Misunderstanding the concept of torque

1. Dec 19, 2012

fawk3s

I think I am misunderstanding something. Because I had this strain of thought and it doesnt 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 lets look at the other half of the lever. Lets 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 lets 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 dont 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, Im 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. Im only interested in how these forces are created and how they balance each other out.

2. Dec 19, 2012

A.T.

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)

3. Dec 19, 2012

Studiot

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.

4. Dec 19, 2012

fawk3s

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

http://img6.imageshack.us/img6/8206/48580036.png [Broken]
http://img6.imageshack.us/img6/8206/48580036.png [Broken]

Though lets observe as the lever has no load anywhere on it. Lets just observe the point where the load is placed in the picture.
Also, lets 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, shouldnt 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.

Last edited by a moderator: May 6, 2017
5. Dec 19, 2012

A.T.

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.

6. Dec 19, 2012

fawk3s

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

7. Dec 19, 2012

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.

8. Dec 19, 2012

A.T.

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

9. Dec 19, 2012

fawk3s

Would one of you mind doing a quick drawing? Im 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.

10. Dec 19, 2012

A.T.

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.

11. Dec 19, 2012

fawk3s

http://img834.imageshack.us/img834/7752/87101201.png [Broken]
http://img834.imageshack.us/img834/7752/87101201.png [Broken]

This is what I could make out of it. If thats 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.

Last edited by a moderator: May 6, 2017
12. Dec 19, 2012

A.T.

Well, it is not really clear what the red and green arrows are. But lets 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.

Last edited by a moderator: May 6, 2017
13. Dec 19, 2012

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?

14. Dec 19, 2012

A.T.

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.

15. Dec 19, 2012

fawk3s

That doesnt litterally mean you cant ask "why" questions, though. You can paraphrase my question to whats 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 cant 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.

16. Dec 19, 2012

A.T.

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.

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.

17. Dec 20, 2012

fawk3s

http://img13.imageshack.us/img13/2144/56032918.png [Broken]
http://img13.imageshack.us/img13/2144/56032918.png [Broken]

Alright, but I still cant 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 doesnt that part get more acceleration now? Confuses me.

Last edited by a moderator: May 6, 2017
18. Dec 21, 2012

A.T.

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

19. Dec 21, 2012

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.

20. Dec 21, 2012

xAxis

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