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#### #1procrastinator

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##### Newton's Laws and probably more questions
« on: September 09, 2011, 03:44:37 pm »
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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?

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##### Re: Newton's Laws and probably more questions
« Reply #1 on: September 09, 2011, 05:33:01 pm »
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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).

#### lukew

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##### Re: Newton's Laws and probably more questions
« Reply #2 on: September 10, 2011, 09:18:22 am »
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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

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##### Re: Newton's Laws and probably more questions
« Reply #3 on: September 10, 2011, 10:12:16 am »
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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.

#### xZero

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##### Re: Newton's Laws and probably more questions
« Reply #4 on: September 10, 2011, 10:21:49 am »
+1
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
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##### Re: Newton's Laws and probably more questions
« Reply #5 on: September 10, 2011, 10:25:13 am »
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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

#### lukew

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##### Re: Newton's Laws and probably more questions
« Reply #6 on: September 11, 2011, 01:29:09 pm »
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Thankyou all, good to know I was on the right track.

#### #1procrastinator

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##### Re: Newton's Laws and probably more questions
« Reply #7 on: September 13, 2011, 05:20:10 pm »
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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?

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##### Re: Newton's Laws and probably more questions
« Reply #8 on: September 13, 2011, 06:54:06 pm »
+1
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).

#### #1procrastinator

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##### Re: Newton's Laws and probably more questions
« Reply #9 on: September 16, 2011, 12:46:10 pm »
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^ 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

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##### Re: Newton's Laws and probably more questions
« Reply #10 on: September 16, 2011, 04:38:39 pm »
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^ 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.
« Last Edit: September 16, 2011, 05:35:44 pm by laseredd »

#### xZero

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##### Re: Newton's Laws and probably more questions
« Reply #11 on: September 16, 2011, 05:01:04 pm »
+2
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.
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##### Re: Newton's Laws and probably more questions
« Reply #12 on: September 16, 2011, 05:33:16 pm »
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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.

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##### Re: Newton's Laws and probably more questions
« Reply #13 on: September 17, 2011, 02:11:34 pm »
+2
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?)

$F \Delta x=\Delta E$

#### #1procrastinator

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##### Re: Newton's Laws and probably more questions
« Reply #14 on: September 25, 2011, 12:04:03 pm »
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@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