# Conservation of Energy and Virtual Work

1. ### MarkFarrell82

16
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. ### tiny-tim

26,053
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. ### MarkFarrell82

16
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. ### MarkFarrell82

16
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. ### 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. ### tiny-tim

26,053
Hi Mark!

(just got up :zzz: …)
Yes, that's right, the displacements in the virtual work equation are the displacements through which the masses might move, in this case vertical.

7. ### MarkFarrell82

16
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. ### tiny-tim

26,053
erm noooo you haven't!
Yes … sometimes physics questions are simple!

9. ### MarkFarrell82

16
How about now for the attachment?

File size:
58.5 KB
Views:
34
10. ### MarkFarrell82

16
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. ### tiny-tim

26,053
oooh yes, that's definitely an attachment!
Yes (except you can't move a fulcrum, can you? )

Sorry, don't know.

12. ### MarkFarrell82

16
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. ### 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. ### MarkFarrell82

16
Thanks Studiot

15. ### 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. ### tiny-tim

26,053
Yes, fulcrum is just another word for pivot.
The virtual work equation always holds.

Have faith!

17. ### 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. ### MarkFarrell82

16
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. ### tiny-tim

26,053
If it doesn't move, ignore it (in applying the virtual work theorem).

20. ### 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.