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Wow, this is a switch! I would have expected Evan to be on "the other side" of a question like this, spewing supporting facts, formulae, and citations.
As I said, I'm playing devil's advocate on this. I want to see somebody else provide the ironclad reason that there is only 50 lbs of tension and the equivalent elastic deformation in the cable.
There isn't any force magnification here. It doesn't even need to be a pully, a frictionless support will do fine.
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Dear Evan,
When I had the exact same problem counterbalancing the weight of my drillpress table, being weak on math, I decided to fit a cable that could take 1501bs anyway.
Hasn't broken yet...
Richard in Los Angeles
The top of the pulley System, the roof, is supporting 100, but the cable is just doing 50. Imagine a 50 lb weight with a cable on it sitting on the floor. Pick it up. What's the tension? 50, duh! Now walk over to a pulley. Put the rope over the pulley. Now, pulling down, it's the same force as it is to hold it up, just in the opposite direction, right? Now back to sans-a-pulley. How much force do you need to hold it up? 50 lbs. In the cable you have 50 lbs pulling each direction, for a tension of 50 lbs, not 100. Going on to the pulley, you laid the cord over the pulley. Now you are pulling Down with 50 lbs, correct? There is still just one weight. Just the 50 lb weight. So now pretend you need something else to hold that weight. Put another 50 lbs on there. Nothing changes. The pulley simply allows you to redirect the force of each weight to cancel each other out (in the rope, excluding it's mounts).
1. The system is at rest - therefore all forces must be balanced.
2. The tension through the length of cable is constant (ignoring the weight of the cable and friction between the cable and the pulley).
If the tensile force is constant through the length of the cable, it must all bear 50 lbs tensile load or there would be an imbalance and the system would no longer be at rest.
See point 2.. Ignoring friction and the incremental weight of the cable itself, the tension in the cable is constant throughout - 0.00001 mm above each weight - between the two pulleys - and... and...
regarding the stretch Q, assuming loads are well under that required for deformation, one 100lb load over the whole length would stretch it 2x as much as 2x50lbs over 50% of the length - with the same modulus of elasticity, stretch is a function of length and and force and is linear. 100lbs x 1length = 2x 2*50lb*.5length
The tension is still the same but now you've picked up a mechanical advantage of 2:1 at the 50 lb loads should you chose to pull down with 50 lbs force to lift the 100 lbs. Distance lifted, of course, is half the downward pull distance.
Sorry Boomer, I somehow missed that. I was a bit busy today improving the security at my shop. We have had a couple of breakins on either side of me so I built a couple of "bait" computers that contain entirely junk dead parts but look nice and are easily identifiable. Hopefully if somebody breaks in they run off with those. I also have a camera that is motion activated as well as bars on the windows.
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