So if it extended outside Earth's atmosphere, only the section of the wood subject to Earth's gravity would be weighed accurately?
The atmosphere is not a magic barrier, but distance from a large mass like the Earth is. I wonder how a scale would measure a hypothetical beam which extended outside the atmosphere. Some of it would be normal, some of it lighter, and at the ends it would be weightless.
There is no such thing as true weightlessness - everything is impacted by gravity, though, sure, in some cases, it's so small as to be imperceptible.
The situation you're thinking of, however, where an object is in orbit isn't true weightlessness, either, but rather a sort of perceived weightlessness. What happens with, say, a space craft, in orbit is not that it's weightless and unaffected by gravity, but rather that it's moving at such a high speed that, as it falls towards the earth (which it is doing all the time), the path along which it falls matches the curvature of the earth. The space station is literally falling in a big circle (or ellipse, if you want to get fussy).
The ends of your 2x4 wouldn't be in orbit like that and would still "weight" something, though it would be somewhat less their weight were they at a lower altitude.
Stand the 4x2 on end and let go for a moment as you take the reading. It will take a few seconds before it starts to tip.
I have also weighed long objects by weighing each end.
If you put one end of the wood on a brick, and the other end on the scales, each end will bear 50% of the load, so just double it. You can check by swapping the scaled from one end to the other.
Its the same as using 2 or even 3 sets of scales on a large object. Add the scale weights together to get the total.
Yup, stand it on end, or suspend it from something and have the scale under or attached to the thing from which you are suspending it.
On a related note, I still work in the entertainment industry a little bit and keep an eye on some of the rigging groups just because I think it's interesting. I recently came across some discussions regarding the distribution of weight across multiple motors lifting a single straight truss and the numbers aren't what I would've guessed:
Rigging Loading Calculations for Multipoint straight truss | HireHop.com
Assuming the weight is evenly distributed across a truss, then putting a hoist at the location of each arrow in the diagram causes those hoists to bear unequal percentages of the total load: