Simple argument that a force applied farther from the rotation axis

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  • #1
aaaa202
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Torque is defined as:

τ = rxF

this means that the farther from the rotation axis of a body a force is applied, the more it will tend to rotate the body.

My question is:

Can anybody give me a simple argument that a force applied farther from the rotation axis should be better at rotating a body than one closer. You can use energy considerations to say that the work done by a force applied at the farthest distance does a greater work, but I am not looking for that since it is not general enough. Because imagine that two tangential forces of equal length but opposite direction are being applied to our body. Then it will still rotate, and energy considerations cant really explain that.

So yeah, please give me an intuitive, yet rigorous reason that it MUST be so greater torque = no equilibrium.
 

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  • #2
Filip Larsen
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Perhaps you can hone your intuition by thinking about what happen when you apply the force directly towards the axis (or the center of mass) of a rigid body. Why don't you get any rotational acceleration in this case and how does that relate to your question?
 
  • #3
aaaa202
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Well in that case you are obviously not speeding up the particles in going around the centre of rotation. They do need a centripetal force but that will be provided by the forces keeping the rigid body together. But how does this relate to my question?
 
  • #4
russ_watters
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Your question is unclear to me: your two tangential forces of equal length but opposite direction: where are they being applied? Opposite sides of the shaft...? Could you draw us a diagram?
 
  • #5
aaaa202
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What I wanted to say with that is that you can't always use an energy consideration to show that force applied at greater distance from rotation centre = bigger acceleration. Because that wont do to explain why our rigid body on the picture STARTS to move in the drawn situation.
 

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  • #6
russ_watters
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Why not? I see an imbalance there: http://en.wikipedia.org/wiki/Couple_(mechanics)
 
  • #7
aaaa202
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Yes yes! I just dont see WHY a couple will tend to rotate our body according to the net torque. What is it that makes a force applied farther from the rotation centre "BETTER" at rotating our object than one applied closer? And do not use an energy consideration, because that wont do. I want a simple argument in terms of the internal forces inside the body.
 
  • #8
russ_watters
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I don't understand: internal forces will not cause a body to move.

I'm not sure if what you are asking can be proven mathematically without using energy conservation and I'm not sure why you object to using energy conservation to analyze a situation that you recognize involves motion. But back to the original question:
Can anybody give me a simple argument that a force applied farther from the rotation axis should be better at rotating a body than one closer.
A simple argument? Sure: it is observed to be true.
 
  • #9
aaaa202
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Okay when you push with a tangential force the force is transmitted through our body to make it rotate. That's what I meant by internal forces - thos transmitting the tangential forces to an overall rotation. I want an argument with forces that shows why a force applied farther away is better at rotating. Because you must be able to prove it. As for energy considerations those will only be valid once the body is actually rotating and the forces are doing work. When it stands still like one the picture you can't use an energy consideration to show that it must start to rotate due to the couple. Understand my problem now? :)
 
  • #10
russ_watters
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Because you must be able to prove it.
If by that you mean you must be able to prove it mathematically, no, I'm sorry but that is not correct. Physics is a mathematical study of reality. The only "proof" required is experimental evidence.
As for energy considerations those will only be valid once the body is actually rotating and the forces are doing work.
No, that isn't correct. Looking at a previous thread you started on a similar subject, you're making the mistake of not counting the effect of inertia in opposing the force, so you're not summing all of the forces and you're not recognizing that the instant you apply the forces, there is an acceleration and a power. Ie, an object that is not moving, but has an unbalanced force (couple) is still accelerating.
When it stands still like one the picture you can't use an energy consideration to show that it must start to rotate due to the couple. Understand my problem now? :)
Yes, I think so: your problem is you misunderstand what "standing still" means. "Standing still" means not moving and not accelerating.

[edit] Put another way: you're asking 'why is there an acceleration?' when you could be asking 'is there an acceleration?', and calculating that it is non-zero.
 
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  • #11
aaaa202
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Okay, I think you are the one misunderstanding me once again.

My problem is NOT that I don't know what standing still means. My problem is that I geometrically can't see why a force applied at greater distance = bigger acceleration = greater tendency to rotate.

Let me try to be more clear: In translational dynamics you learn that if the resultant force on an object is zero then it will not accelerate. This is actually where I have a problem. Because the resultant force on our rigid body is zero. Yet the body tends to rotate. I want to understand why the force applied at greater distance has the property of being able to rotate the body. I have said that in terms of energy it is clear once the body is actually rotating. Then you can see that the force applied at greater distance does greater work etc. etc. But we remain to show that the body will actually start rotating when there is a couple on it. That cannot be shown by energy considerations since in that instant no work will be done. Do you understand me better now? :)
 
  • #12
russ_watters
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I was mistaken about my use of power since you are right that at the moment of the beginning of the acceleration, power is zero. It is change in momentum that is observed at the moment of application of the torque: http://en.wikipedia.org/wiki/Angular_momentum#Conservation_of_angular_momentum

My problem is that I geometrically can't see why a force applied at greater distance = bigger acceleration = greater tendency to rotate.
Ok....
Let me try to be more clear: In translational dynamics you learn that if the resultant force on an object is zero then it will not accelerate.
Ok....but if it is non-zero, it does accelerate, right? So what equation describes that acceleration?
This is actually where I have a problem. Because the resultant force on our rigid body is zero. Yet the body tends to rotate.
The linear force is zero, so the linear acceleration is zero. The 'angular force' (torque/couple) is non-zero, so the angular acceleration is nonzero. I don't see why you see a contradiction between the two examples you gave. They look to be the same, with the same logic and the same outcome if you are consistent in the construction of the examples.
 
  • #13
aaaa202
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I know all about angular momentum, but for me that is all just math and does not necessarily provide an intuition or an understanding of what happens on microscopic level.
And just to show you, I do know that (in vector notation):

dL/dt = Ʃτ

Which again can be regarded as an analogue of F = dp/dt for linear motion.
But that equation does not say much for me.
I want an argument purely in terms of forces and interactions of the particles that make up our rigid body.
Even though linear motion and rotations are different the basic ingredient for any of the two motions is this: A force. Therefore you should be able to show purely through the use of force analysis that a force applied at a greater distance should mean that the moment of our body changes. Shouldn't you?
 
  • #14
russ_watters
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But that equation does not say much for me.
How about f=ma? I find that to be more of a "why" than f=dp/dt, which is more of a definition or description. At least f=ma speaks to me more (not that the universe cares how it speaks to me...)
I want an argument purely in terms of forces and interactions of the particles that make up our rigid body.
You're looking too deep. This problem deals with theoretically perfect bodies that don't have particles. There are no inter-object forces to deal with. You would not ask for such a proof regarding linear motion: f=ma deals with completely solid, monolithic (or point) objects only.
Even though linear motion and rotations are different the basic ingredient for any of the two motions is this: A force. Therefore you should be able to show purely through the use of force analysis that a force applied at a greater distance should mean that the moment of our body changes. Shouldn't you?
I don't know -- you're declaring what seem to me to be relevant points to be unacceptable to you. Do you have similar issues with f=ma? Do you need a proof of it as well?

I guess the point here is that ultimately you're going to have to accept that some parts of what we understand of physics cannot be derived. They are simply observed to be true. May I ask why this is a problem for you? What would you gain by seeing a derivation that speaks to you that you are losing by not seeing it? Do you accept that the physics we currently have works?
 
  • #15
aaaa202
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hmm it is not a problem to me when things are observed rather than derived. I just want to be clear of what can be derived and what will always be an observation. Newtons laws, e.g. F = dp/dt, are for instance no problem to me since they are clearly empirical.
But I am still not sure that you can't show the above stated from Newtons laws, and I don't see why you are so sure either.
Also I don't agree that you can't divide a rigid body into a lot of small mass elements. That is exactly what you do for moment of inertia calculations etc. etc. The important property of a rigid body as I understand it is not that the small mass elements can't interact through forces, but rather that it is undeformable.
 
  • #16
russ_watters
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hmm it is not a problem to me when things are observed rather than derived. I just want to be clear of what can be derived and what will always be an observation. Newtons laws, e.g. F = dp/dt, are for instance no problem to me since they are clearly empirical. But I am still not sure that you can't show the above stated from Newtons laws, and I don't see why you are so sure either.
We've actually had this exact question asked a number of times before and I've never seen an answer for the static case. Indeed, asking for a static answer to a non-static situation appears to me to be a contradiction in terms.

People don't tend to realize that the term "unbalanced force" is at best sloppy and at worst is simply nonexistent: Forces are never really unbalanced, it is just that externally applied forces that don't sum to zero are opposed by the inertia of the object, which does then result in a zero sum of forces.
Also I don't agree that you can't divide a rigid body into a lot of small mass elements. That is exactly what you do for moment of inertia calculations etc. etc. The important property of a rigid body as I understand it is not that the small mass elements can't interact through forces, but rather that it is undeformable.
I'm not saying that you can't divide up an object, just that you don't. Doing so for the purpose of finding a moment of inertia isn't a correct analogy: the different pieces aren't being treated as interacting, they are just being summed to a single number. Ultimately, the arrangement of those specific pieces makes no difference; only the sum matters.
 
  • #17
aaaa202
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Can you elaborate on the point with the inertia of the object. I didn't quite understand that part - probably because my english skills are not always too good.
Anyways, I'm going to bed and will think more about it tomorrow, though I should like to think that you wrapped it up with the points in your last post.
 
  • #18
russ_watters
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Physicists don't tend to like the way I used the term inertia, but for engineers, inertia is the ability of mass to resist an acceleration. So a=f/m quantifies how much acceleration an object can resist under a given force. Or put another way, when you push on an object and it accelerates, it is the inertia that is pushing back at you.

For example, consider the following forces, acting in opposite directions against the same point:

F1 = 5N
F2 = 3N

You might say the force is imbalanced at F1-F2=2N or you might say the force is balanced at F1-F2-ma=0, where ma=-2N.
 
  • #19
ecneicS
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Okay, I think you are the one misunderstanding me once again.
I want to understand why the force applied at greater distance has the property of being able to rotate the body.

Let me try and justify this for you.

Let f_2 be the force acting at greater distance from axis, and f_1 acting from the smaller distance.
K=∫(ƩF dot dr)
=∫((F_2-F_1) dot dr)
=∫(F_2 dot dr)-∫(F_2 dot dr)

Consider both of these forces acting tangentially and for a quarter revolution (arbitrary). Since F_2 acts over an arclength of a_2 and F_1 acts over an arclength of a_1, where a_1 is less than a_2, the first integral is evaluated from 0 to a2 and the second is evaluated from 0 to a1, which results in the net force doing nonzero work over a time interval, even though the forces are equal. Trippy.
 
  • #20
A.T.
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What is it that makes a force applied farther from the rotation centre "BETTER" at rotating our object than one applied closer?
Try to ask the opposite question:

What is it that makes a force applied closer to the rotation centre "WORSE" at rotating the object?

And to answer it, consider the extreme case: A force acting exactly at the rotation center.
 
  • #21
aaaa202
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Let me try and justify this for you.

Let f_2 be the force acting at greater distance from axis, and f_1 acting from the smaller distance.
K=∫(ƩF dot dr)
=∫((F_2-F_1) dot dr)
=∫(F_2 dot dr)-∫(F_2 dot dr)

Consider both of these forces acting tangentially and for a quarter revolution (arbitrary). Since F_2 acts over an arclength of a_2 and F_1 acts over an arclength of a_1, where a_1 is less than a_2, the first integral is evaluated from 0 to a2 and the second is evaluated from 0 to a1, which results in the net force doing nonzero work over a time interval, even though the forces are equal. Trippy.

That wont do, because as we spent half the last page discussing it doesn't explain the moment where the body stands still and begins to rotate due to a couple. In that instant you can't apply a work-consideration since the body is not moving.
 
  • #22
aaaa202
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Try to ask the opposite question:

What is it that makes a force applied closer to the rotation centre "WORSE" at rotating the object?

And to answer it, consider the extreme case: A force acting exactly at the rotation center.

I don't see how this helps other than providing an intuition based on nothing really. Applying a force close and at the rotation centre is two different things.
 
  • #23
A.T.
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Try to ask the opposite question:

What is it that makes a force applied closer to the rotation centre "WORSE" at rotating the object?

And to answer it, consider the extreme case: A force acting exactly at the rotation center.

I don't see how this helps other than providing an intuition based on nothing really.
I thought you are looking for a simple intuitive argument. What exactly do you seek then?
Applying a force close and at the rotation centre is two different things.
Yes, but empirically we know that nature behaves continuously: If you change one variable continuously, the depending variables also change continuously.

- Applying the force at the center has no effect on rotation (This should be clear, which way would it rotate anyway.)

- Applying the force at circumference has an effect on rotation (This we know empirically)

- Because of natures continuity, moving the point of force application continuously from the center to the circumference, will change the effect on rotation continuously.

Of course a universe is conceivable, where there so no such continuity. Where when you move the point of force application just slightly off-center, the effect on rotation immediately jumps from zero to a fixed value that is the same for all distances. But such a "jump" is simply not how our universe works in general.
 
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  • #24
aaaa202
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We've actually had this exact question asked a number of times before and I've never seen an answer for the static case. Indeed, asking for a static answer to a non-static situation appears to me to be a contradiction in terms.

People don't tend to realize that the term "unbalanced force" is at best sloppy and at worst is simply nonexistent: Forces are never really unbalanced, it is just that externally applied forces that don't sum to zero are opposed by the inertia of the object, which does then result in a zero sum of forces. I'm not saying that you can't divide up an object, just that you don't. Doing so for the purpose of finding a moment of inertia isn't a correct analogy: the different pieces aren't being treated as interacting, they are just being summed to a single number. Ultimately, the arrangement of those specific pieces makes no difference; only the sum matters.
Hey Russ, can you try to elaborate on this post or rephrase the things you said. I think this one was the closest I got to new insight on the topic, but unfortunately I didn't quite understand it all. You said something about it being a contradiction going from the static to the non-static scenario right? Did you mean that a completely static case is an idealization?
 
  • #25
A.T.
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Hey Russ, can you try to elaborate on this post or rephrase the things you said. I think this one was the closest I got to new insight on the topic, but unfortunately I didn't quite understand it all. You said something about it being a contradiction going from the static to the non-static scenario right? Did you mean that a completely static case is an idealization?

You can make your scenario static, by making the outer force smaller. Then you ask why the greater inner force cannot win against the smaller outer force.

This is basically the same as your original question, but you don't get into those static vs. non-static issues here, and can justify better why you don't want an explanation based on work done.
 
  • #26
ecneicS
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That wont do, because as we spent half the last page discussing it doesn't explain the moment where the body stands still and begins to rotate due to a couple. In that instant you can't apply a work-consideration since the body is not moving.

Actually it does. If the NET force acting on an object does NONZERO work over a certain time or distance, that is equivalent to saying that the body accelerated. Consider a point mass where two forces are being applied on it, each in opposite direction but equal in magnitude. Using the same methodology I outlined above, you will always get zero work done which gives the intuitive result of a no movement. The rotating case is unique because one of the forces acts over longer distance.

I've read the whole thread, but I still dont see proper justification for you saying that energy considerations will not do.
 
  • #27
A.T.
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I've read the whole thread, but I still dont see proper justification for you saying that energy considerations will not do.
See my post #25. In a balanced scenario there is no work done over time, but you still can ask the question why torque depends on the lever arm.
 
  • #28
chingel
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I think the OP is basically asking about leverage, why it works, how can a force be multiplied? What supplies the extra force? Push with the same force, but depending on the location of application of the force, it's as if you push with twice the force. For example, a rod is attached to a pivot point, you push a distance L from the pivot point, your friend pushes a distance 2L from the pivot. You have to push twice as hard in the opposite direction to balance the torque caused by your friend. Why is that?

Consider two weights on each side of a rod balancing on a pivot, both a distance L away from the pivot. Each weight applies a force F, but the pivot point has to apply a force 2F upwards so that the rod and the weights stay in place, just to hold the weights up. Now consider the previous scenario, one end of the rod is balanced, a force F is applied downward at a distance 2L from the pivot and a force 2F upwards in the middle. Since the rod doesn't move and the forces are exactly identical as with the weights, the third force at the pivot point has to be the same too. Basically you have your friend pushing with a force F at one end and the pivot point also applying a force F at the other end, and you have to apply a force to counteract both, as if you were holding two weights up, one at either end.
 
  • #29
russ_watters
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Hey Russ, can you try to elaborate on this post or rephrase the things you said. I think this one was the closest I got to new insight on the topic, but unfortunately I didn't quite understand it all. You said something about it being a contradiction going from the static to the non-static scenario right? Did you mean that a completely static case is an idealization?
You asked that at the end of last night: I answered in post #18.
 
  • #30
aaaa202
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You can make your scenario static, by making the outer force smaller. Then you ask why the greater inner force cannot win against the smaller outer force.

This is basically the same as your original question, but you don't get into those static vs. non-static issues here, and can justify better why you don't want an explanation based on work done.

So you are saying that in this case you can see why the inner force cannot win. How do you see that?
 
  • #31
A.T.
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So you are saying that in this case you can see why the inner force cannot win. How do you see that?
No, I just wanted to help to you rephrase your question, so it avoids the problems pointed out by Russ.

Currently I don't have a better explanation for you than the one in post #23. But I think the key is to consider a truly static case, where all forces and torques must balance. Keep in mind that:
- There is a 3rd force at the pivot.
- The torques must balance around any point, not just the pivot.

Based on this it might be possible to show, that the torque must depend linearly on both: the (tangential)force and the lever arm.
 
  • #32
Aimless
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When I was first learning torque and moments of inertia, I had some of the same problems with it that the OP did, so I'm going to take a stab at answering this from a different approach than those above. Conceptually, this approach worked for me, so I hope it will help the OP as well.

Imagine a non-rigid rod, but rather one made of a lattice of point particles of some arbitrary mass connected by springs. Prior to the application of a force, suppose that all of the particles are at rest with respect to each other, and all of the springs are in their equilibrium position.

Now, apply a force to a particle at one end of the rod in a direction perpendicular to the axis of the rod. Initially, you will get a deformation in the shape of the rod as that particle begins to move in the direction of the force. Hooke's Law will propagate the force through the rod along the direction parallel to it, but what happens with respect to the perpendicular directions? Those springs will become elongated as the rod deforms, creating an internal restoring force which acts (roughly) along the axis of the rod. However, because our external force is being applied only at one end of the rod, this restoring force is unbalanced, leading to a net acceleration along the axis of the rod in addition to the net acceleration in the direction of the external force. Thus, rotation.

Now, add in a second force, anti-parallel to the first and of equal magnitude, some distance between the end of the rod and the center of mass. The particles at the top of the still feel the same two forces: that generated by external force one and the restoring force. At the point where external force two acts, the same thing happens; the external force causes a deformation, and the deformation generates a restoring force. However, in the case, the restoring force is less unbalanced, because portions of the rod are on both sides of the point where the external force acts. Thus, while this force also wants to generate rotation (in a direction opposite to the first force), it does so less strongly because of the cancellation of some of the restoring force. If the force were acting at the center of the rod, the restoring would cancel completely and no rotation would be generated at all. It's easy to see that the magnitude of the component along the axis of the rod in this picture is roughly proportional to the distance from the center of mass, so at a qualitative level this gives the justification for torque having the same property.

Now, granted, this picture changes as soon as you allow the system to evolve in time, due to the deformation of the rod. However, if you assume the spring constants to be arbitrarily large, then the deformation stays small and this picture holds some validity. Taking the spring constants to be infinite gives you the rigid body limit.
 
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  • #33
aaaa202
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When I was first learning torque and moments of inertia, I had some of the same problems with it that the OP did, so I'm going to take a stab at answering this from a different approach than those above. Conceptually, this approach worked for me, so I hope it will help the OP as well.

Imagine a non-rigid rod, but rather one made of a lattice of point particles of some arbitrary mass connected by springs. Prior to the application of a force, suppose that all of the particles are at rest with respect to each other, and all of the springs are in their equilibrium position.

Now, apply a force to a particle at one end of the rod in a direction perpendicular to the axis of the rod. Initially, you will get a deformation in the shape of the rod as that particle begins to move in the direction of the force. Hooke's Law will propagate the force through the rod along the direction parallel to it, but what happens with respect to the perpendicular directions? Those springs will become elongated as the rod deforms, creating an internal restoring force which acts (roughly) along the axis of the rod. However, because our external force is being applied only at one end of the rod, this restoring force is unbalanced, leading to a net acceleration along the axis of the rod in addition to the net acceleration in the direction of the external force. Thus, rotation.

Now, add in a second force, anti-parallel to the first and of equal magnitude, some distance between the end of the rod and the center of mass. The particles at the top of the still feel the same two forces: that generated by external force one and the restoring force. At the point where external force two acts, the same thing happens; the external force causes a deformation, and the deformation generates a restoring force. However, in the case, the restoring force is less unbalanced, because portions of the rod are on both sides of the point where the external force acts. Thus, while this force also wants to generate rotation (in a direction opposite to the first force), it does so less strongly because of the cancellation of some of the restoring force. If the force were acting at the center of the rod, the restoring would cancel completely and no rotation would be generated at all. It's easy to see that the magnitude of the component along the axis of the rod in this picture is roughly proportional to the distance from the center of mass, so at a qualitative level this gives the justification for torque having the same property.

Now, granted, this picture changes as soon as you allow the system to evolve in time, due to the deformation of the rod. However, if you assume the spring constants to be arbitrarily large, then the deformation stays small and this picture holds some validity. Taking the spring constants to be infinite gives you the rigid body limit.

I think this is exactly the type of argument I have been looking for. It will take me some time to get into your way of thinking though. Is the picture you describe similar to the one I have tried to draw on the attached sketch?
 

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  • #34
ImaLooser
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Torque is defined as:

τ = rxF

this means that the farther from the rotation axis of a body a force is applied, the more it will tend to rotate the body.

My question is:

Can anybody give me a simple argument that a force applied farther from the rotation axis should be better at rotating a body than one closer. You can use energy considerations to say that the work done by a force applied at the farthest distance does a greater work, but I am not looking for that since it is not general enough. Because imagine that two tangential forces of equal length but opposite direction are being applied to our body. Then it will still rotate, and energy considerations cant really explain that.

So yeah, please give me an intuitive, yet rigorous reason that it MUST be so greater torque = no equilibrium.


It's a lever. That's how I think of it. The longer the wrench, the more torque.

Not rigorous, I know. Maybe you already thought of this too.
 
  • #35
A.T.
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I think this is exactly the type of argument I have been looking for. It will take me some time to get into your way of thinking though. Is the picture you describe similar to the one I have tried to draw on the attached sketch?

attachment.php?attachmentid=50091&d=1345641964.png


I think you can simplify this to a King Truss (top left), just ignore the overhang at the sides:

common-trusses.png


Apply 2F downwards in the middle, and F upwards to the 2 sides nodes. It is a static case. Now your question translates to: Why doesn't it start rotating around one of the sides nodes? How does the one F on the other side balance the 2F in the middle?

Aside from the trivial symmetry argument, you can work out the internal forces here.
 
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