DandelionTheory

if i set up a wire loop J with a current I flowing clockwise, and attach wires AB, BC, CD & DG as seen in the picture the current on wires AB, BC, CD & DG i would expect to be 0.5I while the portion of loop J (portion AC and CG of wire loop J)adjacent to wires AB, BC, CD & DG also carries 0.5I my question: are the forces between the current in each wire and current loop J attractive to each other? and are the forces between the currents in the wires repulsive or attractive?

Condescension does not promote confidence. pretend i know the basics about inline masses.

Can we agree my example has nothing inline masses? Right angles buddy Also read the part about rigid connection mentioned, seems you think C would "move closer" to A regardless of the rigid body between them. Am I correct to assume you forgot it wasn't a rope?

How do I do it? If you take out mass A entirely it's opposite still, if you take mass C out instead it's opposite.

got you, so if i calculate over time i would need to account for this. i attempted to use variables in order of magnitude to help with this problem, also i add force at certain times due to the units being in newtons/kg

F1 is done to mass A by mass B, F2 is the opposite reaction mass B feels, F3 is the opposite to F2 due to mass B & C being connected.

may i ask if this partains to the slight thousandths of a unit distance the 3 weight system seem to do?

if you have 3 masses A, B & C A & C are rigidly attached to each other B Pivots around C via a rigid bearing. If a force on any of these masses is to be represented correctly, opposite forces need to be shown for every force applied. so if mass A acts on Mass B(represented by F1 and F2), the work done on mass B will be done oppositely to Mass C (F3)correct? see picture below.

woopsie

Yes, I'll ask you to check which one is the parent object. As the mass rotates relative to the pins position, the proper way to describe "down" to a 180°rotated object is with a negative vector. The idea is to change the Mass's direction and apply force to the center pin, the variables were entered into the computer to be stimulated by a physics engine. So I have to explain the phenomenon in exact words before it is allowed to be posted on a physics forum in speculation? That's absurd, what if I don't know how it works? How would I bring forth the phenomenon to your attention? PS, do some work too.

Now you're getting it. Okay, did you assume I made that mistake? Or did you not read or look at anything I claim? Because you keep bringing up how I'm missing something, yet your argument is with the engine not the method. Elaborate... Opposite forces can be represented with unity if it is set up correctly, right? Are they set up correctly? Or did I miss a reaction force? Tell me. Don't argue language when you don't know the sentence structure. Also, downvotes are the only thing on this forum open for opinion. Or did you forget that

Thank you. Can you be more specific? I do want to point out I wasn't trying to show perpetual motion or some nonsense. It just shifts it's gravity.

Apologies, it's something I've been working on currently due to the need for specificity. https://forum.unity.com/threads/finding-the-center-of-mass-of-a-collection-of-rigidbodies.87578/ showed me how to do it.

I never said it was perpetual motion. I said it made the center of gravity shift.

So solar panels to motor to masses right? I wrote a calculation for the center of mass. I've seen the same kind of motion with one mass being forced upon about the pin in one direction, and another script with equal but opposite force on the center pin. I'm really trying to make it not work... Gotcha, these are swinging weights, one with an opposing direction. Maybe looking at the center of gravity shifting would give a clue.

In the physics engine. I want a good representation of reality I can put numbers into, that won't argue back with opinion.

then how do you do it?

this is an example from the script: Right.gameObject.rigidbody.AddRelativeForce (0,-holder,0); Pin.gameObject.rigidbody.AddRelativeForce(0,holder,0); how else would i represent "equal but opposite"?

Yay, now look to the script and ask yourself how I represented an equal but opposite force. Maybe I should mention the lines connecting the masses are rigid and rotate around the common mass. I hope this also answers swansont's question about what a pin is. A "pin" in this scenario is the common connected mass to all other masses by a rigid body with the pin as the pivot point.

fig1 represents how the mass variable of F1 is not the same as the mass variable in F2. F1 = F2 F1 = 1kg * a F2 = 2kg * a while fig2 represents 3 opposite but equal reactions on 4 bodies. the pin is the wire I1 travels through. per unit length

yes it does. I1 is a current. i said: , but i assumed you knew that I1 meant a current and B1 was its magnetic field. if I1 and I2 were currents, B1 and B2 would represent a possible magnetic field they produce, if thr red box was a core material.

I1 is into the page, B1 is the magnetic field direction (green arrow) due to I1. I2, which is around a magnetic core material(red box), induces B2 in the core material(red box) in the direction indicated by the green arrows within the red box. of course it does, its just some of the masses are not aligned with the force vectors when they are applied, and therefore the application of F=m*a changes based on angle. you didnt watch the video...?

right, with 2 interacting no net motion is seen. with 2 weights and a pin no net motion is seen; but with 3 weights and a pin, the other mass slings around, at some angles in the cycle the center pin and another weight are inline with another weight's interacting force, that would make the interacting masses 2kg and 1kg, but while the line between the initial mass and the pin are 90 degrees from the acting 3rd force the interacting masses are 1kg and 1kg

I1 and I2 acting on eachother. they will attempt to spin around eachother, the pin allows the weight to roll around it in this case. working on a proof of concept...

one weight(I2) and pin(I1) C= center pin, (i made it a while ago)