**Prefer a different theme?**

Just **Click Here** and choose a style.

For an explanation, see **this post**.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter InuyashaX
- Start date

Here is an IEEE document that I just found on the idea. I was pleasantly surprised that someone did some work already on the idea I had.

So, my homework or project is to figure out how to do this with screenshots on FTV.

http://ieeexplore.ieee.org/iel5/9605/30345/01394329.pdf

http://ieeexplore.ieee.org/iel5/9605/30345/01394329.pdf

Bad link daft, for some reason, it wont open for me

It does open a link to another page, but you need a username and password.

It is hard to read this picture and the first one was un-readable!

====================

'''''''''''''''''''''''''''''|

'''''''A-----------------------G

'''''''''''''''''''''''''''''''''''''''''''''''''|

''''''''''''''''''''''H-------------------------B

'''''''''''''''''''''''|

'''''''''''D-----------C

I assume all horizontal strings are of equal length and attached to the centroid (center of mass) of every fish.

The sum of moments on the A-G rod needs to be 0.

The moment generated by A is Ma = Fa * 120 = 3.4*120=408 N*mm.

So the moment at G needs to be Mg = -408 N*mm in order to get Ma + Mg = 0

Just calculate the sum of moments at G to get the weigth of D.

Mg = -408 = Mg - Mb = (Md - Mc) - Mb = ( [Fd*70] - Mc ) - Mb = ...

Calculation of sum of moments is easy and gets you out of nearly any static equilibrium problem. Just make to use the right units. For instance, here notice I used mm everywhere. So the moments are in M*mm. Be carefull not to mix meters in there.

Believe it or not this was fun to do and allowed me to put my mind to something else than the maths I am doing right now.

Just calculate the sum of moments at G to get the weigth of D.

Mg = -408 = Mg - Mb = (Md - Mc) - Mb = ( [Fd*70] - Mc ) - Mb = ...

What??? so what you're saying is

-408 -20*5.4 = D*70 - 20*2.0 -20*5.4 ???

which'd give 1.428571...N ??

(units are not included to prevent further mess and star is used instead of multipication sign.)

Just calculate the sum of moments at G to get the weigth of D.

Mg = -408 = Mg - Mb = (Md - Mc) - Mb = ( [Fd*70] - Mc ) - Mb = ...

Oups, little typo, I meant

Mg = -408 =

But that is wrong. See my comment under here.

-408 -20*5.4 = D*70 - 20*2.0 -20*5.4 ???

Actually, I forgot something. The 60mm rod needs to be accounted for. I calculated the moments about 2 points (G and H). I can't sum them up like that. Scratch what I just said in my above post

If you assume the length of the

That is what I was supposed to do and many suggested. That is, the sum of the moments of right hand side will equal to left hand side. However, I fail to understand this theory.

Are you simply saying:

Ma = Mb + Mc + Md??

Ma = Mb + Mc + Md??

Watch out for the signs in moments calculation. The sign is usally the source of errors.

If you calculate the moment about the attachement to the ceiling, then all point on the left would be positive and all the ones to the right negative.

In other words:

Ma + Md + Mc = Mb

The tricky part to visualize here is that the fish D will, if it goes down, push the road near B wich will in turn make fish A go down even. In other words, fish D tends to make the mobile rotate counter clockwise.

Try imagining the mobile with hard pieces of wood vertically instead of strings.

If you calculate the moment about the attachement to the ceiling, then all point on the left would be positive and all the ones to the right negative.

In other words:

Ma + Md + Mc = Mb

So is that it? does Mb really equal to Ma + Md + Mc?

I thought it was more like

Mb = Ma + Md - Mc

Note how Mb and Mc, both are clock wise and Ma and Md are counter-clock wise... Was I on the right track?

I thought it was more like

Mb = Ma + Md - Mc

Note how Mb and Mc, both are clock wise and Ma and Md are counter-clock wise... Was I on the right track?

If you use the numbers provided for the rods, C is about 10mm left of the center of attachement to the ceiling. So C will tend to make the system rotate counter clockwise around the point of attachement.

B is the only one to the right. So it is the only one that will tend to make it turn clockwise. What makes it really weird is that D and C are attached to a point to the right. Try redrawing it with a single stick and all fishes in it

-----------+-------------

________|_________

A....D..C................B

Does it help to see it? Keep in mind that moments an be moved around. The "path" from where the force is applied and the point about which it is calculated is not important.

You can calculate one equation for each "level". So you will end up with 3 equation but only a single variable. So, basic multi-equation solving, either the same values solves all 3 equations or the problem is not solvable. You have more contrainst (equation) than variable.

Ma/30=B+C+D

and since I was given the B, C, and A, I was able to solve for D. Thanks for the help though

Yes, you are quite right about that one. I guess my physics are more rusted than I thougth they were!

But you again assume the smaller rods are in equilibrium (that their sum of moments is zero). Nothing wrong with this assumption since this is a text book problem. Not a real world one. (If it was a real world one, you could play with the weigth of the other fishes and with the weigth of the rods to make this work

In that case it'd be something like

Md = Mc

Dividing Mc by the distance of D would give you Fd.

[ 20 * 2 ] / 70 = Fd = Weight of D

However, using that method, you'll reach an answer of.. 0.5714, no where near the answer using the other method, which would be 6.2... I know, this is quite tricky

Raster image: usually rectangular grids of pixels making an image look not very smooth when resizing, thus the effects of bluriness in computer images.

Vector image: they are somewhat the opposite of raster images. When creating a vector image, it creates smoother images so that you don't see the rectangular grids of pixels when resizing to any form, size, or shape.

I know it was a very good description, but I hope this helps somehow or clarify things if you didn't understand. By the way, you can try this in photoshop I believe and you can see the difference.