| Topic: |
Science > Physics |
| User: |
"hetware" |
| Date: |
30 May 2007 02:11:49 PM |
| Object: |
Virtual Work /w Unequal-Arm Ballance Problem |
Here's the problem:
"Use the principle of virtual work to establish the formula for an
unequal-arm balance: w1 L1=w2 L2 (neglect the weight of the cross-beam)"
Is the following right?
First we write an expression for the potential energy of the system:
U(L1,L2,a) = (L2 w2 - L1 w1) sin(a)
Next we determine how the potential energy changes with the parameters of
the system. Since we want the system to be in equilibrium, we assume \a
remains fixed.
pd U/pd a =0
Changes in the other two parameters involve these forces:
F1 = pd U/pd L1 = -w1 sin(a)
F2 = pd U/pd L2 = w2 sin(a)
A change in L1 will result in work W1 = F1 DL1. Similarly for L2, W2 = F2
DL2.
W1 = F1 DL1 = -w1 DL1 sin(a)
W2 = F2 DL2 = w2 DL2 sin(a)
We want W1+W2 = 0 so
w1 DL1 sin(a) = w2 DL2 sin(a)
Clearly we can toss sin(a).
w1 DL1 = w2 DL2
Since F1 and F2 are linear with respect to displacements along the beam, DL1
and DL2 give exact values for work. We can, therefore, treat them as
finite displacements of arbitrary magnitude with the requirement that the
equation holds. So we replace them with L1 and L2 respectively.
w1 L1 = w2 L2
--
http://www.vho.org/GB/c/DC/gcgvcole.html
http://www.vho.org/GB/Books/dth/
http://www.germarrudolf.com/
http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm
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| User: "Puppet_Sock" |
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| Title: Re: Virtual Work /w Unequal-Arm Ballance Problem |
30 May 2007 02:49:17 PM |
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On May 30, 3:11 pm, hetware <massl...@nutrino.none> wrote:
Here's the problem:
"Use the principle of virtual work to establish the formula for an
unequal-arm balance: w1 L1=w2 L2 (neglect the weight of the cross-beam)"
Is the following right?
As a homework assignment it does not match the input. You are
supposed to use the principle of virtual work. First, state the
principle, then show how you use it to establish the formula.
First we write an expression for the potential energy of the system:
U(L1,L2,a) = (L2 w2 - L1 w1) sin(a)
Next we determine how the potential energy changes with the parameters of
the system. Since we want the system to be in equilibrium, we assume \a
remains fixed.
[snip]
You are already in deep trouble. Virtual work involves virtual
displacement. What displacement could you have if you
assume a is constant?
Go look up a statement of the principle of virtual work.
Then decide what virtual displacement you can have for
an unequal balance beam. Then show what the work is
for such a displacement. Then see what the principle
of virtual work tells you about an equilibrium condition,
and what that implies about the formula.
Socks
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| User: "hetware" |
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| Title: Re: Virtual Work /w Unequal-Arm Ballance Problem |
30 May 2007 03:22:56 PM |
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Puppet_Sock wrote:
On May 30, 3:11 pm, hetware <massl...@nutrino.none> wrote:
Here's the problem:
"Use the principle of virtual work to establish the formula for an
unequal-arm balance: w1 L1=w2 L2 (neglect the weight of the cross-beam)"
Is the following right?
As a homework assignment it does not match the input. You are
supposed to use the principle of virtual work. First, state the
principle, then show how you use it to establish the formula.
"We use the very small imagined motions to apply the law of conservation of
energy." That's what I did.
First we write an expression for the potential energy of the system:
U(L1,L2,a) = (L2 w2 - L1 w1) sin(a)
Next we determine how the potential energy changes with the parameters of
the system. Since we want the system to be in equilibrium, we assume \a
remains fixed.
[snip]
You are already in deep trouble. Virtual work involves virtual
displacement. What displacement could you have if you
assume a is constant?
I showed clearly what displacements I was imagining?
Go look up a statement of the principle of virtual work.
Then decide what virtual displacement you can have for
an unequal balance beam.
Why do I have to move the whole beam?
Then show what the work is
for such a displacement. Then see what the principle
of virtual work tells you about an equilibrium condition,
and what that implies about the formula.
That's what I did.
--
http://www.vho.org/GB/c/DC/gcgvcole.html
http://www.vho.org/GB/Books/dth/
http://www.germarrudolf.com/
http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm
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| User: "Puppet_Sock" |
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| Title: Re: Virtual Work /w Unequal-Arm Ballance Problem |
30 May 2007 04:35:00 PM |
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On May 30, 4:22 pm, hetware <massl...@nutrino.none> wrote:
[snip]
"We use the very small imagined motions to apply the law of conservation of
energy." That's what I did.
The principle of virtual work is different from the law of
conservation
of energy. That's the point. You didn't apply the POVW.
Start with a statement of the POVW. Look it up.
You didn't really apply conservation of energy either. But that's
another issue.
You took the angle of the beam as constant. But that's the
first thing that gets displaced is the angle of the beam.
Socks
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| User: "hetware" |
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| Title: Re: Virtual Work /w Unequal-Arm Ballance Problem |
30 May 2007 06:56:38 PM |
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Puppet_Sock wrote:
On May 30, 4:22 pm, hetware <massl...@nutrino.none> wrote:
[snip]
"We use the very small imagined motions to apply the law of conservation
of
energy." That's what I did.
The principle of virtual work is different from the law of
conservation
of energy. That's the point. You didn't apply the POVW.
Start with a statement of the POVW. Look it up.
I have consulted several resources, and none give a concise statement of the
principle of virtual work which is applicable to this situation. I will,
therefore, provide my own:
The principle of virtual work states that for a system in equilibrium, the
calculated change in potential energy resulting from a virtual displacement
in the configuration space representing the system will be zero.
In my calculation, I made a virtual displacement of the system's
configuration while imposing the requirement that the potential energy
remain unchanged.
You didn't really apply conservation of energy either. But that's
another issue.
I defined the forces in terms of partial derivatives of the potential energy
with respect to distance and in the directions of the virtual
displacements. I then required that the sum of the changes in potential
energy due to these displacement - i.e., the work - be zero.
You took the angle of the beam as constant. But that's the
first thing that gets displaced is the angle of the beam.
It doesn't get displaced if there is no net component of force acting
perpendicular to the beam. I start with the assumption that the system is
in equilibrium. That means the same thing. I then perform a virtual
displacement of the configuration in such a way that the net change in
potential energy is zero.
--
http://www.vho.org/GB/c/DC/gcgvcole.html
http://www.vho.org/GB/Books/dth/
http://www.germarrudolf.com/
http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm
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| User: "Puppet_Sock" |
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| Title: Re: Virtual Work /w Unequal-Arm Ballance Problem |
31 May 2007 08:50:06 AM |
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On May 30, 7:56 pm, hetware <massl...@nutrino.none> wrote:
[snip]
I have consulted several resources, and none give a concise statement of the
principle of virtual work which is applicable to this situation. I will,
therefore, provide my own:
The principle of virtual work states that for a system in equilibrium, the
calculated change in potential energy resulting from a virtual displacement
in the configuration space representing the system will be zero.
Sigh. The POVW does not involve potential energy.
http://en.wikipedia.org/wiki/Virtual_work
[snips]
You took the angle of the beam as constant. But that's the
first thing that gets displaced is the angle of the beam.
It doesn't get displaced if there is no net component of force acting
perpendicular to the beam.
[snips]
Sigh. You do seem to be insisting on doing this wrong.
I think it's time for me to stop talking to you.
Socks
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| User: "hetware" |
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| Title: Re: Virtual Work /w Unequal-Arm Ballance Problem |
31 May 2007 05:25:40 PM |
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Puppet_Sock wrote:
On May 30, 7:56 pm, hetware <massl...@nutrino.none> wrote:
[snip]
I have consulted several resources, and none give a concise statement of
the
principle of virtual work which is applicable to this situation. I will,
therefore, provide my own:
The principle of virtual work states that for a system in equilibrium,
the calculated change in potential energy resulting from a virtual
displacement in the configuration space representing the system will be
zero.
Sigh. The POVW does not involve potential energy.
Please send your corrections of I:4.2 "Gravitational potential energy" to
the maintainers of the errata:
http://www.feynmanlectures.info/errata.html
"We use the very small imagined motions to apply the law of conservation of
energy."
I now have 2 corrections to my name. And one to my nym.
http://en.wikipedia.org/wiki/Virtual_work
Like the other guy said, that leaves something to be desired. Furthermore,
if I /were/ to use it, I could surely justify my approach through the
definition given.
[snips]
You took the angle of the beam as constant. But that's the
first thing that gets displaced is the angle of the beam.
It doesn't get displaced if there is no net component of force acting
perpendicular to the beam.
[snips]
Sigh. You do seem to be insisting on doing this wrong.
I think it's time for me to stop talking to you.
Socks
I don't see how my approach is wrong. Of course, my first approach to the
problem was to consider increments of work resulting from rotating the beam
by a small change in \a. It seemed to me, however, that there was another
way of thinking about it, and that is what led to the solution presented in
this thread.
--
http://www.vho.org/GB/c/DC/gcgvcole.html
http://www.vho.org/GB/Books/dth/
http://www.germarrudolf.com/
http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm
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| User: "hetware" |
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| Title: Re: Virtual Work /w Unequal-Arm Ballance Problem |
30 May 2007 04:01:20 PM |
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hetware wrote:
Here's the problem:
"Use the principle of virtual work to establish the formula for an
unequal-arm balance: w1 L1=w2 L2 (neglect the weight of the cross-beam)"
Is the following right?
First we write an expression for the potential energy of the system:
U(L1,L2,a) = (L2 w2 - L1 w1) sin(a)
Next we determine how the potential energy changes with the parameters of
the system. Since we want the system to be in equilibrium, we assume \a
remains fixed.
pd U/pd a =0
Changes in the other two parameters involve these forces:
F1 = pd U/pd L1 = -w1 sin(a)
F2 = pd U/pd L2 = w2 sin(a)
A change in L1 will result in work W1 = F1 DL1. Similarly for L2, W2 = F2
DL2.
W1 = F1 DL1 = -w1 DL1 sin(a)
W2 = F2 DL2 = w2 DL2 sin(a)
We want W1+W2 = 0 so
w1 DL1 sin(a) = w2 DL2 sin(a)
Clearly we can toss sin(a).
w1 DL1 = w2 DL2
Since F1 and F2 are linear with respect to displacements along the beam,
Actually, I should have said that they are constant. What I was thinking is
that the potential energy is a linear function of displacement along the
beam.
DL1
and DL2 give exact values for work. We can, therefore, treat them as
finite displacements of arbitrary magnitude with the requirement that the
equation holds. So we replace them with L1 and L2 respectively.
w1 L1 = w2 L2
--
http://www.vho.org/GB/c/DC/gcgvcole.html
http://www.vho.org/GB/Books/dth/
http://www.germarrudolf.com/
http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm
.
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| User: "blackhead" |
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| Title: Re: Virtual Work /w Unequal-Arm Ballance Problem |
31 May 2007 04:53:38 PM |
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On May 30, 10:01 pm, hetware <massl...@nutrino.none> wrote:
hetware wrote:
Here's the problem:
"Use the principle of virtual work to establish the formula for an
unequal-arm balance: w1 L1=w2 L2 (neglect the weight of the cross-beam)"
Is the following right?
First we write an expression for the potential energy of the system:
U(L1,L2,a) = (L2 w2 - L1 w1) sin(a)
Next we determine how the potential energy changes with the parameters of
the system. Since we want the system to be in equilibrium, we assume \a
remains fixed.
pd U/pd a =0
Changes in the other two parameters involve these forces:
F1 = pd U/pd L1 = -w1 sin(a)
F2 = pd U/pd L2 = w2 sin(a)
A change in L1 will result in work W1 = F1 DL1. Similarly for L2, W2 = F2
DL2.
W1 = F1 DL1 = -w1 DL1 sin(a)
W2 = F2 DL2 = w2 DL2 sin(a)
We want W1+W2 = 0 so
w1 DL1 sin(a) = w2 DL2 sin(a)
Clearly we can toss sin(a).
w1 DL1 = w2 DL2
Since F1 and F2 are linear with respect to displacements along the beam,
Actually, I should have said that they are constant. What I was thinking is
that the potential energy is a linear function of displacement along the
beam.
DL1
and DL2 give exact values for work. We can, therefore, treat them as
finite displacements of arbitrary magnitude with the requirement that the
equation holds. So we replace them with L1 and L2 respectively.
w1 L1 = w2 L2
--http://www.vho.org/GB/c/DC/gcgvcole.htmlhttp://www.vho.org/GB/Books/dth/http://www.germarrudolf.com/http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm- Hide quoted text -
- Show quoted text -
The Wikipedia link is terrible. Try this link for an explanation of
virtual work:-
<http://ocw.mit.edu/NR/rdonlyres/Aeronautics-and-Astronautics/
16-61Aerospace-DynamicsSpring2003/D453E02B-5218-4154-8531-DB35ECD76A6C/
0/lecture9.pdf>
The principle states that for a system of particles in equilibrium
where the forces of constraint do no work, then the total work done by
the applied forces in moving through an arbitary, virtual displacement
satisfying the constraints is zero.
So in this problem, you have the constraints that the particles of the
lever must maintain their distance wrt one another, and you can only
rotate the lever about the support.
Rotating the lever an angle dtheta about the support gives rise to two
displacements at F1 and F2 respectively dl1 = l1dtheta, dl2 =
l2dtheta.
The principle then gives F1.l1dtheta = F2.l2.dtheta, so F1.l1 =
F2.dl2
i.e. W1xL1 = W2xL2
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| User: "hetware" |
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| Title: Re: Virtual Work /w Unequal-Arm Ballance Problem |
31 May 2007 07:34:45 PM |
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blackhead wrote:
On May 30, 10:01 pm, hetware <massl...@nutrino.none> wrote:
hetware wrote:
Here's the problem:
"Use the principle of virtual work to establish the formula for an
unequal-arm balance: w1 L1=w2 L2 (neglect the weight of the
cross-beam)"
Is the following right?
First we write an expression for the potential energy of the system:
U(L1,L2,a) = (L2 w2 - L1 w1) sin(a)
Next we determine how the potential energy changes with the parameters
of
the system. Since we want the system to be in equilibrium, we assume
\a remains fixed.
pd U/pd a =0
Changes in the other two parameters involve these forces:
F1 = pd U/pd L1 = -w1 sin(a)
F2 = pd U/pd L2 = w2 sin(a)
A change in L1 will result in work W1 = F1 DL1. Similarly for L2, W2 =
F2 DL2.
W1 = F1 DL1 = -w1 DL1 sin(a)
W2 = F2 DL2 = w2 DL2 sin(a)
We want W1+W2 = 0 so
w1 DL1 sin(a) = w2 DL2 sin(a)
Clearly we can toss sin(a).
w1 DL1 = w2 DL2
Since F1 and F2 are linear with respect to displacements along the
beam,
Actually, I should have said that they are constant. What I was thinking
is that the potential energy is a linear function of displacement along
the beam.
DL1
and DL2 give exact values for work. We can, therefore, treat them as
finite displacements of arbitrary magnitude with the requirement that
the
equation holds. So we replace them with L1 and L2 respectively.
w1 L1 = w2 L2
--http://www.vho.org/GB/c/DC/gcgvcole.htmlhttp://www.vho.org/GB/Books/dth/http://www.germarrudolf.com/http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm-
Hide quoted text -
- Show quoted text -
The Wikipedia link is terrible. Try this link for an explanation of
virtual work:-
<http://ocw.mit.edu/NR/rdonlyres/Aeronautics-and-Astronautics/
16-61Aerospace-DynamicsSpring2003/D453E02B-5218-4154-8531-DB35ECD76A6C/
0/lecture9.pdf>
The principle states that for a system of particles in equilibrium
where the forces of constraint do no work, then the total work done by
the applied forces in moving through an arbitary, virtual displacement
satisfying the constraints is zero.
So in this problem, you have the constraints that the particles of the
lever must maintain their distance wrt one another,
No I don't!
and you can only
rotate the lever about the support.
That is one way of viewing the problem; the Newtonian method. To insist
that the length of the rod remain fixed, and that only the angle is subject
to change is to miss the meaning of all of analytical dynamics.
Rotating the lever an angle dtheta about the support gives rise to two
displacements at F1 and F2 respectively dl1 = l1dtheta, dl2 =
l2dtheta.
The principle then gives F1.l1dtheta = F2.l2.dtheta, so F1.l1 =
F2.dl2
i.e. W1xL1 = W2xL2
--
http://www.vho.org/GB/c/DC/gcgvcole.html
http://www.vho.org/GB/Books/dth/
http://www.germarrudolf.com/
http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm
.
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| User: "blackhead" |
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| Title: Re: Virtual Work /w Unequal-Arm Ballance Problem |
31 May 2007 08:21:06 PM |
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On Jun 1, 1:34 am, hetware <massl...@nutrino.none> wrote:
blackhead wrote:
On May 30, 10:01 pm, hetware <massl...@nutrino.none> wrote:
hetware wrote:
Here's the problem:
"Use the principle of virtual work to establish the formula for an
unequal-arm balance: w1 L1=w2 L2 (neglect the weight of the
cross-beam)"
Is the following right?
First we write an expression for the potential energy of the system:
U(L1,L2,a) = (L2 w2 - L1 w1) sin(a)
Next we determine how the potential energy changes with the parameters
of
the system. Since we want the system to be in equilibrium, we assume
\a remains fixed.
pd U/pd a =0
Changes in the other two parameters involve these forces:
F1 = pd U/pd L1 = -w1 sin(a)
F2 = pd U/pd L2 = w2 sin(a)
A change in L1 will result in work W1 = F1 DL1. Similarly for L2, W2 =
F2 DL2.
W1 = F1 DL1 = -w1 DL1 sin(a)
W2 = F2 DL2 = w2 DL2 sin(a)
We want W1+W2 = 0 so
w1 DL1 sin(a) = w2 DL2 sin(a)
Clearly we can toss sin(a).
w1 DL1 = w2 DL2
Since F1 and F2 are linear with respect to displacements along the
beam,
Actually, I should have said that they are constant. What I was thinking
is that the potential energy is a linear function of displacement along
the beam.
DL1
and DL2 give exact values for work. We can, therefore, treat them as
finite displacements of arbitrary magnitude with the requirement that
the
equation holds. So we replace them with L1 and L2 respectively.
w1 L1 = w2 L2
--http://www.vho.org/GB/c/DC/gcgvcole.htmlhttp://www.vho.org/GB/Books/d...
Hide quoted text -
- Show quoted text -
The Wikipedia link is terrible. Try this link for an explanation of
virtual work:-
<http://ocw.mit.edu/NR/rdonlyres/Aeronautics-and-Astronautics/
16-61Aerospace-DynamicsSpring2003/D453E02B-5218-4154-8531-DB35ECD76A6C/
0/lecture9.pdf>
The principle states that for a system of particles in equilibrium
where the forces of constraint do no work, then the total work done by
the applied forces in moving through an arbitary, virtual displacement
satisfying the constraints is zero.
So in this problem, you have the constraints that the particles of the
lever must maintain their distance wrt one another,
No I don't!
By lever, I meant cross-beam. So you don't want to maintain the
distances between the particles that make up the cross-beam?
If you don't, then it means having to know what the internal forces
are. The internal virtual work is zero if you maintain their distance
relative to one another.
and you can only
rotate the lever about the support.
That is one way of viewing the problem; the Newtonian method.
What is the Newton method? This is the Principle of Virtual Work
method.
To insist
that the length of the rod remain fixed, and that only the angle is subject
to change is to miss the meaning of all of analytical dynamics.
I think you will find that the Principle of Virtual Work and the way
it exploits workless constraints lies at the foundations of analytical
mechanics.
I'm not insisting anything, just solving your problem as easily as
possible. How else do you propose to do it in a few lines like I did?
Rotating the lever an angle dtheta about the support gives rise to two
displacements at F1 and F2 respectively dl1 = l1dtheta, dl2 =
l2dtheta.
The principle then gives F1.l1dtheta = F2.l2.dtheta, so F1.l1 =
F2.dl2
i.e. W1xL1 = W2xL2
--http://www.vho.org/GB/c/DC/gcgvcole.htmlhttp://www.vho.org/GB/Books/dth/http://www.germarrudolf.com/http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm- Hide quoted text -
- Show quoted text -
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| User: "hetware" |
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| Title: Re: Virtual Work /w Unequal-Arm Ballance Problem |
31 May 2007 11:07:41 PM |
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blackhead wrote:
On Jun 1, 1:34 am, hetware <massl...@nutrino.none> wrote:
blackhead wrote:
So in this problem, you have the constraints that the particles of the
lever must maintain their distance wrt one another,
No I don't!
By lever, I meant cross-beam. So you don't want to maintain the
distances between the particles that make up the cross-beam?
That's correct.
If you don't, then it means having to know what the internal forces
are.
The total force on any point in a system in equilibrium is zero. It is
assumed that the constraining forces do not change under virtual
displacements.
The internal virtual work is zero if you maintain their distance
relative to one another.
I'm not talking about compressing a physical beam. I am talking about
studying the changes in the system which take place due to virtual
displacements.
and you can only
rotate the lever about the support.
That is one way of viewing the problem; the Newtonian method.
What is the Newton method? This is the Principle of Virtual Work
method.
What I mean by the Newtonian method is that changes to the system are
treated as physical changes. For example, if the length of the beam is
changed there would have to be some force compressing the beam.
To insist
that the length of the rod remain fixed, and that only the angle is
subject to change is to miss the meaning of all of analytical dynamics.
I think you will find that the Principle of Virtual Work and the way
it exploits workless constraints lies at the foundations of analytical
mechanics.
Exactly my point.
I'm not insisting anything, just solving your problem as easily as
possible. How else do you propose to do it in a few lines like I did?
I could certainly reduce the number of lines in my solution by combining
some fairly obvious steps.
--
http://www.vho.org/GB/c/DC/gcgvcole.html
http://www.vho.org/GB/Books/dth/
http://www.germarrudolf.com/
http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm
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| User: "blackhead" |
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| Title: Re: Virtual Work /w Unequal-Arm Ballance Problem |
03 Jun 2007 10:00:41 AM |
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On Jun 1, 5:07 am, hetware <massl...@nutrino.none> wrote:
blackhead wrote:
On Jun 1, 1:34 am, hetware <massl...@nutrino.none> wrote:
blackhead wrote:
So in this problem, you have the constraints that the particles of the
lever must maintain their distance wrt one another,
No I don't!
By lever, I meant cross-beam. So you don't want to maintain the
distances between the particles that make up the cross-beam?
That's correct.
If you don't, then it means having to know what the internal forces
are.
The total force on any point in a system in equilibrium is zero. It is
assumed that the constraining forces do not change under virtual
displacements.
Yes, but I'm use the fact that constraints do no work to end up with
just the applied forces and displacements.
At point P1 where F1 is applied, there will be the forces of
constraint F2,...,Fn. How are you going to eliminate these if you
don't ensure the particles maintain their distance relative to one
another?
The internal virtual work is zero if you maintain their distance
relative to one another.
I'm not talking about compressing a physical beam. I am talking about
studying the changes in the system which take place due to virtual
displacements.
You have said that don't want to maintain their relative distances, so
at the very least you are compressing and expanding parts of the beam.
and you can only
rotate the lever about the support.
That is one way of viewing the problem; the Newtonian method.
What is the Newton method? This is the Principle of Virtual Work
method.
What I mean by the Newtonian method is that changes to the system are
treated as physical changes. For example, if the length of the beam is
changed there would have to be some force compressing the beam.
It's choosing the displacements so that the forces of constraint are
eliminated. This is why the principle is so useful. It's true that you
can move each particle independently of one another, and the total
virtual work is zero. It's just not a useful strategy in solving
static problems because yoy have unknown forces of constraint.
To insist
that the length of the rod remain fixed, and that only the angle is
subject to change is to miss the meaning of all of analytical dynamics.
I think you will find that the Principle of Virtual Work and the way
it exploits workless constraints lies at the foundations of analytical
mechanics.
Exactly my point.
The rod length remains fixed, and the angle changes so that I end up
eliminating the internal forces of constraint, and that at the support
of the cross beam. This is the whole point of using the Principle of
Virtual Work.
I'm not insisting anything, just solving your problem as easily as
possible. How else do you propose to do it in a few lines like I did?
I could certainly reduce the number of lines in my solution by combining
some fairly obvious steps.
I'd be interested to see your simplified method so that perhaps I can
learn from it.
http://www.vho.org/GB/c/DC/gcgvcole.htmlhttp://www.vho.org/GB/Books/dth/http://www.germarrudolf.com/http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm
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| User: "hetware" |
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| Title: Re: Virtual Work /w Unequal-Arm Ballance Problem |
04 Jun 2007 12:03:10 AM |
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blackhead wrote:
On Jun 1, 5:07 am, hetware <massl...@nutrino.none> wrote:
blackhead wrote:
Yes, but I'm use the fact that constraints do no work to end up with
just the applied forces and displacements.
At point P1 where F1 is applied, there will be the forces of
constraint F2,...,Fn. How are you going to eliminate these if you
don't ensure the particles maintain their distance relative to one
another?
The constraints of the virtual system are that the weights are positioned at
fixed distances from the ends of the beam. It is implied that the
positions of the weights relative to the ends of the beam are in the
downward vertical direction. The beam is assumed to be straight. The
fulcrum is a fixed point, around which the beam can rotate in the plane of
the diagram.
Any change in the virtual system assumes the existence of workless
constraining forces which maintain that configuration.
You have said that don't want to maintain their relative distances, so
at the very least you are compressing and expanding parts of the beam.
No. I am changing the configuration in virtual space. The statement of the
problem tells me that the lengths l1 and l2 are variable.
It's choosing the displacements so that the forces of constraint are
eliminated. This is why the principle is so useful. It's true that you
can move each particle independently of one another, and the total
virtual work is zero. It's just not a useful strategy in solving
static problems because yoy have unknown forces of constraint.
A physical rotation of the beam is not a virtual displacement. That is a
physical displacement. You can talk about virtual displacements which
correspond to such physical displacements, but virtual displacements are
not confined to physical displacements.
To insist
that the length of the rod remain fixed, and that only the angle is
subject to change is to miss the meaning of all of analytical
dynamics.
I think you will find that the Principle of Virtual Work and the way
it exploits workless constraints lies at the foundations of analytical
mechanics.
Exactly my point.
The rod length remains fixed, and the angle changes so that I end up
eliminating the internal forces of constraint, and that at the support
of the cross beam. This is the whole point of using the Principle of
Virtual Work.
In variational mechanics you are comparing configurations. These
comparisons do not necessarily correspond to physical changes in a real
system. For example, if we have a truss, and we want to know the lateral
force on a particular member, we can virtually lengthen the member and
calculate the resulting change in potential energy of the system. In doing
so, we are implicitly assuming all of the welds holding the truss together
are worklessly adjusting their angles to the new configuration.
I'm not insisting anything, just solving your problem as easily as
possible. How else do you propose to do it in a few lines like I did?
I could certainly reduce the number of lines in my solution by combining
some fairly obvious steps.
I'd be interested to see your simplified method so that perhaps I can
learn from it.
I could eliminate the step of defining the forces, and simply use the
derivative of potential energy directly in the expressions for work. There
is really no need to define W1 and W2 before I setup the work equation. I
could also omit the elimination of sin(a) as a stated step.
Here's the problem:
"Use the principle of virtual work to establish the formula for an
unequal-arm balance: w1 L1=w2 L2 (neglect the weight of the cross-beam)"
The energy of the system is given by:
U(L1,L2,a) = (L2 w2 - L1 w1) sin(a)
Since we want the system to be in equilibrium, we have the condition
(pd U)/(pd a) = 0
[I'm surprised that on one complained that I failed to explicitly require
0<|a|<Pi/2]
as well as
w1 DL1 sin(a) = w2 DL2 sin(a)
Since dy/dx is constant along the beam we replace the variations DL1 and DL2
with L1 and L2 respectively.
w1 L1 = w2 L2
--
http://www.vho.org/GB/c/DC/gcgvcole.html
http://www.vho.org/GB/Books/dth/
http://www.germarrudolf.com/
http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm
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| User: "Bob Cain" |
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| Title: Re: Virtual Work /w Unequal-Arm Ballance Problem |
01 Jun 2007 01:09:49 AM |
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hetware wrote:
The total force on any point in a system in equilibrium is zero.
Say that with your head in a vise. :-)
Bob
--
"Things should be described as simply as possible, but no simpler."
A. Einstein
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| User: "hetware" |
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| Title: Re: Virtual Work /w Unequal-Arm Ballance Problem |
01 Jun 2007 02:07:09 AM |
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Bob Cain wrote:
hetware wrote:
The total force on any point in a system in equilibrium is zero.
Say that with your head in a vise. :-)
Bob
Sounds like the voice of experience. Perhaps if you had worked out the
theoretical part _before_ conducting the experiment, you would understand
the result. Kind of a catch-22 thing, eh?
--
http://www.vho.org/GB/c/DC/gcgvcole.html
http://www.vho.org/GB/Books/dth/
http://www.germarrudolf.com/
http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm
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| User: "Eric Gisse" |
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| Title: Re: Virtual Work /w Unequal-Arm Ballance Problem |
01 Jun 2007 03:50:40 AM |
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On May 31, 5:34 pm, hetware <massl...@nutrino.none> wrote:
[...]
Check out section 9.1, page 361 of Symon for an adequate discussion of
virtual work.
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| User: "hetware" |
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| Title: Re: Virtual Work /w Unequal-Arm Ballance Problem |
01 Jun 2007 04:43:37 AM |
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Eric Gisse wrote:
On May 31, 5:34 pm, hetware <massl...@nutrino.none> wrote:
[...]
Check out section 9.1, page 361 of Symon for an adequate discussion of
virtual work.
I find the treatment in Joos's _Theoretical Physics_ more applicable to this
situation. Symon doesn't really state a "Principle of Virtual Work", per
se. Neither does Joos, but he states a "Principle of Virtual
Displacements" which I take to be the same thing. It appears evident that
I am interpreting this differently than are some other people.
--
http://www.vho.org/GB/c/DC/gcgvcole.html
http://www.vho.org/GB/Books/dth/
http://www.germarrudolf.com/
http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm
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