Simple argument that a force applied farther from the rotation axis

[A.T.]

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.

 

[Aimless]

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.

 

[aaaa202]

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?

 

[ImaLooser]

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 can't 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.

 

[A.T.]

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?

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

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.

 

[Aimless]

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?

Yes; that's about what I was trying to describe. This view of things is oversimplified, but at a qualitative level I think it captures the dynamics of how internal stresses on a microscopic level can translate into rotation on a macroscopic level.

From a teaching perspective, it might be worth it to construct a numerical simulation of a model along these lines (with tunable string constants) and allowing it to evolve in time. It shouldn't be all that difficult; certainly nothing compared to, say, a DFT calculation.