Misunderstanding the concept of torque

[fawk3s]

I think I am misunderstanding something. Because I had this strain of thought and it doesn't really want to logically up. By my logic anyways.

Ok, here it is:
Let us have a standard lever/seesaw and a fulcrum at a point. Let us apply a force to the very end of the lever (which is the farthest away from the fulcrum).
Now let's look at the other half of the lever. Let's observe a point on it closer to the fulcrum than the force we applied. Since torque
M = F*l
we see that the point closer to the fulcrum has to have a higher force applied to it (since its closer to the fulcrum) in order to have the same moment, the moment the whole lever has.
So we got that a point closer to the fulcrum has a higher force applied to it.

Now let's look at the movement of the lever, we can tell its circular motion, where fulcrum is the centerpoint. Say the force we applied was perpendicular to the lever at that moment. As force results in an acceleration by Newton's second law, we get that an acceleration perpendicular to the lever, aka tangential acceleration, is higher closer to the fulcrum than away from it.
And this is the part that I don't understand... As the lever has an angular acceleration, we can tell that the tangential acceleration of a point would get higher as we move farther away from the fulcrum. So why do we get a higher force closer to the fulcrum?

P.S. At this point, I am only interested in the forces. Please do NOT bring in the conservation of mechanical energy/work, saying something like "the farther you are from the fulcrum the longer distance you have to travel, ergo closer to the fulcrum point there has to be higher force" or likewise. I do not care about that at this moment. I am only interested in how these forces are created and how they balance each other out.

Thanks in advance

 

[A.T.]

As the lever has an angular acceleration, we can tell that the tangential acceleration of a point would get higher as we move farther away from the fulcrum. So why do we get a higher force closer to the fulcrum?

Acceleration depends on the NET force. The acceleration of a lever-part depends on the sum of:
- external forces applied to the lever at this part
- internal forces from adjacent lever parts (which you ignored)

 

[Studiot]

Let us have a standard lever/seesaw and a fulcrum at a point. Let us apply a force to the very end of the lever (which is the farthest away from the fulcrum).
Now let's look at the other half of the lever. Let's observe a point on it closer to the fulcrum than the force we applied. Since torque
M = F*l
we see that the point closer to the fulcrum has to have a higher force applied to it (since its closer to the fulcrum) in order to have the same moment, the moment the whole lever has.
So we got that a point closer to the fulcrum has a higher force applied to it.

I am having trouble picturing this. Can you draw a diagram?

It is important to distinguish whether you mean a lever or a see saw since they place the forces on opposite sides of the fulcrum and this makes a difference.

Further it is not clear whether you mean your lever to be in equilibrium or moving, that also makes a difference.

 

[fawk3s]

Acceleration depends on the NET force. The acceleration of a lever-part depends on the sum of:
- external forces applied to the lever at this part
- internal forces from adjacent lever parts (which you ignored)

Could you bring out those internal forces? Because I am not sure I follow.

I am having trouble picturing this. Can you draw a diagram?

It is important to distinguish whether you mean a lever or a see saw since they place the forces on opposite sides of the fulcrum and this makes a difference.

Further it is not clear whether you mean your lever to be in equilibrium or moving, that also makes a difference.

http://img6.imageshack.us/img6/8206/48580036.png
http://img6.imageshack.us/img6/8206/48580036.png

Though let's observe as the lever has no load anywhere on it. Let's just observe the point where the load is placed in the picture.
Also, let's observe both the static/equillibrium (as if there was a counter force/torque, so say that in one case there is a load placed on it) and in other case, the lever would be moving.
We see that higher force is achieved closer to the fulcrum. Yet if the seesaw has an angular acceleration, shouldn't the part closer to it have a lower acceleration (and therefore force) than the part which is further away?
I am interested in how the higher force is achieved closer to the fulcrum. And maybe even in how the torque equation M = F*l actually represents the turning force.

 

[A.T.]

Could you bring out those internal forces? Because I am not sure I follow.

Consider a small part of the lever, where the external force is applied. Aside from that external force, there are also internal forces from the adjacent lever parts acting on that small part of the lever. The net acceleration of that small part of the lever is determined by the sum of all those forces.

 

[fawk3s]

Consider a small part of the lever, where the external force is applied. Aside from that external force, there are also internal forces from the adjacent lever parts acting on that small part of the lever. The net acceleration of that small part of the lever is determined by the sum of all those forces.

But what kind of forces are we talking about here? Be clear. Consider the lever practically massless.

 

[jbriggs444]

If the lever is practically massless, it follows that small pieces of the lever are practically massless.

If the lever is not accelerating dramatically it follows that the NET force on EVERY small piece of the ruler is close to ZERO.

Consider the small piece of the lever on which an external force is applied. It follows that this small piece of the lever is experiencing a total force from the neighboring small piece(s) that is nearly equal and opposite to the applied external force.

This applies at each of the three points where the lever connects to the outside world -- effort, load and fulcrum.

Despite the fact that the net force on every small piece is near-zero, the force from each piece on the next piece will be similar in magnitude to the forces at the effort, load and fulcrum.

 

[A.T.]

But what kind of forces are we talking about here? Be clear. Consider the lever practically massless.

Massless or not. In order to work as a lever, it must be able to transmit forces internally.

 

[fawk3s]

Would one of you mind doing a quick drawing? I am a little confused and not sure how it relates to larger forces being applied to the load closer to the fulcrum. Would be very helpful.

 

[A.T.]

Im a little confused and not sure how it relates to larger forces being applied to the load closer to the fulcrum.

It relates to your question regarding acceleration of a lever part. It you want to analyze lever parts, you have to split the lever in parts in your free body diagram. Then there are forces between those parts.

 

[fawk3s]

http://img834.imageshack.us/img834/7752/87101201.png
http://img834.imageshack.us/img834/7752/87101201.png

This is what I could make out of it. If that's what you were pretty much talking about, it describes the picture I posted about the tangential acceleration earlier. Doesnt really explain the force leverage on lower radius.
If it is not, is it really that hard to do that little sketch? No disrespect. Drawings are helpful.

 

[A.T.]

http://img834.imageshack.us/img834/7752/87101201.png
http://img834.imageshack.us/img834/7752/87101201.png
Drawings are helpful.

Well, it is not really clear what the red and green arrows are. But let's say:
red : force from next segment away from fulcrum
green : force from next segment towards fulcrum

The vector sum of red & green (net force on a segment ) gets smaller towards the fulcrum, and so does the tangential acceleration of the segments.

 

[fawk3s]

Why is the force from the segment closer to fulcrum (green) smaller than the force exerted by the segment farther away (red) and why does their sum vector get smaller toward the fulcrum?

Thanks in adv

 

[A.T.]

Why ...

Physics is not about "why?" but about "how much?". If the forces weren't as I said, the lever would not move as we assumed it moves. The internal forces in a rigid body always provide whatever support is necessary.

 

[fawk3s]

The internal forces in a rigid body always provide whatever support is necessary.

That doesn't litterally mean you can't ask "why" questions, though. You can paraphrase my question to what's the cause behind the forces getting smaller towards the centerpoint in a circular motion. I mean, I can intuitively see the reason, as for velocity, the parts of the circle which are farther away from the center must travel longer distances than those closer to it in the same timespan. But the reason for the forces to decrease is not that clear to me. As that last sentence of yours seems like a laconic answer to it, could you elaborate, please?

Besides that, I can't link the previous discussion to the force leverage we get with the lever when the load is close to the fulcrum and the force is applied farther than the load.

 

[A.T.]

But the reason for the forces to decrease is not that clear to me.

If it accelerates less, then there is less net force. Keep in mind than when dealing with acceleration & forces you cannot really assume the lever is massless. That would give infinite acceleration.

As that last sentence of yours seems like a laconic answer to it, could you elaborate, please?

It's not laconic. It is what internal forces in a rigid body do. Otherwise it would not be a rigid body. They oppose any deformation and adjust to ensure it moves as a rigid body.

 

[fawk3s]

http://img13.imageshack.us/img13/2144/56032918.png
http://img13.imageshack.us/img13/2144/56032918.png

Alright, but I still can't deal with how we get more net force on that load nearer to the fulcrum. We just cleared that the net force/acceleration on the parts closer to the fulcrum accelerate less than those farther away. So what changes? We obviously add mass, but does it really relate to that? I understand that in order for the lever to be in equillibrium, we must exert less force farther away from the fulcrum in order to get the same counter torque where that load lies, but what doesn't that part get more acceleration now? Confuses me.

 

[A.T.]

I understand that in order for the lever to be in equillibrium, we must exert less force farther away from the fulcrum in order to get the same counter torque where that load lies, but what doesn't that part get more acceleration now? Confuses me.

Because acceleration of a part depends on the net force acting on a part, and not just on the force exerted externally.

 

[Studiot]

Please note that you cannot apply the equations and arguments of equilibrium to an object that is not in equilibrium.

An object that is accelerating is not in equilibrium.

In particular when an object is not in equilibrium there is never any need to 'balance' anything.

This is a very common error.

Hope this general comment helps.

 

[xAxis]

..Let us have a standard lever/seesaw and a fulcrum at a point. Let us apply a force to the very end of the lever (which is the farthest away from the fulcrum).
Now let's look at the other half of the lever. Let's observe a point on it closer to the fulcrum than the force we applied. Since torque
M = F*l
we see that the point closer to the fulcrum has to have a higher force applied to it (since its closer to the fulcrum) in order to have the same moment, the moment the whole lever has.
So we got that a point closer to the fulcrum has a higher force applied to it.

No. A point closer to the fulcrum needs a higher force applied to it. It does not have it applied.

 

[fawk3s]

Because acceleration of a part depends on the net force acting on a part, and not just on the force exerted externally.

Then explain to me, how do we "win" with force in this situation when most of it goes into balancing internal forces?

 

[jartsa]

Here is a little exercise:

When we apply a force of 1 N to the end of a lever, the lever exerts a force of 10 N on a mass near the fulcrum. The force 10 N accelerates the mass.

Now we add a second identical mass to the end of the lever, I mean that end where the force is applied.

Question: how does the force exerted on the mass near the fulcrum change, when the force applied to the end of the lever is still 1 N?My answer:

New force is about one hundreth of the old force.

 

[A.T.]

Then explain to me, how do we "win" with force in this situation when most of it goes into balancing internal forces?

The internal forces over the entire lever add to zero, so they are irrelevant for the force balance of the external forces (functionality of the lever). But to get the correct acceleration of a lever part, you obviously have to consider them. You are confusing yourself by being inconsistent about how to divide the system into parts.

 

[eudo]

Ok I'll try my hand at this. First a little elaboration on internal forces.

Imagine two situations: you're out in the middle of space and in the first case you see a rigid rod rotating around and around with no external forces or acceleration. In the other case you see a rope floating around in space and you give it a little spin, but because it's not rigid, it bends around a bit. What's the difference between these two cases? It's the intramolecular forces keeping the rod together and keeping the rope together. In general you would need quantum mechanics to understand these forces and explain why one material is rigid and another is not. No one before the 20th century understood them, so you wouldn't expect that a first year classical physics class would be able to explain them. Instead we simply assume that whatever forces are needed to keep the rod rigid are at work when talking about a lever. If they weren't then we'd be talking about a rope and not a rigid lever and we'd need to use other equations.

Now let's look at your case. As others have alluded to, you're trying to use a static equation for a dynamic situation. If you apply a torque M to the lever on one side, then it will take a force F=M/l to balance the torque so the lever won't rotate. So far so good. But then you wanted to start talking about acceleration of a point on the lever. That requires a bit more work.

If you simply apply a torque M to one end of the lever, then the angular acceleration is M/i, where i is the moment of inertia of the lever system. So the linear acceleration of a test mass placed on the other side a distance of r from the fulcrum is rM/i. (Keeping in mind that the moment of inertia is for the lever AND the test mass).

Assuming that the test mass doesn't affect the moment of inertia much, then the acceleration is indeed proportional to the distance r.

In the other extreme when the lever is massless, then the size of the test mass matters. If the test mass has mass m, then the moment of inertia is m*(r squared) and the linear acceleration is M/(rm). In this case as you slide the mass closer to the fulcrum, the acceleration increases. That's simply because as you move it closer to the fulcrum it gets easier to turn the lever and if you keep the torque constant then the lever will accelerate more. Finally when the test mass is sitting right on the fulcrum then the tiniest torque will cause an infinite angular acceleration and the lever will just spin around and around because the lever system has zero moment of inertia.

 

[HallsofIvy]

It might be better to look at energy rather than force or acceleration. There is a 'conservation of energy' law, but no 'conservation of force'. If we have an object we want to move "x" m from the fulcrum, and apply force f Newtons at a point "y" m from the fulcrum, moving the lever through an angle $\theta$ radians, we have applied the force, f, through distance $x\theta$ meters and so have done $fx\theta$ Joules of work. That must be the same energy applied at the other end. If we call the force applied there "F", have moved the lever through $y\theta$ meters and so we must have $Fy\theta= fx\theta$ and so we must have $F/f= x/y$.