If you had a long bar floating in space, what would be the compressive force at the centre of the bar, due to the self-weight of both ends?
Diagram - what is the force at point X in the middle of the bar?:
<----------------------L--------------------->, total mass M =======================X====================== <- the bar F---> X <---F
You should be able to simplify by cutting the bar into pieces, but that gives a different answer depending on how many pieces you use (see below). So the simplification must be wrong - but why?
Split bar in two
So, one approximation would be to cut the bar in half - two pieces of length L/2, mass M/2:
(M/2)<-------L/2------->(M/2) #1 X #2 <- bar approximated as blobs #1 and #2
Force at X is G(M1.M2)/(R^2) = G (M/2)^2 / (L/2)^2 = G M^2 / L^2
Or Fx / (G. M^2 / L^2) = 1
But is that really valid? If so, shouldn't you get the same answer if you split the bar into four pieces?
Split bar into four
(M/4)<-L/4->(M/4)<-L/4->(M/4)<-L/4->(M/4) #1 #2 X #3 #4
My assumption is that the force at X is the sum of the attractions of each blob on the left to every blob on the right.
Force at X = #1<>#3 + #1<>#4 + #2<>#3 + #2<>#4
('<>' being force between blobs #x and #y).
Fx / (G.M^2 / L^2) = (2/4)^-2 + (3/4)^-2 + (1/4)^-2 + (2/4)^-2 = 1.61
This is bigger than the previous result (1.61 vs 1).
Split bar into six
Similarly, if you split into 6 blobs, the total force comes out as:
Fx / (G.M^2 / L^2) = (3/6)^-2+(4/6)^-2+(5/6)^-2 + (2/6)^-2+(3/6)^-2+(4/6)^-2 + (1/6)^-2+(2/6)^-2+(3/6)^-2
Fx / (G.M^2 / L^2) = 2.00
So what's wrong with my approach? And what is the real answer?
So it seems the more pieces we split the bar into, the larger the result gets. There's clearly something wrong with my assumptions! - but what?
I'd be very glad if someone here could explain this. Thanks!
EDIT As Peter Shor pointed out, my calculations had some dodgy algebra and I'd calculated $$L^2/M^2$$ values rather than $$M^2/L^2$$. I've now corrected that - the value still increases as you divide into more masses.
I'll do a bit more work with more divisions and see if this leads to convergence or not.