I found a very easy and intuitive way to solve this by considering the constant relation:
Atotal = Aforce + Atorque
Aforce = F / m
Atorque = F * d * d / I
now it is easy to work out the force at the contact point
I understand placing one of the forces at CM - that is a great trick :-)
But I don't understand your above statement.
Surely an off-center torque is always equivalent to a torque about CM PLUS a translation through CM? As the CM is accelerating?
And isn't it possible that the forces in a...
take a rigid body with CM at (0,0)
there are equal 2 torques: 1 about (4, 1) and 1 about (5, 1)
how would I determine the single torque (about 0,0) and the single translation (though 0,0) ?
take a rigid body with CM at (0,0)
there are equal 2 torques: 1 about (4, 1) and 1 about (5, 1)
how would I determine the single torque (about 0,0) and the single translation (though 0,0) ?
hi
I have multiple torques acting at different points on a rigid body
I want to find the resultant force and torque (about the body CM)
I think I can see how to "convert" an off-center torque to a torque + translation using the parallel axis theorem - but am wondering if there is a...
all this is deprecated now, as explained by this PDF:
http://syrte.obspm.fr/iau/iauWGnfa/wallace_com8_ga06.pdf"
the reason I could find no answer - is because there is no answer
Hi
I can't find a clear answer to this anywhere.
Can anybody confirm that the ecliptic plane for J2000.0 intersects these 3 points:
a) center of mass of sun @ J2000.0
b) center of mass of Earth @ J2000.0
c) center of mass of Earth @ 20 March 2000 07:35 UTC
Experiments with Celestia...
I think the OP's confusion stems from a typo:
O - longitude of ascending node, of the intersection of the ORBITAL plane and the plane of the ecliptic
this should read:
O - longitude of ascending node, of the intersection of the EQUATORIAL plane and the plane of the ecliptic
the...
AFAIK all browsers have this feature - what key binding they use I don't know
CTRL-F5 works on IE and FF for Windows
http://support.mozilla.com/en-us/kb/keyboard+shortcuts
this option alters the HTTP request to force intermediate caching proxies to pass on the request (including...
Thanks for all your help - my understanding is improving :-)
As for finishing - this is just the start - I am writing a vehicle simulator
but both my maths and physics are weak - luckily I do at least know how to write software
I bought a Classical Mechanics book which I am working...
ok I think I am beginning to see
I was confused before because the unknown force at the pivot was a reaction to the external force
but because the action was exerted on the pivot - and the reaction was exerted on the rod
it becomes part of the total force on the rod
and I was confused...
trying to flush out latex bug on the forum
\sqrt{dog}
\sqrt{dog}
\sqrt{dog}
\sqrt{dog}
\sqrt{dog}
\sqrt{dog}
\sqrt{dog}
[edit] ok that worked // there is a bug in the latex preview // it caches old, stale latex renders and reuses them inappropriately // to fix just enter lots of tex sections...
understood - I was doing a check to match the output of my simulator
can I take it that my workings are correct?
that confuses me because ... we started with a single external force of 10N
and we end up with a resultant linear force of 12.974N and some torque
LaTeX editing on this...