Let's say there's a lever with 3 forces acting on it:

1 2 3
V_________________________V____________V
/\

let F1, F2, & F3 be the forces and L1, L2, & L3 be the distances of
those forces from the fulcrum.

Then
F1*L1 = F2*L2 + F3*L3

Simple enough. Now, let's say that F1 is a fixed force (e.g., a weight
sitting on the lever). Then F2 & F3 could be any combination of forces,
as long as it balances.

Now say that points 2 & 3 are fixed. I.e., the lever is "tied down" at
those points. What are the forces acting on those tie downs? Is the
torque (moment of inertia, if you will) attributable to each equal, so
that the forces are inversely proportional to the distances? I.e.:
F2*L2 = F3*L3 (= 1/2 F1*L1)

It seems intuitively so, but if it is, how do you show that it is?

This has been bugging me for days, so any resolution that you can offer
will be greatly appreciated.

Bob

On Fri, 05 Jul 2013 17:47:20 -0400, Bob Engelhardt
wrote:

Let's say there's a lever with 3 forces acting on it:

1 2 3
V_________________________V____________V
/\

let F1, F2, & F3 be the forces and L1, L2, & L3 be the distances of
those forces from the fulcrum.

Then
F1*L1 = F2*L2 + F3*L3

Simple enough. Now, let's say that F1 is a fixed force (e.g., a weight
sitting on the lever). Then F2 & F3 could be any combination of forces,
as long as it balances.

Now say that points 2 & 3 are fixed. I.e., the lever is "tied down" at
those points. What are the forces acting on those tie downs? Is the
torque (moment of inertia, if you will) attributable to each equal, so
that the forces are inversely proportional to the distances? I.e.:
F2*L2 = F3*L3 (= 1/2 F1*L1)

It seems intuitively so, but if it is, how do you show that it is?

This has been bugging me for days, so any resolution that you can offer
will be greatly appreciated.

Bob

If 2 and 3 are fixed the problem is "statically indeterminate." In
other words it can't be solved with simple algebra; the elastic
properties of the lever must be considered. In a structural context
the lever would be considered a continuous beam.

--
Ned Simmons

"Bob Engelhardt" wrote in message
...
Let's say there's a lever with 3 forces acting on it:

1 2 3
V_________________________V____________V
/\

let F1, F2, & F3 be the forces and L1, L2, & L3 be the distances of
those forces from the fulcrum.

Then
F1*L1 = F2*L2 + F3*L3

Simple enough. Now, let's say that F1 is a fixed force (e.g., a
weight sitting on the lever). Then F2 & F3 could be any combination
of forces, as long as it balances.

Now say that points 2 & 3 are fixed. I.e., the lever is "tied down"
at those points. What are the forces acting on those tie downs? Is
the torque (moment of inertia, if you will) attributable to each
equal, so that the forces are inversely proportional to the
distances? I.e.:
F2*L2 = F3*L3 (= 1/2 F1*L1)

It seems intuitively so, but if it is, how do you show that it is?

This has been bugging me for days, so any resolution that you can
offer will be greatly appreciated.

Bob

A real beam will bend between F1 and F2, and pull away from F3.

http://en.wikipedia.org/wiki/Statically_indeterminate

jsw

On 7/5/2013 4:47 PM, Bob Engelhardt wrote:
Let's say there's a lever with 3 forces acting on it:

1 2 3
V_________________________V____________V
/\

let F1, F2, & F3 be the forces and L1, L2, & L3 be the distances of
those forces from the fulcrum.

Then
F1*L1 = F2*L2 + F3*L3

....

Now say that points 2 & 3 are fixed. I.e., the lever is "tied down" at
those points. What are the forces acting on those tie downs? Is the
torque (moment of inertia, if you will) attributable to each equal, so
that the forces are inversely proportional to the distances? I.e.:
F2*L2 = F3*L3 (= 1/2 F1*L1)

It seems intuitively so, but if it is, how do you show that it is?
....

It's a moment and force balance, yes, but torque isn't moment of inertia.

To help visualize, redraw simply replacing F2/F3 w/ ropes tied to the
ground, say, and their resulting tensions...

|
| 1 2 3
V_______________________________________
/\ | |
/ \ | |
----------------------V----------V-
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX


Besides the above, there's the balance of forces that

F1 = F2+F3

Substitute for F1 above gives

(F2+F3)*L1 = F2*L2 + F3*L3
F2L1 + F3L1 = F2L2 + F3L3

collecting terms in F1, F3

F2(L1-L2) = F3(L3-L1)

F2/F3 = (L3-L1)/(L1-L2)

So their ratio is the ratio of the overall spacing...

--

"Bob Engelhardt" wrote in message
...
Let's say there's a lever with 3 forces acting on it:

1 2 3
V_________________________V____________V
/\

let F1, F2, & F3 be the forces and L1, L2, & L3 be the distances of those
forces from the fulcrum.

Then
F1*L1 = F2*L2 + F3*L3

Simple enough. Now, let's say that F1 is a fixed force (e.g., a weight
sitting on the lever). Then F2 & F3 could be any combination of forces,
as long as it balances.

Now say that points 2 & 3 are fixed. I.e., the lever is "tied down" at
those points. What are the forces acting on those tie downs? Is the
torque (moment of inertia, if you will) attributable to each equal, so
that the forces are inversely proportional to the distances? I.e.:
F2*L2 = F3*L3 (= 1/2 F1*L1)

It seems intuitively so, but if it is, how do you show that it is?

This has been bugging me for days, so any resolution that you can offer
will be greatly appreciated.

Bob

The answer is that there is nothing that fundamentally sets the ratio of
tensions at the two tie-downs. By lengthing one tie-down a smidgen, you
reduce the tension at that point and increase it at the other.

So then you will ask, "what if the lever is infinitely stiff as are both
tie-downs, and both tie downs have exactly equal length?" This is the same
class of question as "what happens when an infinite force meets an infinite
mass?" When you talk about a non-real world, you have to make up your own
rules. Depending on the rules of physics you create for infinite stiffness,
you might say there is no tension at 3 since it has all been relieved at
point 2.

On 7/5/2013 6:00 PM, Ned Simmons wrote:

If 2 and 3 are fixed the problem is "statically indeterminate." ... the elastic
properties of the lever must be considered....

OK. Or, make the tie downs less absolute. See my reply to myself. Bob

On 7/5/2013 6:09 PM, Jim Wilkins wrote:

A real beam will bend between F1 and F2, and pull away from F3.
...

Oh yeah - not what you'd first expect. Thanks. Bob

On 7/5/2013 6:13 PM, dpb wrote:
[another drawing that didn't quote right]

Besides the above, there's the balance of forces that

F1 = F2+F3
....

I don't think so ... consider if L2 & L3 were much greater than L1.
Then F2 & F3 would be much less than F1 & still balance.

Thanks anyhow,
Bob

On 7/5/2013 6:24 PM, anorton wrote:

The answer is that there is nothing that fundamentally sets the ratio of
tensions at the two tie-downs. ...

So then you will ask, "what if the lever is infinitely stiff as are both
tie-downs, and both tie downs have exactly equal length?" This is the
same class of question as "what happens when an infinite force meets an
infinite mass?"...

Thanks - that makes sense. Instead of making up rules, I need to
re-state the problem less abstractly. More at 11:00. Thanks, Bob

Having been enlightened by knowledgeable replies, let me re-state the
problem.

Instead of points 2 & 3 being perfectly fixed points, let them be
springs, with equal spring constants. So the forces will be
proportional to the deflections at those points. If I still assume a
perfectly rigid lever, then the deflections will be proportional to the
distances and it's solvable (statically determinable, if that's the way
it's put).

If A (alpha) is the angular deflection and k is the spring constant then:

F2 = k * L2 * sinA (for really small A)
F3 = k * L3 * sinA
and F2 = F3*L2/L3
etc

Which is exactly opposite of my original intuition that the forces would
be _inversely_ proportional to the distance.

As k gets really large (and A small, but k * sinA finite, it starts to
look like my original problem of points 2 & 3 fixed. But I suppose that
the limit of k*sinA is not determinate and it's not _exactly_ the same case.

I await your destruction of my "thesis",
Bob

"Bob Engelhardt" wrote in message
...
On 7/5/2013 6:24 PM, anorton wrote:

The answer is that there is nothing that fundamentally sets the
ratio of
tensions at the two tie-downs. ...

So then you will ask, "what if the lever is infinitely stiff as are
both
tie-downs, and both tie downs have exactly equal length?" This is
the
same class of question as "what happens when an infinite force
meets an
infinite mass?"...

Thanks - that makes sense. Instead of making up rules, I need to
re-state the problem less abstractly. More at 11:00. Thanks, Bob

Make a physical model with fishing scales and experiment. I think you
will find that the total of F2 and F3 force* moment is a constant but
either force can be any value between zero and the total. When you
pull one scale harder the other one will back off proportionally. As I
pointed out F3 can even become negative.

If you hang weights from the lever you have an infinite combination of
L2 and L3 that combine to balance F1. Weights don't change their pull
for small lever deflections, making them theoretically simple, the
limiting case of infinite compliance. Limiting cases can be extremely
useful to simplify a problem.

The practical problem with rigid supports is that the force
distribution changes very rapidly and unpredictably as they and the
lever yield. That can be handled to some extent by assuming every
connection has been stressed until it yields enough to pick up its
share, which the designers of the I-35 bridge failed to do correctly.
http://en.wikipedia.org/wiki/File:Gusset.jpg

Here is a theoretical basis:
http://www.cs.cmu.edu/~rapidproto/mechanisms/chpt4.html
Basically you need one and only one restraint on each degree of
freedom to keep the math simple.

In your last post F2*L2 and F3*L3 equal F1*L1, not each other.
jsw

On 07/05/2013 04:41 PM, Bob Engelhardt wrote:
Having been enlightened by knowledgeable replies, let me re-state the
problem.

Snip worse-than-before description of problem

I await your destruction of my "thesis",

You already had your answer the first time you posted this silly problem.

"Then F2 & F3 could be any combination of forces, as long as it balances."

F2 and F3 are any combination that balances the stupid thing. One can
even be weight applied above, while the other is tension from below.


Why don't we make it interesting by turning it all upside down? Then you
might see that there are 4 forces acting on the lever, not 3.

F1, F2, and F3 are floats. The lever is a boat, and you're sitting in it
at the fulcrum.

On Fri, 05 Jul 2013 19:09:24 -0400, Bob Engelhardt
wrote:

On 7/5/2013 6:00 PM, Ned Simmons wrote:

If 2 and 3 are fixed the problem is "statically indeterminate." ... the elastic
properties of the lever must be considered....

OK. Or, make the tie downs less absolute. See my reply to myself. Bob

That *may* make the solution easier, but doesn't make the structure
statically determinate. A bit easier if, for example, you make the
supports springs with a known rate and assume the lever is
sufficiently stiff in comparison with the springs that it can be
treated as a rigid body. Which is the converse of the usual structural
situation where the supports of an elastic beam are assumed to be
perfectly rigid. If neither of those assumptions applies, in other
words, you have to account for the elasticity of both the lever and
supports, the problem is somewhat more complicated than your original.

--
Ned Simmons

"Ned Simmons" wrote in message
...
That *may* make the solution easier, but doesn't make the structure
statically determinate. A bit easier if, for example, you make the
supports springs with a known rate and assume the lever is
sufficiently stiff in comparison with the springs that it can be
treated as a rigid body. Which is the converse of the usual
structural
situation where the supports of an elastic beam are assumed to be
perfectly rigid. If neither of those assumptions applies, in other
words, you have to account for the elasticity of both the lever and
supports, the problem is somewhat more complicated than your
original.
Ned Simmons

A valid simplification to demonstrate this is to make L2 = L3 and the
springs fishing scales, the ones you keep in the vehicles to check
weight at the scrapyard. You can determine the spring constant with
the built-in tape measure. Mine extend 1" at 45 lbs.
http://assets.academy.com/mgen/10/10...jpg?is=500,500

jsw