$
$V_{Brelative_A} = V_B - V_A = 450 - 200 = 250$

$p=(152)(250)=F*1$

$F=38000N=38kN$

Same approach for the rest - I'll leave it up to you. I reckon all you did wrong is to forget to adjust your values for the fact that the velocities you were given were relative.
Title: Re: Newton's Laws and probably more questions
Post by: xZero on September 10, 2011, 10:21:49 am
since we are using air as our reference frame, we must convert some velocity
14. the mass would be 150+2=152 and speed relative to air is 450-200=250. so f*1=152*250=38kN
15. intaking air can be seen as air resistance and q14 gives you thrust so f=38-30=8kN
16. f=8000,v=200 so p=8000*200=1.6*10^6
Title: Re: Newton's Laws and probably more questions
Post by: Lasercookie on September 10, 2011, 10:25:13 am
since we are using air as our reference frame, we must convert some velocity
14. the mass would be 150+2=152 and speed relative to air is 450-200=250. so f*1=152*250=38kN
15. intaking air can be seen as air resistance and q14 gives you thrust so f=38-30=8kN
16. f=8000,v=200 so p=8000*200=1.6*10^6
Off-topic: Congratulations on 666 posts :P
Title: Re: Newton's Laws and probably more questions
Post by: lukew on September 11, 2011, 01:29:09 pm
Thankyou all, good to know I was on the right track.
Title: Re: Newton's Laws and probably more questions
Post by: #1procrastinator on September 13, 2011, 05:20:10 pm
So the knife is accelerating at 20m/s^2, and that times its mass gave the force of 10N

And 10N is acting upon the cheese, so it reacts with 10N, take into account its mass and you get its acceleration of -10m/s^2

So they're accelerating at each other?

What does this mean when you push something though? If it's accelerating at you, then how's it also accelerating away (same direction as initial force)? Is thisrelated to the inertia thing?

----

1) With mass and weight, scales are calibrated to measure mass? It takes into account the (average?) gravitational pull?
Title: Re: Newton's Laws and probably more questions
Post by: Lasercookie on September 13, 2011, 06:54:06 pm
What does this mean when you push something though? If it's accelerating at you, then how's it also accelerating away (same direction as initial force)? Is thisrelated to the inertia thing?
This is how I understand it: The object is not accelerating positively and negatively at the same time. One acceleration refers to the action set of forces and the other refers to the reaction set of forces. They act on different bodies. They are seperate.

Ignore friction and throw a ball while on roller blades. Let's say you throw it with a force of 50N. You would be pushed back with a force of 50N. Similar situation with gun recoil.

Remember that it's a third law pair. The forces are applied to different bodies. (this is a fact that is easily overlooked)
One pair is the 'action', the other is the 'reaction' (arbitrary terms really). They are separate parts that comprise an entire interaction.
The 'action' force is the gun firing the bullet. i.e. the gun applies a force to the bullet. The bullet accelerates with a positive acceleration.
The other force is the 'reaction' i.e. the bullet applying a force on the gun.  The gun accelerates with a negative acceleration.

So, the action force is a separate force to the reaction force. The gun applies a force to the bullet and the bullet applies a force to the gun. Then why doesn't the gun fly off as far as the bullet does (ignoring the fact that the person is holding it - that would get into inertia)?

Simple, the acceleration the gun applied to bullet was greater than the acceleration bullet applied to the gun.
But wait, aren't they meant to be equal and opposite? No.

Yes, it's true that the are forces are equal and opposite. But acceleration depends on Force and Mass.
The bullet is a lighter mass, and therefore applies a smaller acceleration on the gun.
The gun is a heavier mass, and therefore it can apply a greater acceleration on the bullet.

I think that explanation made sense. Wikipedia has a page on reactions (discusses common misunderstandings as well): http://en.wikipedia.org/wiki/Reaction_(physics)

Yes, this is an un-intuitive concept. Also remember you have have to pay attention to wording carefully in physics.
Quote
1) With mass and weight, scales are calibrated to measure mass? It takes into account the (average?) gravitational pull?
I don't know the answer. I'd imagine it'd vary according to the type of scale (electronic vs. those manual ones that operate with a spring vs. the other numerous types of scales).
Title: Re: Newton's Laws and probably more questions
Post by: #1procrastinator on September 16, 2011, 12:46:10 pm
^ I think I'm starting to get it now...so then the equation F=ma gives the forces acting ON the object and not the force it produces when it accelerates?

---

For the kinematic equations, can you derive them without using a graph? e.g.

x = ut + (1/2)at^2

In the textbook, they got that from the velocity-time graph (if I remember correctly), but I personally don't like deriving equations/formulas from graphs. I prefer to find them using intuition (or algebraically first) . It's the (1/2)at^2 in that equation that I don't quite get - why is it half (ignoring graph). I get that if it goes from 0m/s to 10m/s at 10m/s/s, in that second, it can't have travelled 10m, but why does it have to be the constant half? Why not 1/4 or something? I wrote out a table with time, distance and velocity for reasons I'll state later (I forgot what I was looking for, sorry! haha)

And the work formula, I've only briefly read that part, but I think the equation they derive for work is also based on a graph...necessary or is there another way?

Thanks a lot for the replies by the way
Title: Re: Newton's Laws and probably more questions
Post by: Lasercookie on September 16, 2011, 04:38:39 pm
^ I think I'm starting to get it now...so then the equation F=ma gives the forces acting ON the object and not the force it produces when it accelerates?
I really don't understand this question. Can you rephrase it.

F=ma gives the net force acting on the object. A force is acted upon one object to another object (e.g. A on B).
This tells us the acceleration is due to the forces acting upon it. You could also look at the other way, a = F/m, acceleration produces a force.See xZero's post below

Quote
For the kinematic equations, can you derive them without using a graph? e.g.
Yes, you can. Calculus gives a pretty elegant (well I think it is anyway) derivation, but you can also do it without calculus. I'm not very good with the latex, so I'll write it out and scan it in (possibly later tonight, if not then sometime tomorrow).

I've got the non-calculus and calculus derivations that I did earlier in the year in my stack of notes.
They're kind of messy, I'll write them out neatly for you.

When I get around to scanning it in, I'll attach it to my post and send you a PM.

edit: crossed out my mistake as pointed out by xZero.
Title: Re: Newton's Laws and probably more questions
Post by: xZero on September 16, 2011, 05:01:04 pm
This tells us the acceleration is due to the forces acting upon it. You could also look at the other way, a = F/m, acceleration produces a force.
no no, a net force on an object causes an acceleration, not acceleration causes a net force. If you have either net force/accele it implies that accele/net force exists but the cause-relationship can't be switched.
Title: Re: Newton's Laws and probably more questions
Post by: Lasercookie on September 16, 2011, 05:33:16 pm
This tells us the acceleration is due to the forces acting upon it. You could also look at the other way, a = F/m, acceleration produces a force.
no no, a net force on an object causes an acceleration, not acceleration causes a net force. If you have either net force/accele it implies that accele/net force exists but the cause-relationship can't be switched.
My bad, thanks for the correction.
Title: Re: Newton's Laws and probably more questions
Post by: Lasercookie on September 17, 2011, 02:11:34 pm
I can't get the scanner to work at the moment.
I did find these for the equations of motion:

You should also realise you can derive pretty much all the equations from F=ma.

http://physics.info/motion-equations/
http://physics.info/kinematics-calculus/

I don't think that showed the Work formula. This is how you can do it (saw this in a textbook):
$F=ma$

$F \Delta x=ma \Delta x$ (multiply both sides by x)

$F \Delta x=ma \Delta x$

We know: $\Delta x = ut + \frac{1}{2}at^2$

$F \Delta x=ma v_i t + \frac{1}{2}at^2$

$F \Delta x=ma v_i t + \frac{1}{2}ma^2t^2$

$a=\frac{\Delta v}{t}$ and $\Delta v = v - v_i$

$F \Delta x=m\frac{\Delta v}{t} v_i t + \frac{1}{2}m(\frac{\Delta v}{t})^2t^2$

$F \Delta x=m\frac{(v - v_i)}{t} v_i t + \frac{1}{2}m(\frac{(v - v_i)}{t})^2t^2$

$F \Delta x=m\frac{(v - v_i)}{t} v_i t + \frac{1}{2}m(\frac{(v - v_i)}{t})^2t^2$
You can simplify out time (t*1/t = 1, t^2 * 1/t^2 = 1)
$F \Delta x=mv_i(v - v_i) + \frac{1}{2}m(v - v_i)^2$

Expand and simplify:
$F \Delta x=mvv_i - mv_i^2 + \frac{1}{2}m(v^2 - 2vv_i + v_i^2)$

$F \Delta x=mvv_i - mv_i^2 + \frac{1}{2}mv^2 - mvv_i + \frac{1}{2}mv_i^2$

$F \Delta x=\frac{1}{2}mv^2 + \frac{1}{2}mv_i^2 - mv_i^2$

In case you're wondering: $\frac{1}{2}mv_i^2 - mv_i^2 = -\frac{1}{2}mv_i^2$. I'm too lazy to type out the in between step there.

$F \Delta x=\frac{1}{2}mv^2 - \frac{1}{2}mv_i^2$ (look familiar?)

$F \Delta x=\Delta E$

Title: Re: Newton's Laws and probably more questions
Post by: #1procrastinator on September 25, 2011, 12:04:03 pm
@laseredd

'I think I'm starting to get it now...so then the equation F=ma gives the forces acting ON the object and not the force it produces when it accelerates?'

i mean like i can't say an object of mass 2kg moving at 2m/s^2 would exert a force of 4N on whatever it hits, i can only say it's acceleration*mass is equal to the sum of forces acting on it (which brings to mind, weight, that's mass*acceleration is it not?)

Thanks for the derivation
Title: Re: Newton's Laws and probably more questions
Post by: xZero on September 25, 2011, 12:53:03 pm
i mean like i can't say an object of mass 2kg moving at 2m/s^2 would exert a force of 4N on whatever it hits, i can only say it's acceleration*mass is equal to the sum of forces acting on it (which brings to mind, weight, that's mass*acceleration is it not?)
you can say that an object of mass 2kg moving at 2ms^-1 has a momemtum of p=mv=4 and will exert a force on impact base on the other object's momemtum and the time interval of the collosion. In short, you use momemtum and impluse to work out the force exerted on objects in a collosion
Title: Re: Newton's Laws and probably more questions
Post by: #1procrastinator on September 25, 2011, 09:33:04 pm
^
interesting, i'm not up to momentum yet lol. everyone else is - i'm waay behind

what about weight? you can use mass * g to find that - is that the only exception?
Title: Re: Newton's Laws and probably more questions
Post by: Lasercookie on September 25, 2011, 09:35:05 pm
It's not an exception. 'g' is the acceleration due to gravity.
Weight (or force of gravity on object)
$F_g=ma=mg$

$W=mg$

Same thing really.
Title: Re: Newton's Laws and probably more questions
Post by: Kiro Kompiro on September 27, 2011, 02:00:34 pm
@procrastinator:  I found wording Newton's Laws this way helpful:

first law: for an object to change its speed or direction, an UNBALANCED force must act on it.  to get the unbalanced force and see if it is more than zero, add all the forces up taking into account their DIRECTIONS.

law 2:  The UNBALANCED force equals the body's mass times its acceleration.  The 'F' in F=ma refers to the UNBALANCED force.  To work this out you add up all the forces acting on the object, taking into account their direction.  I always use positive for forces acting to the right and negative for forces acting to the left.  When talking about vertical motion, I default to positive if the force acts upward, and negative if it acts downward.  If an object is thrown up, gravity acts down so I make down negative because it acts in the opposite direction to the initial movement. But if the object is thrown down, then gravity acts in the same direction so it then is positive.

law 3:every action has an equal an opposite reaction.  Deceptively simple.  It usually refers to the forces on objects that are in contact but not moving RELATIVE to each other, on the contact surfaces. eg you are standing on a concrete floor.  The earth's gravity exerts a force on you downward, yet you do not move through the floor because the floor pushes up on you with an equal but opposite force.  To test this, take a sheet of card board, lay it across two chairs, and stand on it.  What happens?  You fall through the cardboard because the upward force of the cardboard is insufficient to balance your weight.

Eg that uses all three laws:  two square blocks are in contact and sit on a frictionless floor.  Their masses are 2kg and 5 kg.  A push of 21 N is applied to the small block, and the blocks move together.  What is their acceleration?

Ans: unbalanced force is 21 N, total mass is 2+5=7 kg, a=F/m= 3 m/sec sq

What is the force of the little block on the big block?

Ans: the big block accelerates at 3 m/sec-sq.  It must have an unbalanced force of 5x3=15 N.  This is the force the little block exerts on the big block.

What is the force of the big block on the little block?

ans: Since there is no relative movement between the blocks, the force the big block exerts on the little block must also be 15 N in the opposite direction.

Does this make sense?

Draw the forces acting on the little block and add them up:

21N push acts on the side of the little block, -15N acts in the opposite direction from the big block, unbalanced force on the little block is 21-15=6 N, mass is 2 kg so a=F/m=3 m/sec-sq, which is what we had at the start.

Similarly, the unbalanced force on the big block comes from the little block which is 15N, m=5 kg, a=F/m=3 m/sec-sq
Title: Re: Newton's Laws and probably more questions
Post by: #1procrastinator on October 03, 2011, 12:57:55 pm
It's not an exception. 'g' is the acceleration due to gravity.
Weight (or force of gravity on object)
$F_g=ma=mg$

$W=mg$

Same thing really.

so as long as you know there's only one force in one direction you can use m*a to find it

thanks kiro, it's coming to me now albeit slowly

---

in physics, is it always possible to visualise what the equations mean physically? e.g. squared constants/variables, i can't imagine. is it just an abstract thing that describes the physical thing but you can't imagine the individual terms and what they mean physically?

how is coulomb's law derived?
Title: Re: Newton's Laws and probably more questions
Post by: Lasercookie on October 03, 2011, 01:35:36 pm
in physics, is it always possible to visualise what the equations mean physically? e.g. squared constants/variables, i can't imagine. is it just an abstract thing that describes the physical thing but you can't imagine the individual terms and what they mean physically?
Apparently you can't with really visualise string theory, but with the stuff we do in VCE, yeah definitely.

I'll use the example of Coulomb's Law:
Here's the fancy definition: "The magnitude of the Electrostatics force of interaction between two point charges is directly proportional to the scalar multiplication of the magnitudes of charges and inversely proportional to the square of the distances between them."

I haven't found a derivation of it that's really simple for it yet (didn't really care about derivations in units 1&2 lol). I've seen it derived from Maxwell's equation. http://planetmath.org/encyclopedia/DerivationOfCoulombsLawFromGaussLaw.html. I think there was also a method about something to do with electric fields.

$\int_S {E_n dA = } \frac{q}{{\varepsilon _0 }}$ (copied this from some equation database, don't know if it's correct)

Gauss's Law states (in fancy terms): "The electric flux through any closed surface is proportional to the enclosed electric charge."

I have also seen this even fancier version: ""the divergence of the electric field equals charge density divided by ε_0" That fancier one is just the equation spelled out in words.

Also this fancy definition kind of combines the two: "The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity."

So you can see that equations are really just a fancy way of describing something that sounds even fancier. However, this even fancier thing (the words), really just describes an idea that's fairly straight forward a lot of the time.

Flux is sort of a measure of 'flow' through a surface. What we've been doing in Unit 4 is magnetic flux. A magnetic field can be modeled using magnetic field lines. If you take an area (e.g. a rectangular piece of paper), magnetic flux is the measure of how many magnetic field lines pass through the area. So flux can be considered to be the "amount of field in an area".

So in less fancy words, Gauss's Law states that "The amount of electric field in an area" is proportional to the "electric charge in that area".  That's what I understood of it anyway.

Damn, I was supposed to be talking about Coulomb's Law. Flux is a concept that clicked for me when the teacher demonstrated it using a piece of paper. I assume you know what electric charge is. I guess it still shows my approach though, just start off with the fancy definitions you are given and then boil it down to the simpler terms. I have a feeling I didn't answer your question entirely.

Maybe this is kind of relevant (probably not, I just like xkcd): http://xkcd.com/895/. I like the mouse-over text: "Space-time is like some simple and familiar system which is both intuitively understandable and precisely analogous, and if I were Richard Feynman I'd be able to come up with it."
Title: Re: Newton's Laws and probably more questions
Post by: xZero on October 03, 2011, 01:54:02 pm
$\int_S {E_n dA = } \frac{q}{{\varepsilon _0 }}$
I'll continue from here, Coulomb's law is for a point charge yeh? so essentially we can think of it as a sphere, the integral above is the surface integral and if you follow the planetmath link, it converts to spherical co-ordinate and use triple integral and what it finds is the amount of electric field going through the surface of the sphere. After integrating it just rearrange it a bit and you'll get Coulomb's law, what does this imply? It implies that for a sphere with uniformly distributed electrons (a conductor for example) it acts as a point charge. (This stuff is way beyond VCE btw)
Title: Re: Newton's Laws and probably more questions
Post by: #1procrastinator on October 03, 2011, 02:10:21 pm
so for example, what is c^2 physically? i'm talking about the equations that can't really be derived intuitively, like finding distance is simply speed * time, it's obvious, but some of the other kinematic equations for example, they have to be derived mathematically and not using intuition

i'm just reading your reply right now, a bit loss, but that's a fine looking equation. it's crazy how they come up with those equations. have you heard of abstrusegoose/

that sucks for me cause it bothers me when i'm not getting the whole picture or learning from first principles

Title: Re: Newton's Laws and probably more questions
Post by: #1procrastinator on October 04, 2011, 04:19:43 pm
this is probably more a math question but i'll ask anyway. if f is proportional to q1q2 and also proportional to 1/r^2, why does multiplying them both give a common constant?

if you find the constants individually and combine them, i notice i always get f^2 (so far)
Title: Re: Newton's Laws and probably more questions
Post by: #1procrastinator on October 24, 2011, 08:43:17 pm
Is 10J/sqm/s the same as 10J/ms?  metres-second doesn't make much sense to me
Title: Re: Newton's Laws and probably more questions
Post by: Lasercookie on October 24, 2011, 09:18:18 pm
Is 10J/sqm/s the same as 10J/ms?  metres-second doesn't make much sense to me
square metres is m^2. m^2 and m is not the same as you know.

10 Joules per m^2/s (Energy is changing with acceleration?)
10 Joules per metres per second - (Energy is changing with velocity?)
Can't think of the relevant equation for those units.

Wait "ms" - milliseconds? Energy over time? That's power: $P = \frac{\Delta E}{\Delta t}$
Title: Re: Newton's Laws and probably more questions
Post by: #1procrastinator on October 25, 2011, 10:01:40 am
ah yes, sorry, i forgot the ^2. 10J per square metre per second is what i'm trying to get at

Title: Re: Newton's Laws and probably more questions
Post by: #1procrastinator on November 02, 2011, 06:18:20 pm
Why does a pool appear shallower if you look at the bottom from directly above? what does the speed of light have to do with this? If you're looking from directly above, wouldn't there only be a change in speed and no change in angle?
Title: Re: Newton's Laws and probably more questions
Post by: #1procrastinator on November 03, 2011, 11:56:09 am
also, when they (the heinemann book) says a mechanical wave involves the passing of a vibration through an elastic medium, does that just mean any physical medium? it doesn't explicitly define 'elastic medium'
Title: Re: Newton's Laws and probably more questions
Post by: Lasercookie on November 03, 2011, 04:41:07 pm
I'll have an attempt at trying to answer your questions. I don't think any of my answers here are fully satisfactory, but better than nothing hopefully :P

also, when they (the heinemann book) says a mechanical wave involves the passing of a vibration through an elastic medium, does that just mean any physical medium? it doesn't explicitly define 'elastic medium'
Elastic medium I think is one that can retain it's own form after a disturbance. I think that sort of makes sense with my understanding of a mechanical wave. The energy can be carried by the wave through the medium, but the medium is not permanently affected. A guitar string vibrating is a mechanical wave - it goes back to it's regular form. Sound waves - the air particles are not permanently changed.

"Net transfer of energy without net transfer of matter" if you want the fancy wording.

I don't ever recall coming across the term elastic medium in this context - but doing some googling, there are a few websites that use the terminology. They all enjoy neglecting the actual definition as well - it might be defined with 'elastic' being something to do with energy conservation or something. I think the definition I gave is more for springs and all that.

Why does a pool appear shallower if you look at the bottom from directly above? what does the speed of light have to do with this? If you're looking from directly above, wouldn't there only be a change in speed and no change in angle?
I have forgotten a lot about this :/ but it's do with refraction. The speed of light stuff comes into when you're talking about speed of light in different mediums (i.e. the refractive index?). I don't think that helped much, it's probably just restating what you already know.

I still don't quite know the answer to that question you had about units either. But http://en.wikipedia.org/wiki/Joule gives a few other definitions. Perhaps doing a bit of arranging with those definitions might get you to what you want to find out. Whether or not those two units you asked about earlier are equal - I don't really know.

edit: wrote out the definition from wikipedia out wrong, so I've cut it out. I made a typo and subsequently all the stuff I figured out didn't work. Typing 10J per square metre per second into google tells you that it's equivalent to kg/s though.

This relates to that other question you had, about derivations and such not making sense to you. My entire answer to that is unsatisfactory (and such I've decided to delete most of it) and that's usually the reason why I'll leave one of your questions unanswered. I can say that perhaps that the for the constant accel. equations, graphical derivations might make more sense to you.

I will also show you this post by paulsterio - which I thought that as well being a very nice explanation, is relevant. If you need to, reading his post before may help make more sense of what he is talking about Re: Vincezor's Unit 4 Physics Questions~ [this is one of the things we cover in Unit 4 btw].

He's showing where the formula $F=nILBsin \theta$ (force created with a current in a wire placed in a magnetic field) came from - and also the right hand slap rule. I think you may be familiar with the right hand slap rule. I remember covering it in Year 9 science. His post should show you another example of how a lot of intuition and logic actually is involved in getting to a conclusion that might seem pretty random. I'm not expecting you to understand the mathematics he's talking about (I don't really understand it that well either, I only currently know the general gist of vector cross product etc.) - it's irrelevant to the argument though - just trying to show you an example of how starting from something simple and then a few steps will get you to the conclusion (even if it does seem odd).
If you were interested though, the formula and the sin(theta) part come from the vector cross product. The length of the wire is a scalar quantity, so it doesn't matter, same as the number of wires, n, but the two vector quantities, the current, I, and the field, B, are multipled using a vector cross product (there are two ways of multiplying vectors, the cross product (which you'll do in Specialist) where you get a scalar, and the cross product (which you'll do in First Year Uni Maths), where you get a vector) - and the vector cross product is defined as the Product of the Magnitudes of the two vectors by the sin of the angle between them in the direction perpendicular to both of them.

So the Vector Cross Product explains the formula F = nIBL sin(theta)  as well as the Right-Hand slap rule - finding the direction of the Force (direction perpendicular (orthagonally) to the Current and Field)

That's the maths behind it :)
Title: Re: Newton's Laws and probably more questions
Post by: xZero on November 03, 2011, 04:50:46 pm
Why does a pool appear shallower if you look at the bottom from directly above? what does the speed of light have to do with this? If you're looking from directly above, wouldn't there only be a change in speed and no change in angle?
I think this diagram answers your question, the observed depth Y is less than the actual depth X but this doesnt happen when you're looking straight down the pool from above.
Title: Re: Newton's Laws and probably more questions
Post by: #1procrastinator on November 03, 2011, 04:57:29 pm
yep, thanks for putting up with these questions lol

ah that makes a bit more sense. so, an ocean wave - that would be an elastic medium? i think in the book it said none of the molecules that crash on the beach would have been at the source of the wave so there's no transfer of matter

---

About the units, that' was a question from the alternative energy topic (something like that). I don't remember it exactly but basically there's a solar panel and you had to work out the rate of energy absorbed per sqm…

So 10J per square metre per second is 10J/m^2/s and when you simplify it mathematically , you get 10J/m^2 s which reads 10J per square metre second which don't make sense. I guess basically what I wanted to know if it's still considered J per metre squared per second or does metre squared second mean the same thing?

@xZero - what about when you're looking from directly above? i can't find anything in my textbook about that but in the handouts, it says  a pool appears shallower when looking from directly above (no refraction)
Title: Re: Newton's Laws and probably more questions
Post by: Lasercookie on November 03, 2011, 05:09:26 pm
Quote
Yeah, I found that as well.

Quote
So 10J per square metre per second is 10J/m^2/s and when you simplify it mathematically , you get 10J/m^2 s
$m^2/s = m^2*s^{-1}$ not $m^2*s$

Quote
@xZero - what about when you're looking from directly above? i can't find anything in my textbook about that but in the handouts, it says  a pool appears shallower when looking from directly above (no refraction)
Snell's law was it? If you plug in the angle 0 for incident angle, what is the refracted angle? It's 0 is it not. (I hope I used the correct formula).

http://en.wikipedia.org/wiki/Snell's_law

I used the one off Wikipedia, if I remember correctly, angle 1 is incident, angle 2 is refracted. n2/n1 = 1.33 for air into water (rounded off air to 1). If I did mix up my angles, either way, they will both be zero.
Title: Re: Newton's Laws and probably more questions
Post by: #1procrastinator on November 04, 2011, 12:05:25 pm
So you treat the $m^2/s$ as a fraction and not $m^2$ and and $10J$ together? This is what I did:

$\frac{10J}{m^2} \times\frac{1}{s}$

Would everything together be

$10J m^-^2s$

?
Title: Re: Newton's Laws and probably more questions
Post by: Lasercookie on November 04, 2011, 04:39:12 pm
So you treat the $m^2/s$ as a fraction and not $m^2$ and and $10J$ together? This is what I did:

$\frac{10J}{m^2} \times\frac{1}{s}$

Would everything together be

$10J m^-^2s$

?
Oh yes, you're (almost) correct, I see my mistake.

I assumed: $\frac{J}{(\frac{m^2}{s})}$, but since it's Rate of "energy absorbed per sqm", it would be:
$\frac{(\frac{J}{m^2})}{s}$ which simplifies to: $\frac{J}{m^2 s}$ (this is what you had)
This further simplifies to:
${J m^{-2} s^{-1}}$ (Joules per sq. metre per second)

Note that $\frac{1}{s} = s^{-1}$ (I'm guessing you simply forgot the type up the -1 at the end in that post)

A bit of beating around the bush, but I think I can confirm that $\frac{\frac{J}{m^2}}{s} = \frac{J}{m^2 s}$
Title: Re: Newton's Laws and probably more questions
Post by: #1procrastinator on November 04, 2011, 05:38:19 pm
That was based on your last post :p

so does that read 'joules per metre squared per second', or 'joules per metre squared second' (what didn't make sense to me, unless it means the same thing...i want to make sure i got this lol)
Title: Re: Newton's Laws and probably more questions
Post by: xZero on November 04, 2011, 05:53:10 pm
they are different, joules per metre squared per second is $Js/m^2$ and joules per metre squared second is $J/m^2s$
Title: Re: Newton's Laws and probably more questions
Post by: Lasercookie on November 04, 2011, 05:57:00 pm
they are different, joules per metre squared per second is $Js/m^2$ and joules per metre squared second is $J/m^2s$
Hmm, well that makes a lot more sense.

I am not entirely sure what I've been doing wrong though (somewhere in my mathematics probably, I'm guessing probably with my initial assumptions).
Title: Re: Newton's Laws and probably more questions
Post by: #1procrastinator on November 04, 2011, 07:10:52 pm
they are different, joules per metre squared per second is $Js/m^2$ and joules per metre squared second is $J/m^2s$

So $\frac{J}{1}$ over $\frac{m^2}{s}$...how did you know?
Title: Re: Newton's Laws and probably more questions
Post by: #1procrastinator on November 10, 2011, 05:32:57 pm
How come it isn't J/m^2 and s/1?

And I just noticed you mention kg/s^3 laseredd, that makes no sense to me haha (i think i know how the calculator got that though)

If a jet fires a missile, it pushes back against it right (if not released first), but what if the missile was fired at a low acceleration so it gradually increased it's velocity to match the aircraft's before being released, would this act as a boost for the jet? There would be no push back, yes?

EDIT: How would one go abut deriving the work formula?
Title: Re: Newton's Laws and probably more questions
Post by: xZero on November 10, 2011, 05:43:21 pm
they are different, joules per metre squared per second is $Js/m^2$ and joules per metre squared second is $J/m^2s$

So $\frac{J}{1}$ over $\frac{m^2}{s}$...how did you know?
replace per with /
joules per metre squared per second = joules/m^2/s = Js/m^2

joules per metre squared second = j/m^2s

If a jet fires a missile, it pushes back against it right (if not released first), but what if the missile was fired at a low acceleration so it gradually increased it's velocity to match the aircraft's before being released, would this act as a boost for the jet? There would be no push back, yes?
jet usually release missile before it accelerates, letting it accelerate before release is just wasting fuel, prob run out of juice before you can fire it
Title: Re: Newton's Laws and probably more questions
Post by: #1procrastinator on November 10, 2011, 05:48:00 pm
joules/m^2/s , i don't get why m^2 and second are together in a fraction and not joules and m^2

hypothetically :p

it's the acceleration that pushes it back though, right? if it was increasing velocity slowly, there would be no (or very little) push back?
Title: Re: Newton's Laws and probably more questions
Post by: xZero on November 10, 2011, 05:59:30 pm
idk, joules per metre squared per second is just a badly phrased sentence, its like the divide sign argument on facebook. I interprete it as joules/m^2/s=Js/m^2