Going back to the English definition from your textbook (rather than a mathematical one), I want to stress that "effectiveness" is not an all or nothing affair. If I hit something such that the vector of my force does not go through the center of mass of the object, some of the force will go to torque, some will go to acceleration. That is what is meant by "effectiveness" in this case. How much goes to torque? The answer is R cross F.

One can set up a systems so that all the force goes to torque (turning a nut, for example) or that none does (all goes to acceleration). Or anything in between.

At the risk of redundancy, you are not limited to torque=all or torque=none.


_________________
Jeff Schwarz
__________________________________________________
I have Invader's blood marching through my veins
like giant radioactive rubber pants! The pants
command me! Do not ignore my veins!
--Invader ZIM

<font size=-1>[ This Message was edited by: Geo3gh on 2002-07-17 16:40 ]</font>

No. R cross F gives us the torque vector for a force (F) acting on a point that is some radius (R) away from the center of mass.

The point of origin of that vector (tau) is indeed the center of mass of the object being torqued, but that is simply how vector cross products work.

This is true by definition.

__________________
Jeff Schwarz
__________________________________________________
Argh!! They booby-trapped their sun!!****--Invader ZIM

I'm sending a bitmap, "2force_torque.bmp" to Gary, so he can post it on his website (as he has kindly offered).

What we have here is a rod, 2.0 meters in length, of uniform density, sitting on a frictionless surface. We have two lines of negligible mass attached to each end, each pulling with a force towards the right side of the page.

Each line is attached 1.0 meter from the center of mass. Line 1 is pulling with F₁ = 1.0 N. Line 2 is pulling with F₂ = 1.1 N. Each is pulling at 90 degrees to the orientation of the rod.

So, two forces should give us to torques.

tau₁ = R₁ cross F₁

tau₁ = 1.0 m * 1.0 N * (sin 90)

tau₁ = 1.0 Newton-meter

tau₁'s direction is into the screen, or negative.


tau₂ = R₂ cross F₂

tau₂ = 1.0 m * 1.1 N * (sin 90)

tau₂ = 1.1 Newton-meter

tau₂'s direction is out of the screen, or positive.


Since this is one object, we need to add the torques to see how much total torque there is in this system.

tau_T = tau₁ + tau₂

tau_T = (-1.0 Newton-meter) + 1.1 Newton-meter

tau_T = +0.1 Newton-meter


Therefore, the rod will rotate counterclockwise.

The force that did not go into torque goes into accelerating the rod to the right, along vector A. (I think at A = 2.0 N / m).


Now, I could just as easily have only one line. If I just have Line 1, then the rod is torqued -1.0 Newton-meters, and rotates clockwise, and does not linearly accelerate.

__________________
Jeff Schwarz
__________________________________________________
Argh!! They booby-trapped their sun!!****--Invader ZIM

Geo3gh

Quote:

--------------------------------------------------------

I'm sending a bitmap, "2force_torque.bmp" to Gary, so he

can post it on his website (as he has kindly offered).

--------------------------------------------------------

I changed the 164 k bmp to a 2.9 k .gif



2force_torque .gif





Gary

Many thanks.

I didn't have a utility for that--I just whipped it up in Paint.

__________________
Jeff Schwarz
__________________________________________________
Argh!! They booby-trapped their sun!!****--Invader ZIM

Right. At least until the rod is horizontal, and then it will begin decelerating and head back in the clockwise direction, after which it will oscillate around the horizontal position. All this while, it will accelerate to the right.

A = 2.1/m N / kg (where m in this case is mass, not the unit "meters"). The net force on the rod divided by its mass gives the net acceleration, regardless of torque.

I'm not sure it's useful to talk about the force going to do this or that. Energy sloshes around into different motions, force does not.

No, it will still accelerate linearly to the right, at A=1.0/m N/kg, as long as the constant force is applied. The motion will be somewhat complicated (these frictionless situations are often more counter-intuitive than you'd think), because as the thing spins, the angle of the string to the rod will change, thus changing the angle at which the force is applied to the rod and introducing vertical accelerations. This is why a stick at the end of a rope dangles so crazily. In that case, you're holding the rope steady, and gravity is pulling the rod in the opposite direction, producing an analogous effect. I say analogous because in that case, your pull is balancing gravity, so there is no net force to produce linear acceleration. Your example more like if the stick were on the ground and you pulled up on the string while the ground vanished. Then you'd get your upward linear acceleration as long as you were pulling with more than 9.8m N of force. The torque would cause the dangling, oscillatory motion, while the overall upward acceleration would be at a=(F/m-9.8) N/kg.

Hope that helps,

Don

Oh, just occured to me. I was assuming that the force on the lines would remain pulling to the right. If you meant that the force would move so that it always stayed 90 degrees to the rod, then it would be more complicated. There would still be linear acceleration, but it would be in a direction that was always changing, as well. So I think the rod would spin around its center of mass, and it would move in a ever-increasing spiral, too.

<font size=-1>[ This Message was edited by: DoctorDon on 2002-07-18 01:00 ]</font>

<font size=-1>[ This Message was edited by: DoctorDon on 2002-07-18 01:03 ]</font>

What do you all have me on the pay no mind list? It's obvious your not going to accomplish anything in this post after four months. This guy just doesn't get it and he never will. Anyway I'm sick of reading that darn thread title so please wrap this up and make your conclusions. I bid you good day.
Mongo

Well, I've done it. Here is the step-by-step, no-holds-barred, nothing-glossed-over PROOF that the moon torques the Earth. No calls to authority, no slight of hand, no quoting of textbooks. Just the math, laid out for you to follow it yourself.

If that doesn't satisfy you, Gary, I don't know what else to do.

Unless there are specific questions about or criticisms of my proof (if someone can find a flaw, please let me know!), I am going to try very hard to stay out of this thread from now on. For all its longevity, it doesn't seem to be getting anywhere.

Yours,

Don

I just went back to my physics book, and see that I misremembered that part. Thanks.

This made me think of a possible point to Gary. The reason one would need two forces on an object is when you want torque but no linear acceleration.

For instance, going back to my drawing, let's set Line 2 to be directed to the left at 90 degrees to the rod. Then both Line 1 and Line 2 give an instantaneous, 0.5 newton "tug". The two forces cancel each other out, as far as linear acceleration is concerned, but it gets a 1.0 N-m torque.

There are instances where having two forces is useful. The nut is going to get torqued either way, but it is counter-productive to have you nuts and bolts skittering off to the right when you try to fasten something down.

Just remember, all the definition of torque says is that torque changes an object's angular momentum. It says nothing about its linear momentum.

__________________
Jeff Schwarz
__________________________________________________
Argh!! They booby-trapped their sun!!****--Invader ZIM

I apologize profusely I did say the books lie. That was never my intent. The books are wrong due to ignorance. That was truly a moment of weakness and perhaps frustration on my part.
Sorry.

Geo3gh,
Jeff I think I understand what you tried to do in your drawing.
I'm afraid your idea was wrong.
You failed to give your rod a mass, when you give it mass it tries to stay put. Inertia is not considered a force, but it must be dealt with nevertheless.
I show a way to deal with a portion of it, in this 3force.gif drawing.



If you suspend your rod from the two lines then gravity can serve as an inertia substitute.
If you accelerate one end of the rod faster than the other rotation starts to take place. Along with that rotation, the center of mass is forced to move as well. The motion of the center of mass is along a diagonal indicated by the blue arrow in the drawing. If at the instant of time shown by the right hand drawing we cut loose the lines and turn off gravity, the rod would spin off into space at the angle shown with the speed and angular momentum it has it that instant.



There is more to it than that though.



If you only accelerate one end as you do at the end of your description, the other end will move backward just as the meter stick does when you whack it. The initial direction of the mass center will be diagonally downward and to the right, toward the stronger or only force. As I keep saying put your meter or yard stick on the floor and give it a whack, if you like you can tie a string to it and pull, but the result is the same.



Gravity is a singular field of varying magnitude, not 2 lines of force. Any torque that can be imagined is internal and is in equal and opposite pairs. A body will not start or stop rotating in a gravitational field unless a second force either suspends, supports, or impacts on it. If you turn a bicycle upside down, then balance the front wheel, it will not start turning just because the Earth has gravity. If you give that wheel a spin it does not stop just because the Earth has gravity. Gravity does not torque by itself.


If you can perfect that perpetual motion thing, let me in on it.
If the Earth was supported or somehow suspended from its center of momentum/mass or from its self induced center of gravity, and then if it was acted upon by the Moon's gravity, then yes the "R cross F" formula would locate the point at which the F=GMm/r^2 force would be applied. Then an effective amount of torque could be calculated.

But, that torque would only be effective until the center of gravity was moved to a point under the point of support.


Gary

I purposely did not specify the mass; I kept it as "m", as you can see when I talk about the linear acceleration. The rod has a mass, I just didn't give it a number. Just like I didn't give its moment of inertia (I) a number. I don't care what the rate of rotation or the acceleration is, just that it exists and is non-zero. Thus, I didn't specify I or m.

__________________
Jeff Schwarz
__________________________________________________
Argh!! They booby-trapped their sun!!****--Invader ZIM

That would be different than what I said in the decription of the drawing. In the drawing, the rod is on a frictionless surface. My floor is not frictionless. They are not equivalent, and so what I see on my friction-ful floor will mislead me as to what I should see on a frictionless floor.

You could approximate the frictionless suface with a lightweight ruler on a air-hockey table. But I don't have one. Anyone have an airhockey table?

__________________
Jeff Schwarz
__________________________________________________
Argh!! They booby-trapped their sun!!****--Invader ZIM

But . . . if your bicycle wheel is oval-shaped instead of perfectly circular, the Earth's gravity will eventually stop it, with the long axis aligned with the gravitational pull, won't it?

So if the Earth and Moon themselves are not perfect spheres that means . . . they can torque each other, can't they?


_________________
SeanF


<font size=-1>[ This Message was edited by: SeanF on 2002-07-18 14:56 ]</font>

No, not to my knowledge. The effect of gravity when the bottom nose of the oval is just before bottom-center, is offset symmetrically by the effect of gravity when it is just past bottom-center.

It's not the fact that it is rotating, nor is it because it is oval--it's because the oval bulge of the Earth is moving and encountering friction.

Gary

Do you now understand that the gravity would act at the center of gravity, but the object would rotate (or "be torqued") about the center of mass?

DoctorDon
Quote:**
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If that doesn't satisfy you, Gary, I don't know what else to do.
---------------------------------------------------------------
Thank you so very much for your effort. I know how much time goes into something like that. As you well know I was not able to follow every detail, but I believe I got most of it.
The pictures and description seem nearly identical to the explanation Jeff gave for the "R cross F" earlier. I understand that if the Earth were a sphere the Moon's gravitational pull would fall along the line between the centers of mass so there would be no torque.

In your third page drawing of the Earth and Moon you show "r" and "R" meeting at a point that I would interpret as the Earth's self induced center of gravity. That is what I got from Jeff's explanation, and yours as well.
You know it's not really the center of mass, but the sum of the gravitational forces of the Earth's mass equals zero at that point.
Then you show "r" and "a" meeting in the box "dm" that I understand to be a little chunk of mass (tidal bulge).

It appears to me, that the "R cross F" calculations locate the point at which the Moon's gravitational pull (effectively) focuses itself, that point being somewhere between the dm box and the "r R" point. I call that point, the center of gravity as caused by the Moon.

Is that correct?

We can calculate the force the Moon "effectively" places on that point using the formula F=GMm/r^2.

So we know where the Moon pulls on the Earth, we know the angle from which it pulls, and we can calculate the force with which it pulls.

Is that correct?

Don,
I'm sure that you have heard me say this before, where is the point of support or suspension?
For torque to be effective it needs to work around a point of support, suspension, or an axis.
Here in my physics book it says: "The magnitude of the torque about an axis is defined as the product of the distance from the axis and the component of the force which is applied perpendicular to the point at that distance. The component of the force perpendicular to the line is given by F1 = F sin theta."

To calculate the actual torque of the Moon on the Earth (in foot pounds or whatever) we need to know how far it is from the point where the force is applied to the point of support, suspension, or axis, and in which direction or at what angle that point lies from the point of application. With the sphere the angle was zero so the force is along a line that cannot cause a turning torque.

Is that correct?

For instance, to torque a bolt you use a torque wrench. Those tools have a length between the handle and the drive end. They also are calibrated with a spring or torsion bar to read in inch pounds, foot pounds, gram centimeters, or kilogram meters.
The distance between the input and output is fixed in the tool. The one thing the mechanic must do is apply the force at 90 or 270 degrees to ensure the proper tightness with the least wasted force. If the force were applied at 0 or 180 degrees then no torque would take place the "effectiveness" would be zero just as in the sphere.

So we need to know how far it is from the applied force to the point of suspension and the angle between the force and the resistance so that we can calculate the effectiveness and actual amount of torque.

We need to know where this point is!

Don, please, you must tell us where is the Earth suspended or supported from, where is the axis?!?!?!


Gary

It's the center of mass.

Hi,

I feel like we're getting close here. In reading this message, I had an idea of how I might be able to get at the underlying misunderstanding. If you don't want to get bogged down in nitpicky details, skip right to the part where you quote from your physics book. That's the important part. Let's see if I can pull this off at this late hour...
I apologize if my brain is overtired.

Great! So far, so good.

No, it's the center of mass.

Yes, it is.

It's still not clear to me why you find this a useful concept.

Right, although it's just a coincidence that I drew it as being in the bulging part of the earth. When I take the integral, I am adding up the contributions from all little chunks of mass in the entire Earth.

Maybe. Again, I don't find this "center of gravity" you've defined to be a useful concept.

Wait, I think I see what you are trying to do. You are trying to pick two points that will let you treat this problem as a two-body (i.e. two point masses) problem, right? It can't be done. The whole problem only exists because the bodies in question are not point masses. You're trying to say there is some point for the Earth and some point for the moon which you can pick, and then treat the system as if it were the gravitational interaction between those two points, right?

Have I understood you correctly, here?

That can't be done. Not in this kind of asymmetrical situation. With the right kind of symmetries, you could do that, but not in this problem. In any case, even if the shortcut worked, if it's correct, it should give you the same answer as doing it all out the long way, which is what I did.

No. You can't define a point "where the Moon pulls on the Earth" when you're trying to calculate torque. You need to do what I did: consider the gravitational interaction between the moon and each individual chunk of mass in the Earth, and then add them up. If you try to take a short cut, you will get lost, because the necessary symmetries do not exist in this problem.

Yes, I've heard you say that before, and I keep trying to tell you there isn't one.

No support necessary, no suspension necessary, and the axis is defined by the forces involved.

Okay, I think here is where the misunderstanding is originating. What you've quoted there is a simplification! It's an application of a general formula to a specific situation. In the case you cite here, the axis is defined in advance, and the question the problem is about how the force interacts with that axis. This simplification was clearly done so you wouldn't have to worry about all three components of the vector. If you know in advance that there are external constraints that will force the system to rotate around a particular axis (e.g. the point of suspension of which you speak), then you can disregard all the other components and just look at the one that will produce a torque along that axis. This is as if I had said, in my three-page file, "I know that the earth rotates around the z axis (as I defined the coordinates), so I don't need to worry about the x or y components of the torque." I didn't want to make that assumption, I wanted to show you that the x and y components went to zero. Another example: if I am torquing a bolt with a wrench, and I press down as well as to the side, all the force with which I push down is irrelevent to the rotation of the bolt, because the hole I've put the bolt in is preventing it from turning around any other axis than the one down its length. Only the crosswise force will contribute in that direction, and that's where the sin(theta) comes from in the formula you quote. The magnitude of a cross product is proportional to sin(theta)! This is r cross F!!!! Just one part of it, rather than the general, three-dimensional case.

If there's no axis defined in advance, you have to consider all three components of the force, and figure out from the context of the situation which axis the thing will end up rotating around. To go back to the bolt, if it were not in a hole, but just floating in space, and I pushed on it with the same force (sideways and down), it would still be torqued (despite having no point of support or suspension), but the axis of rotation would be tilted with respect to the "normal" rotation axis. It would be perpendicular to the direction of my push.

You can't take this formula and apply it to a situation in which no axis is defined in advance. What you've done in your argument is taken this specific application and demanded that the general formula have the same conditions. You will note that your quote just says "about an axis". It says nothing about suspension, it doesn't say that axis is fixed. It just says "about an axis". You can pick any axis you want, and as long as you apply the formula correctly, you will get the right answer. In the case of the bolt in the hole, if I pick any other axis but the one around which the bolt will turn, the force against the wall of the hole will cancel out my pushing force, yielding a net F=0, and hence Fsin(theta)=0. No torque. Without some mechanism like the hole, you have to try all possible axes.

If you don't know in advance what the axis of rotation is, you need to calculate r cross F for each mass that feels a force in the problem. When you sum up all those vectors, the final vector you get is the axis around which the system will turn.

No. This is only true if there is a point of suspension. You can't apply it to a situation where there is none, and then turn around and demand that there must be one.

But if the bolt were just sitting on a table, it would still turn if you applied force at 90 degrees, wouldn't it? Even with no point of suspension.