Lifting one end of a barbell--how much weight did I lift?

This question seems kind of obvious, but I’ll throw it out there and see what happens.

Take a look at this video: Hip Belt Squats - YouTube

Assume these facts: (i) the bar weighs 45 lbs and is 7 feet long and (ii) his belt is attached 1 foot from the end of the bar.

If he puts a 45 lb weight on the bar in its natural spot (so bar plus weight is 90 lbs), how much weight is he lifting with each squat?

What if he uses spring clips to move the 45 lb weight out as far as possible to the end of the bar? Does that affect the amount of weight lifted?

(The reason I’m asking is that I’m having back pain from regular squats and thinking of switching to dip belt squats using the setup in the video. I’m wondering what weight to start with to keep on track with what I’ve been doing.)

Thanks.

I think you take the sine, but I’m not sure.

If the weight hung under him and the belt attached straight down and the bar didn’t weigh anything (“in a perfect world” type stuff) he’d be lifting as much as the weights, well, weigh. No different then if he just hung them from a belt of chain around his waist (or held dumbbells).
But he didn’t. The weights were in front of him which means you have to account for the distance from the fulcrum. Also The point at which he was lifting wasn’t under his center of gravity and that’s going to make everything wonky (and it looked really awkward and hard on his back) Because he’s not just lifting UP but up and back so he doesn’t fall forward. Two vectors are more complicated then one and in this case, less efficient when he’s trying to lift something off the floor. The bar has weight all along it’s length which needs to be dealt with.
The math can be done, it’s not all that hard, but the easiest thing to do would be to get a spring scale and pick the weight up, at the belt attachment point, with it and just see what it says. The way he’s got that set up, it’s going to read higher then what the weights, well, weight.

The further out he puts them, the heavier it’ll get. The further back (towards the pivot/fulcrum) he attaches the belt, the heavier it’ll get.
OTOH, if he moves the weights towards the fulcrum/pivot it’ll get lighter and easier to lift.

The magic word to all of this is “leverage”

Well, he’s using a third class lever since his input is between the fulcrum and the weight lifted. If he moved the weight out to the end of the bar and kept the lift point 1 foot from the end, he’d be lifting with a mechanical advantage of about 0.86, which means a resistance force of about 51.43 pounds. The force needed to lift the thing will increase the closer the belt is to the fulcrum and decrease the closer it is to the weight in this type of lever. Where the weight is in the video would reduce the force needed to lift it compared to having at the end (if the belt and weight are both 1’ from the end, the ideal calculated force would be whatever the weight alone is, but reality demands you have to lift the bar’s weight as well).

If you want to take the math out of it, do it like this. Add up the weights and you’ll know exactly what you were working with. If you want to get really technical, toss the harness and chains on a scale and add that to your total.

Just a quess on my part, but I think I would find the center of mass on the bar and weights combined. Find out how far the center of mass moved and compare that to how far the lifter moved. Say 90# moved 22" lifter moved 19" so 90X22/19

Center of gravity of the bar is in the middle, and the middle moves about half the height, so he’s lifting close to half the weight. A lot of the weight is at the end, which is beyond the attachment point, but half is close enough for an estimate.

The attached weight is at the attachment point so you just add that. He’s lifting at least 67.5 pounds.

Or you buy one of these MH Big Game Spring Scale :: Luggage Scales :: East High Scales Shop On-line or rig something to put the force from the attachment point on your bathroom scales.

Or do the exact math for how far the center of gravity of each component moves compared to the distance of the lift.

The math part of it is anything but obvious to me, and it is very interesting to see how it is approached, but the practical solution as a matter of exercise is obvious.

Lift the amount you can lift, hitting the limit of completing with good form the number of reps you are aiming for. You can do trial and error for a few sets, putting on or taking off some weights, in a fraction of the time it will take to figure out the math.

If he was at the very end of the 45 lb bar, he’d be lifting half of its weight. Since he’s not at the end, he’s lifting a bit more, but only a little since he’s close to the end. It seems to me this is one of those calculus problems where the result winds up being trivial, but don’t remember and probably too rusty to work it out. But, I’d say it’s about 25 lbs.

If the 45 lbs weights were suspended directly below him, that would add 45 lbs. They’re farther out on the lever, though, so he’s lifting a bit more. The math for this one is simple. If he’s say 9/10 of the way between the fulcrum and the load, the load would be 10/9 heavier. Note that this assumes the weight bears on just one point, but it’s close enough.

For practical purposes, figure it’ll be a little more than half the bar weight plus the sum of the weights. So in this case, it’s more than 22.5 + 45 = 67.5, but probably less than 25 + 50 = 75 lbs.