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seen Sep 30 '13 at 8:58

Apr
18
comment How far does a trampoline vertically deform based on the mass of the object?
I don't think a small angle approximation is valid for a trampoline. E.g. a typical circular garden trampoline has height ~R; so a heavy bounce would dip at least R/2 -> 30 degrees.
Apr
18
comment Basic explosion physics - determining force
Just a quick note re angles - there's nothing wrong with the angle being negative. sin & cos are defined for any angle from $-\inf$ to $+\inf$, with a period of 360 degrees (or 2pi radians).
Mar
31
comment Calculate stainless steel pole necking limit
... [cont]: If you're interested in this kind of thing, there's plenty more to read on the internet. E.g. the Singapore tunnel collapse from a few years ago was caused in part because a certain detail was changed by the contractor without properly considering stability (buckling), and this subsequently failed. scribd.com/doc/6231559/DESIGN-AND-FAILURE-OF-SINGAPORE-TUNNEL
Mar
31
comment Calculate stainless steel pole necking limit
@Anna: Sorry for the late reply. The Wikipedia page on buckling is a reasonable introduction. Or a degree in structural engineering ;-). The main point I meant to get at is that for a complex setup like this, it's important to consider every failure mode. It's rare to find a real-world scenario that requires a simple stress calculation. Sadly also most buckling calculations are very complex (except for very simple structures) so it's not normally possible to calculate for them directly in a situation such as this. ...
Mar
10
comment Calculate stainless steel pole necking limit
@ Dave: (without knowing the full details of the joint) - I think the weakest point in the system won't be the joint between the arm and bracket, but between the bracket and pole. Clearly the arm is designed to be bolted to the wall, so whatever bolts would be used for that must be strong enough to make the equivalent join to the bracket. Assuming the holes in the bracket are same / similar distance apart as those in the bracket - if they're significantly closer together the forces will be larger.
Mar
9
comment Calculate stainless steel pole necking limit
Down-voting for the concerns Georg raises in comments to his answer. There's no explanation (or reason as far as I can see) to go from $strain = stress/Young modulus$ to $r = pressure rating / m.g$. The answer makes no mention of buckling and doesn't look at how the arm is attached to the pole.
Nov
19
comment Imagine a long bar floating in space. What force does it exert on itself in the middle due to gravity?
You're right - thanks. I've corrected this in the original question now.
Nov
17
comment Imagine a long bar floating in space. What force does it exert on itself in the middle due to gravity?
[cont] My previous approach using integrals assumed R << L, hence r << z, so your expression for V(z) simplifies to mG int{ (1/(z-z'))dz} (where m = M/L = 2pi.R.rho = mass/unit length). This fails though because you end up with V(0) = [-ln(z')]{0,L/2} = ln(L/2) + infinity. Still not entirely sure why this doesn't work - I suspect because r << z is not true when z -> 0.
Nov
17
comment Imagine a long bar floating in space. What force does it exert on itself in the middle due to gravity?
That's a great answer, thanks. I'm still surprised to see R in the final answer but glad to see it matches what you'd intuitively expect - ie for R << L small changes in R make little difference to the final answer (ln(x) is fairly flat for large x).
Nov
16
comment Imagine a long bar floating in space. What force does it exert on itself in the middle due to gravity?
@Kenny - you're right that the gravity field at the end of the rod is infinite, for nonzero linear density. And clearly in the real world atoms are separate in space, so the 'infinite' problem doesn't arise. But the question remains how to approximate the force/pressure in the middle of this bar. To avoid the infinite problem (linear density) I've instead broken my bar into discrete pieces - but as the number of pieces increases (Ndiv = 2, 4, 6 -> large), the answer keeps increasing which doesn't make sense.
Nov
16
comment Imagine a long bar floating in space. What force does it exert on itself in the middle due to gravity?
@Mark - yes, but I'm a little rusty. Eg with Kenny's explanation, I don't see where the original integral comes from, but I can follow the rest (I wouldn't be able to solve the integral myself but can happily accept how to go from one stage to the next).
Nov
16
comment Imagine a long bar floating in space. What force does it exert on itself in the middle due to gravity?
Thanks. Clearly the pressure does tend to infinity as the area on which it acts tends to zero. But what I have in mind is a "long, thin" bar - ie where the thickness of the bar is small compared to the length. What happens to your formula if you ignore the bar thickness (assume e << L), and try to calculate instead (stress * area) at the centre?
Nov
16
comment Imagine a long bar floating in space. What force does it exert on itself in the middle due to gravity?
Clarification: as TheMachineCharmer correctly points out, the net force in the middle of the bar is zero. But what I'm trying to find is the pressure exerted on you at the middle of the bar, due to the weight of both halves of the bar. Or to ask another way, if you cut the bar in half and put a spring between the two halves, how much would the spring contract?