Attraction forces through an object and torque

neufneufneuf

Hi,

I would like to know if attraction forces through an object change the torque ? I place a vacuum object (or gas under very low pressure) in a liquid, helium for example. All attraction forces from helium to helium is canceled by itself. Except when there is an object between atoms, I think:

Gravity is perpendicular to the screen.










Atom attract another atom: f1 and f2 forces
These forces give forces on support fs1 and fs2
Support reply and give -fs1 and -fs2
Sum1 is the sum of forces on atom 1
Sum2 is the sum of forces on atom 2

Maybe forces are very small but it's possible to imagine bigger force with repulsive magnets. So what is the force that prevent the object to turn alone ?

Blackpanther95

I remember a little know assumption of rigid body mechanics, which is states that if the forces act along the lines joining the particles they cancel. Remember the position vector goes from the center of mass to both particles. You've drawn in radius 1 and 2 but thats irrelevant.

neufneufneuf

Could you draw forces for explain please ?

Blackpanther95

I'm not sure if its possible to draw. (i hope you know what the cross product is)
according to you (and you are correct) the rate of change of the angular momentum is
r1 x F12 + r2 x F21 where F12 is the force 1 exerts on 2 (similarly for F21) and r1 is the position of 1 relative to the centre of mass
and x denotes the cross product
according to Newton's third law, F12 = -F21, so we have
r1 x F12 - r2 x F12 = (r1-r2) x (F12) now if you know much about the cross product, a x b = 0 , precisely when a is parallel to b ,
so going back to the original problem if the force between 1 and 2 acts along the line that joins 1 to 2 , then the cross product
(r1-r2) x (F12) = 0, hence the torque caused by certain internal forces is negligible (and most of the time zero). To get a non-zero impact on the net torque you would need the force on 1 by 2 to be at an angle (which doesn't happen here)
(if you didn't know what the cross product is think of it as like vector multiplication, that produces another vector perpendicular to the first two)

neufneufneuf

the cross product is a.b.sin(a,b) here sin(a,b) is not 0 for me, could you explain please how you have sin = 0 ?

Blackpanther95

neufneufneuf wrote: »

the cross product is a.b.sin(a,b) here sin(a,b) is not 0 for me, could you explain please how you have sin = 0 ?

it is, a is r1-r2, i.e. a vector which starts at 2 and ends at 1, and b is a vector which is F1, think about the angle between these. its even in your drawings,
basically the force goes through 1 and 2 so it is parallel to r1-r2

neufneufneuf

I don't understand, if you look my drawing, if drawing is right, the sum of torque from f1 and f2 (vector) is fs1*r2-sum2*r1 (not in vector), could you explain how this can be at 0 ?

Blackpanther95

I don't understand the fs's and the SUM in the diagram, but if you consider the torque as the some of each torque's my derivation is correct. If there are two particles and as long as the electrostatic force between them acts along the line that joins them together (true) then the net torque is zero. Please explain what the sum's are, also do not try to add the two forces together, as this has a non-trivial impact on net torque.

neufneufneuf

I drawn 2 atoms (gravitationnal attraction for example). Atom2 attract atom1: this is force f1, atom1 attract atom2 this is the force f2. White part is vacuum object (theoretical). Atom1 give force fs1 to wall, atom2 give fs2 to wall. fs1 give torque but fs2 not because it's a radial force. Sum of forces on atom1 is sum1, sum of forces on atom2 is sum2. Sum1 don't give torque (radial) but sum2 give torque (in other direction than fs1). So for me the sum of torques is fs1*r2-sum2*r1.

Blackpanther95

ah, so for two isolated particles you understand the net torque is zero, but what you're saying is the wall is the problem, well I guess the answer would then be yes, the torque is affected, but not because of the electrostatic (or gravitational forces) but because of the wall (an external force) so if you want to consider only internal forces omit the wall. On the other hand if the wall is just a particle in the rigid body, then our problem is alot harder, we have to consider 4 particles not just 2 in which case we have to consider the torque on the wall which you appear not to consider. Please let me know if the "wall" is part of the rigid body or is external to the rigid body. Either way I showed that the net torque from internal forces was zero as long as those forces were along the lines joining the 2 particle. SO with the four particles we must simply show that: the six forces between the particle A,B,C,D where B and D are part of the "wall" act along one of six lines AB,AC,AD,BC,BD,CD. You see all you have to do is build up from two particles, in general people consider the impact on net torque caused by Fij and Fji (two forces) which act on two particles: i and j. If (as i did above earlier) we can show that this is zero, then it doesn't matter how we build up our particles so long as the forces act along the lines. What you need to do is label your four atoms and then draw all the forces between (in simplest form i.e don't add them up or put them in component form) them and try to find a force not parallel to the lines. Here is an example, if a man is running in a hollow ball in space, he may rotate the ball, this is because the friction between his shoes and the ball acts perpendicularly and not parallel to the lines joining his shoes and the ball. Hope that answers the question. (REMEMBER: IS THE WALL PART OF THE RIGID BODY)

neufneufneuf

oh yes, with 2 isolated particles (no walls or other particles) I'm agree with you: no torque.

the torque I want to compute is for all the system, no external force in this case. I don't understand with your lines and 4 particles.

Maybe it's easier to understand if we take big balls and theoretical springs (simulate attraction) and walls like perfect, in this case it must be easy to find the solution, no ?

Blackpanther95

Yes looks like I misunderstood from the beginning. Let's clarify
1) The wall is the black line and it alone is the rigid body we are analysing
2) the white is the vacuum inside the rigid body
3) the outside is helium molecules, which have attractive and repulsive (external) forces on the rigid body
4)the rigid body is rotating around a fixed body and not its own axis (although this is irrelevant to the question)

If 3 is accepted the question is boring, so maybe we should assume that they cancel out or are negligible or something cos otherwise they just change the torque no big mistery.

neufneufneuf

3/ where repulsive forces ?

Blackpanther95

neufneufneuf wrote: »

3/ where repulsive forces ?

wikipedia explains it better than i can http://en.wikipedia.org/wiki/Intermolecular_force

neufneufneuf

ah ok, yes

so I'm agree with 3/ but even the torque is small, in helium with no friction this would say the vacuum object turn alone, it's not possible.

I don't find the question is boring, it's very interesting and imagine big balls with theoretical springs, in this case the torque is not negligeable, and the problem easier to resolve, no ?

neufneufneuf

ah ok, yes

so I'm agree with 3/ but even the torque is small, in helium with no friction this would say the vacuum object turn alone, it's not possible.

I don't think the question is boring, it's very interesting and imagine big balls with theoretical springs, in this case the torque is not negligible, and the problem easier to resolve, no ?

Blackpanther95

No i'm saying the resolotion is boring, the Helium can provide an external force, hence obviously a torque QED. However I though you were asking if the Helium didn't provide an external force then is it possible for a torque to be caused by only particle internal to the rigid body question, this question is very good. The answer is No so long as the forces are electrostatic or gravitational in origin. If you think of why a football can 'curve', the ball is kicked so it has angular and linear momentum, in a vacuum it would carry on in a straight line, but in air, the air exerts a force on the ball due to its spinning, this force causes the ball to change velocity continuously. Here the air has provided a torque, but this is trivial because the force is EXTERNAL so it doesn't violate any principles, I hope your queries are resolved.

dlouth15

What happened to the attached image in the first post of this thread? Is there a timeout on attachments?

neufneufneuf

Sorry, I suppressed my account on wireshack and all images had destroyed. I don't knew. I post image in attached mode like that it would in forum forever.