For those who think they understand Newton's Second Law

steveb

Joined Jul 3, 2008
2,436
Meanwhile you might like to look up centrifugal force, centripetal force/acceleration and D'Alembert on Wikipedia. They certainly support my view there.
I'm still not sure how your view is different from anyone elses. If you are saying that care must be taken when applying Newton's Laws, who is disagreeing with you?

This D'Alembert method seems to convert a dynamic problem into a static problem. OK, that's fine - you need to use a non-inertial reference frame and put the "ficticious forces" created by the accelerating reference frame back in. If you use an inertial frame, then you must put the real forces in.

This is all standard Newton's Law principles used all the time. Whether or not we know the name "D'Alembert" doesn't really matter. If we solve Newton's Law's on the surface of the earth, then we put in the "ficticious" Coriollis force because our reference frame is accelerating as the earth spins. Nothing new here as far as I can see.

As far as Centrifugal force non being real, I can quote a reference too.

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

This talks about the ficticious centripetal force in a rotating reference frame and the real reactive centripetal force in an inertial frame. Centrifugal force is just the reaction against centripetal force. If you want to acknowledge one, you need to acknowledge the other.
 

Thread Starter

studiot

Joined Nov 9, 2007
4,998
I'm still not sure how your view is different from anyone elses.
I don't disagree with professors of mechanical engineering.

The student is particularly warned not to mix the methods, as this is a frequent source of error.
Did you read the posted extract?

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

Well this is the same link I posted back in post44 in the originating thread.

I offered D'Alembert's Principle as it is easier to cope with than the modern interpretation of inertial 'forces' which disappear in any reference frame not accelerating relative to the body in question.

The fact remains that there has to be a resultant for an acceleration to exist. Therefore there is no 'balancing force'. You cannot have both.
 

steveb

Joined Jul 3, 2008
2,436
The fact remains that there has to be a resultant for an acceleration to exist. Therefore there is no 'balancing force'. You cannot have both.

I guess I need to read more carefully. I didn't realize that fact was in contention. Of course I agree with this.
 

Thread Starter

studiot

Joined Nov 9, 2007
4,998
And that resultant is Tsinθ, the horizontal component of the tension in the string.

Which means that there can be no outward force (on the ball)

Which was the conclusion I have been trying to steer everyone towards

And the reason I originally wrote

Tsinθ = ?

I think fig 4.6 on page 156 of my zipped file sums it all up beautifully.

If you are saying that care must be taken when applying Newton's Laws, who is disagreeing with you?
When I said this, did you follow the example about the rocket propulsion? Herein lies another common misconception.

Bill Marsden was quite right that Heinlein referred to this conundrum in several of his books.
 

steveb

Joined Jul 3, 2008
2,436
Which means that there can be no outward force (on the ball)
Yes, I agree with that. The ball wants to keep moving outward due to it's inertia, but there is no outward force on the ball. However, there is an outward force on the string which is the centrifugal force exerted by the ball as a reaction force to the centripetal force the string is exerting on the ball. (OK now my head is spinning as fast as the ball)

When I said this, did you follow the example about the rocket propulsion?
Yes! Of course I did.
 

steveb

Joined Jul 3, 2008
2,436
In the real world the centre of mass of the system does not accelerate when a rocket flys.
Since you brought this up again, I thought I would make a minor correction to this statement. This is essentially correct. However, when a rocket takes off from earth, the ground and the air interact with the rocket propellent and the net center of mass (of the rocket and expelled propellent) does accellerate upward. It is not until the rocket leaves the atmosphere that you can really say this statement is valid.

This is analogous to a person jumping off the ground, which does not require rocket fuel. A similar analogy is a bird flapping it's wings to lift vertically with no wind (not gliding mind you! that's a different principle of course). It also relates to the reason why a hovercraft might get off the ground, but may not be able to rise more than few inches off the ground. Solids, liquids and gasses are all materials on which a reaction force can be obtained.

It's just a minor point and does not take away from your good example. However, it is a minor point that directly relates to the ideas in this thread.

EDIT: Although, if you expand your definition of the word "system" to include the rocket, fuel, atmosphere and the earth itself, then the statement is valid.
 
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Thread Starter

studiot

Joined Nov 9, 2007
4,998
Yes correction duly accepted.

I think the point was really that no object, rockets included, can propel themselves through space, by themselves. This is an extension on the comment that no object can exert a force on itself.

This uncomforable fact is often overlooked by Vice Presidents and Sci Fi writers.

Rockets, of course, work by shedding part of their mass.

Aircraft interact with surrounding matter to generate lift and thrust.
 

DrNick

Joined Dec 13, 2006
110
If you draw a free body diagram at a single instance in time when the ball is at a specific theta, and a specific angular velocity you find:

There is a tension force acting on the body vertically and horizontally (facing in from the rotational path), the force of gravity acting down, and a force due to the centripetal acceleration (facing in from the rotational path) caused by the angular motion of the body.

So if you look at the free body diagram there are two forces summing in-words. This tells us that the body CAN NOT be at equilibrium. Even in a perfect world with no friction, the balls orbit will settle to some strange elliptical orbit...i think...

This is quite an interesting problem..time to pull out the nonlinear phenomina book hehe.

Does this reasoning seem to make sense?
 

Thread Starter

studiot

Joined Nov 9, 2007
4,998
There is a tension force acting on the body vertically and horizontally (facing in from the rotational path), the force of gravity acting down, and a force due to the centripetal acceleration (facing in from the rotational path) caused by the angular motion of the body.
Not quite, I have emboldened the incorrect bit.

There is no force due to the centripetal acceleration. It is the other way round. The ball requires application of force by the string to deviate from straight line motion. Because this force is always directed towards a centreline it is called 'centripetal'.

But the centripetal force is a real force provided by the horizontal tension in the string, it is not additional to this tension.

Centripetal acceleration is a consequence of this force.

Nor is there any need to invoke elliptical motion. The device is called a conical pendulum. Such a device was the first mechanism to demonstrate and measure the rotation of the Earth. As such it is known as Foucault's pendulum.

You are correct in your deduction that the ball in not in static equilibrium however.
 

Nanophotonics

Joined Apr 2, 2009
383
So, my idea of a force causes an acceleration and not the other way round, is it still "non-sense"? (Actually, I didn't refer to F=ma when I said that in the other thread) - I found a nice article in the last issue (No. 9 May/June09) of the IET magazine UK on "Timing - The secret of good engineering". Hope you guys get to read it.
 

Thread Starter

studiot

Joined Nov 9, 2007
4,998
If you really must, apply it to both the string and the ball
In fact if we consider the string some interesting things happen.

Now taking the string as the usual 'light string' ie of negligable mass we find that

F=zero x acceleration = zero

This can only happen if the string experiences neither an inward or outward force/acceleration.

Reaction centrifugal force comes from the same principle: "Every force has an equal and opposite reaction force". If the string is applying an inward force to the ball, then the ball is applying an equal and opposite (i.e. outward) force to the string.
A free body diagram bears this out. A string cannot support transverse forces. So the only forces acting are along the line of the string. These are the tension in the string and the equal and opposite reaction of the ball on the string, by Newton's third Law, which correctly sum to zero.

So this apparantly trivial problem contains hidden depths.
 

Thread Starter

studiot

Joined Nov 9, 2007
4,998
a force causes an acceleration and not the other way round, is it still "non-sense"?
Somewhere I did say I thought you were actually closer than most.

However the questions arise can there be a force without acceleration & can there be acceleration without force?

If either can be shown this surely casts doubt on causality.

Further we should examine the notions that have been expressed about writing

F=ma in other forms

If we write m = F/a, and are given F and a, we can calculate m. Or can we?

what if a=0?

We should always beware of the possibility of division by zero.

It is also instructive to note that Newton himself did not express his law in the form of an equation.
In his time they did not work with equations as we know them. Most maths was presented in terms of proportions - that is how he developed calculus. This was done to avoid division by zero and to avoid limits actually 'tending to zero' as we now say.

So (with modernised spelling) Newton wrote:

1) Every body continues in its state of rest, or uniform motion in a right line, unless it is compelled to change its state by forces impressed upon it.

2) The change of motion is proportional to the motive force impressed, and is made in the direction of the right line in which the force is impressed.

3) To every action there is always opposed an equal reaction; or the mutual actions of two bodies upon each other are always equal and directed to contrary parts.

So clearly Newton thought that the acceleration was a result of the force.
 
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Nanophotonics

Joined Apr 2, 2009
383
1) Every body continues in its state of rest, or uniform motion in a right line, unless it is compelled to change its state by forces impressed upon it.

2) The change of motion is proportional to the motive force impressed, and is made in the direction of the right line in which the force is impressed.

3) To every action there is always opposed an equal reaction; or the mutual actions of two bodies upon each other are alwys equal and directed to contrary parts.

So clearly Newton thought that the acceleration was a result of the force.
Bingo! (I've noticed that many engineers tend to forget their Physics - I'll try not to :p)
 

regrehan

Joined May 27, 2009
3
I am in serious danger of making a complete fool of myself, here, but what the heck ...

Earlier, "studiot" wrote:
"To exert a force on another object you require an external agent -- either another object, such as a string, or a field such as gravity. There is no other external agent available in this case."

But there is. Or, rather, was. An initial force set the ball swinging in the circular motion in the first place. It's not like the effects of that initial force just evaporated from the system because the finger that pushed the ball isn't still pushing it. Looked at from an energy perspective, the energy imparted into the system from the initial push was converted into the kinetic energy of the ball's motion plus the potential energy of the ball's plane of rotation "rising up" a bit from hanging vertically.

In any case, that initial force accelerated the ball, and thereby imparted momentum to the ball. It is this momentum that keeps the ball moving in a circular orbit ... rather than just giving up and hanging motionless.

Focus on just the horizontal plane: The component of tension -- your Tsin(theta) -- applies an inward force to the ball. The ball wants to travel in a straight line, on account of its momentum. The Tsin(theta) pulls it inward ... the ball reacts with an equal and opposite force as it resists this change to its momentum.

It seems to me that this illustration could be simplified if one ignores the vertical and concentrates only on the horizontal. In that case, the arrangement is analogous to a ball sitting on frictionless ice, attached by a string to some center post. The ball is positioned at the full extent of the string. A force is applied at a right-angle to the string, and the ball glides in an eternal circle around the center. If we could attach a massless tension-meter to the string, we would see that something is trying to accelerate the string away from the center post -- something is pulling on the string with a measurable force. Is it any wonder that we would conclude -- erroneously, maybe -- that there is an outward force on the string ... and hence the ball? Similarly, we know that the ball's momentum compels it to travel in a straight line, yet we see it describing a circle. Would we not infer that there is a force pulling the ball toward the post?

Then, we have a massless alien climb aboard the ball as it swings around, and we ask him to direct his attention to the point at which the string is attached to the ball. We ask him to measure the distance from that point to the center post, as well as the distance from that point to the center of mass of the ball. We request he do it many times. In all cases, we find that those distances do not change. Would we be amiss if we thought that whatever force pulls the string outward is balanced by whatever force pulls the ball inward?

I'll do this whole discussion one better, because (and forgive me, "studiot", if I misread you) it appears that "studiot" is saying that this outward force is a mathematical sleight-of-hand.

Imagine again the ball, swinging in a circle, this time out in space. It is attached to a magical center point that -- in respect to a sphere of several light years -- does not move. In addition, the ball is hollow and opaque. We place our alien in the ball, with a couple of marbles. We don't tell him that the ball is swinging in a circle.

You ask him: "Mr. Alien. Can you detect any forces at work in this ball?"
He puts down a marble, and notices that it rolls to a point on the inside of the ball opposite where the ball is attached to the string (though he doesn't know the attachment location). He tries the second marble, and the same thing happens.
He points to the marbles and says: "There appears to be a force pulling all objects with mass in that direction."
"Yes, well," you say, "that is a mathematical fiction."
The alien hits you...

--regrehan
 

Mark44

Joined Nov 26, 2007
628
After the hand applied the force to the ball, and thereby accelerated it in some way, that force is done. That initial force accelerated or decelerated the ball to a certain velocity. If there were no string, and the ball happened to be in a place with no gravity, it would continue in a straight line at whatever velocity it was accelerated to.

The ball on the string is no different in that respect. After it is set in motion, the centripetal force applied by the string applies a force that changes the direction of motion, and therefore accelerates the ball (acceleration is, after all, the time rate of change of velocity).

For the thought experiment of the alien in the hollow ball -- the alien perceives a force acting outward from the center of rotation. This is a pseudoforce. The actual force is due to the hollow ball travelling along a curved path and exerting a force toward the center of rotation on the alien and the marbles he (she? it?) is playing with.
 

Thread Starter

studiot

Joined Nov 9, 2007
4,998
the ball reacts with an equal and opposite force as it resists this change to its momentum.
Any reaction by the ball cannot be applied to the ball itself, only be applied to another object - in this case the string. I have already stated this, but I will do so again.

The reaction by the ball on the string is equal and opposite to the inward pull of the string on the ball, due to the tension in the string. This reaction is applied to the string, not the ball.

Your ice example is essentially the same as example 1 in my post #39.

I note that although there have been 716 views of this thread only 4 have bothered to view this example.

This is a great shame as I hoped it would dispel a good deal of misconception.

In particular post contains free body diagrams
 

regrehan

Joined May 27, 2009
3
"studiot":

1. My apologies for not reading the .zip file posts in #39 first. When I first perused this thread, I had not yet registered, and was not allowed to download the .zip.

2. When you say "Any reaction by the ball cannot be applied to the ball itself" ... what do you mean by that? Do you mean that the reactive force (if I can call it that) of the ball cannot be measured anywhere within the substance of the ball? That it can only be measured on the string? Or something else?

--
 

Thread Starter

studiot

Joined Nov 9, 2007
4,998
The string acts on the ball - it pulls.

The ball reacts on the string - it pulls back.

When you stand on the floor your body pushes down on the floor

The floor pushes back.

A free body diagram of yourself shows gravity acting downwards (=your mass x g) and the push from the floor acting upwards. Nothing else. Your push against the floor is not included.

A free body diagram of the floor shows your push acting downwards and the push from the foundations acting upwards. Again nothing else. the push form the floor to you and to the foundations are not included.

Action and reaction are equal and opposite - Newtons third law - also in the postings in this thread.

When you consider a system you consider a 'free body diagram' which means that you isolate an object from the rest of the universe and only consider forces acting across the boundary which are impressed upon that body by the rest of the universe. You do not include forces impressed by the body on the rest of the universe.

Most people seem happy to get this right in the vertical direction but some have trouble with the horizontal in this example.
 

BillO

Joined Nov 24, 2008
999
Imagine again the ball, swinging in a circle, this time out in space. It is attached to a magical center point that -- in respect to a sphere of several light years -- does not move. In addition, the ball is hollow and opaque. We place our alien in the ball, with a couple of marbles. We don't tell him that the ball is swinging in a circle.

You ask him: "Mr. Alien. Can you detect any forces at work in this ball?"
He puts down a marble, and notices that it rolls to a point on the inside of the ball opposite where the ball is attached to the string (though he doesn't know the attachment location). He tries the second marble, and the same thing happens.
He points to the marbles and says: "There appears to be a force pulling all objects with mass in that direction."
"Yes, well," you say, "that is a mathematical fiction."
The alien hits you...

--regrehan
Actually, the marbles will follow a curved path (curving in a direction opposite to the direction of rotation) to the inner wall of the ball, then once there suddenly move back along the direction of rotation toward a point (p) opposite where the ball is tied to the string. If this environment has friction it will eventually come to rest at p, otherwise it will oscillate around p.

If the alien has any knowledge of the physics of the motion of mass, he will quickly realize the situation and claim the marbles were in free flight until acted upon by a force exerted by the wall of the ball.

Edit: BTW, if the alien is massless, his hit would be rather ineffectual.
 

ELECTRONERD

Joined May 26, 2009
1,147
Might I also imply that mass and inertia are related. Here is a simple experiment:

1. Get a cup, any one will do.
2. Also get a quarter and index card.
3. Place the index card on top of the cup.
4. Then place the quarter on top of the index card.
5. Get your finger ready, and flick the index card.

You should see the quarter fall inside of the cup. There was some friction but the quarter had too much inertia for it to stay with the index card.

Have fun! :p
 
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