Conservation of Energy and Virtual Work

1. Aug 26, 2011

MarkFarrell82

Hi,

I've been reading chapter 4 in the Feynman Lectures on the conservation of energy and I've bought a book off amazon called Exercises in Introductory Physics which accompanies the series. Having looked at some of the questions I feel well out of my depth because essentially I haven't ever really done any problem solving. I have got A-level maths but that was years ago. The question I'm stuck with is the first one on the CoE. It goes like this:

1. Use the principle of virtual work to establish the formula for an unequal-arm balance:
W2L2 = W1L1Neglect the weight of the bar.
The diagram is basically a see-saw with a weight hanging from either end of the bar. The distance to the fulcrum of L2 is > L1's.

How do people approach things like that? What are your thought processes? I thought the answer would be the formula stated and you just rearrage to find an unknown.

I think I understand the theory and when I read the notes I follow everything and understand it but as soon as I get a problem I just go blank and get frustrated!

Hope someone can help set my mind free for problem solving.

Thanks
Mark

2. Aug 26, 2011

tiny-tim

Hi Mark!
The question wants you to write the principle of virtual work as an equation (it involves distances, doesn't it, which are not L1 and L2?),

and then to apply that equation to the see-saw.

3. Aug 26, 2011

MarkFarrell82

Hi tiny-Tim,
Thanks for the response. I think I've missed the point completely. If it's asking me to write an equation, then it's asking me to write down a general formula for every case surely? But isn't that what the W2L2=W1L2 formula is? If you make up numbers to put into the formula it's obviously really easy to work out.

The idea as far as I'm aware is to make the see-saw balance. In the examples in the text he where there's a couple of weights (weights given) on a bar balanced by another weight hooked up to a simple pulley system. The idea is to work out what the weight on the pulley is in order to balance the bar so the other two weights done slide off. He makes up the distance the weight on the pulley is and from that works out how much it should weigh. I followed the logic perfectly fine. This question the distances from the fulcrum are just stated as L2 and L1. How come you say they aren't distances? Is it just because they haven't got a quantity or should they be labelled something else in the case of virtual work?

Hopefully you'll see how my minds going. It's late so sorry if it's a bit rambley but I really what to know where my thinking is falling down.

Thanks again
Mark

4. Aug 26, 2011

MarkFarrell82

Just a thought, the equation I'm supposed to write down, should it be something like the sum of the weights x heights should be equal to zero or is that too simplistic?

5. Aug 27, 2011

Studiot

Tiny Tim didn't day that L1 & L2 are not distance he said they are not the distances. By which he meant they are not involved in the virtual work equation.

The system W1L1W2L2 is in equilibrium which menas that a (small) virtual displacement (or roatation) of the whole system does no work.

Tiny Tim should perhaps have said one (virtual) displacement, not plural.

Try applying a virtual displacement and consider the work done by all forces acting.

6. Aug 27, 2011

tiny-tim

Hi Mark!

(just got up :zzz: …)
Yes, that's right, the https://www.physicsforums.com/library.php?do=view_item&itemid=378" in the virtual work equation are the displacements through which the masses might move, in this case vertical.

Last edited by a moderator: Apr 26, 2017
7. Aug 27, 2011

MarkFarrell82

Thanks again for the responses. It's funny how you can lose the plot with things sometimes by focusing on one little but. The whole sub title of the section is Gravitational Energy so the I guess the clue is in the title. I've attached a copy of the section of text I've been mulling over trying to solve this.

Is it as simple as this: W2.H2-W1.H1=0?

From this if you plug numbers in, you can work out the various heights etc each arm has to move for the total to equal zero even if you move the fulcrum about and also obeying the conservation og energy laws.

Anywhere near?!

8. Aug 27, 2011

tiny-tim

erm noooo you haven't!
Yes … sometimes physics questions are simple!

9. Aug 27, 2011

MarkFarrell82

How about now for the attachment?

Attached Files:

• Virtual Work 3.JPG
File size:
58.5 KB
Views:
176
10. Aug 27, 2011

MarkFarrell82

Also Tim, thanks for all your guidance. I prefer to be led than told.

I notice your background is Mathematics. I feel mines rusty, are there any texts you'd recommend from a Physics perspective to help me brush up?

11. Aug 27, 2011

tiny-tim

oooh yes, that's definitely an attachment!
Yes (except you can't move a fulcrum, can you? )

Sorry, don't know.

12. Aug 27, 2011

MarkFarrell82

Fulcrum is the pivot point, right? Or the point around which the forces act? If you change the diagram so the fulcrum was furhter over to the left for example, the equation would still hold?

13. Aug 27, 2011

Studiot

As regards applications textbooks the main users of Virtual Work are Structural Engineers so you will find at least a chapter on the subject and even whole structural engineering texts devoted to the subject.

There is a useful section in Green and Gliddon

General Degree Applied Mathematics

go well

14. Aug 27, 2011

MarkFarrell82

Thanks Studiot

15. Aug 27, 2011

Studiot

What's to remember is that when you do a problem by virtual work you must also apply compatibility.

That is you must select the virtual displacements from those that are possible ie compatible with the geometry of the system.

In the case of your balance a virtual rotation about the fulcrum leads to virtual vertical displacements of the weights, H1 and H2

You must apply compatibility (geometry) to obtain the relationship between the unknown distances H1 and H2 and the known distances L1 and L2

Does this help?

16. Aug 27, 2011

tiny-tim

Yes, fulcrum is just another word for pivot.
The virtual work equation always holds.

Have faith!

17. Aug 27, 2011

Studiot

Not quite.

In order to fully answer the question proceed as follows

Let the fulcrum reaction be R.
Let there be a virtual rotation d$\theta$ (radians) about the fulcrum
So that W1 moves upwards H1 and W2 moves downwards H2
The fulcrum does not move (up or down)

So considering the virtual work done

-W1H1 + W2H2 + R(0) = 0

With a light inextensible support beam and making the approximations H1 = L1d$\theta$ and H2 = L2d$\theta$
since d$\theta$ is small

substituting for H1 and H2

-W1L1d$\theta$ + W2L2d$\theta$ + R(0) = 0

does this help?

18. Aug 27, 2011

MarkFarrell82

Yeah thanks for the solution. It's really helpful. Cheers

"Let the fulcrum reaction be R." Will this fulcrum reaction always be zero in virtual work problems where the overall answer is zero? Is the fulcrum reaction in there to make the answer more complete?

I follow all the working out below and why it's all there, the thing that bothers me is that I would never have thought of that as a route to the answer..

19. Aug 27, 2011

tiny-tim

If it doesn't move, ignore it (in applying the virtual work theorem).

20. Aug 27, 2011

Studiot

Remember this is a simple example to illustrate the principle. It is clearly easier by other methods.

Remember you have to consider all forces acting and all displacements undergone. Just as judicious taking of moments about a particular force eliminates it from the enquiry, clever application of the virtual load or displacement reduces the work.
No the reaction work is not always zero - sometimes this is what we actually want to calculate.
In this case it is expedient to pick virtual rotation that makes the reaction work zero.
Sometimes the reaction work is zero if you can pick a direction that is always at right angles to it, for instance doing inclined plane problems by virtual work.

21. Aug 29, 2011

codelieb

Mark,

(I am cross-posting this from The Feynman Lectures Forum, for the benefit of the users of this forum. The orginal post can be found http://feynmanlectures.info/forum/index.php?topic=46.msg85#msg85" [Broken].)

First, you need to understand what is being asked. The problem asks you to use the principle of virtual work to demonstrate that for the unequal-arm balance shown in the figure, W2*L2 = W1*L1, The answer to this is not a formula or an equation, but a demonstration, or proof that W2*L2 = W1*L1.

Second, you need to understand the principle of virtual work. When applied to a system such as this one, that is in static equillibrium, the principle of virtual work says that if you make a small displacement to the system, the net work done will be zero.

So, imagine that the see-saw starts level and balanced (and not moving). The only forces on the system are from gravity. These forces are vertical, with magnitudes W1 = M1*g and W2 = M2*g, where M1 and M2 are the masses of the respective weights and g is the acceleration due to gravity. (Note: a "weight" in physics is generally a force = mass time gravitational acceleration, and not a mass). Recall that the work done in displacing a mass a displacement d against a force F is F.d. Here F and d are vectors, and F.d is their dot product - if the force and displacement happen to be in the same direction, this just equals the product of their magnitudes F*d.

So, let us imagine that we make a small displacement of the system. Say we pull down on W1 so that it goes down a small distance -h1, while W2 is lifted up a distance h2 (with h1>0 and h2>0). The work done in displacing W1 downward is W1*(-h1) = -W1*h1. It's negative, which in this case (because of the sign conventions I have adopted) means that the force of the (falling) weight does some work - it is falling, so it gives up its gravitational potential energy (where does that energy go?). Meanwhile, the work done in displacing W2 upward is W2*h2, which is positive, meaning that some work has to be done against gravity to lift the weight, thereby giving it some additional gravitational potential energy (where did it come from?). The total work done in making these displacements is the sum of the work done on each weight, which is h2*W2 - h1*W1. The principle of virtual work tells us that

(1) h2*W2 - h1*W1 = 0.

Now... let us look at the _geometry_ of the system. You can see (by similar triangles) that

(2) h1/L1 = h2/L2,

or h1 = h2*(L1/L2), which we substitute into (1):

(3) h2*W2 - (h2*(L1/L2))*W1 = 0.

Dividing (3) by h2 and rearranging gives you

(4) W1*L1 = W2*L2, which is what we are trying to show.

QED

Last edited by a moderator: May 5, 2017
22. Aug 29, 2011

Studiot

So is your see-saw hanging in mid air by means of magic instead of a support reaction?

The similar triangles version is a good alternative proof however, possibly better than mine.

go well

23. Aug 29, 2011

codelieb

I consider the fulcrum to be part of the see-saw system, so that the reaction force between the fulcrum and the beam and those between the beam and the weights, are not forces on the system, but between parts of it.

24. Aug 29, 2011

Studiot

That's just playing with words.

The system, however you describe it, must be resting on something.

Without the reaction force you cannot claim the system is in vertical equilibrium.

It is always necessary to consider all forces acting before discounting those that do no work.

25. Aug 29, 2011

tiny-tim

i'm with codelieb here

you might as well say that the longitudinal tension in the see-saw also has to be considered …

there undoubtedly is tension, or the see-saw would fall apart

but nobody does that, since it's an internal force (or as i usually prefer to say, the result of a geometrical constraint) …

why should the reaction at the fulcrum (the result of the geometrical constraint that the fulcrum is stationary relative to the Earth) be treated any differently?