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# How to calculate using counter balance to lift a mass from undersea to above seawater. Answered

A huge beam planted into sea and the height 15meter above seawater. Total length 30m. Using pulley system, both side left and right hung 2 equal weight drum. The whole picture should be similar to Elevator method

Question: How counterbalance and pulley system calculation will be? Do I need to consider buoyancy formula and also F=ma.

What kind of formula do I have to consider?

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## 11 Replies

See attached. Do you want it like A? - a counterweight like on an elevator. Only problem is if you take your load (marked at the bottom) off then you have to lift the counterweight out of the water. In ex. C you use an airbag, not attached to the load to float the load to the surface, then deflate the airbag to get the cable back to the bottom. Or is it like B, where...its completely different?

frollard (author)2009-03-29

if the counterbalance stays out of the water, then you need the buoyant 'weight' (mass of object minus mass of water displaced) as a weight Your counterbalance will weigh its normal weight out of water Once in water - the counterbalance will weigh its mass minus the weight of the water it displaces. (you need to know the volume of your counterbalance)

zoohunter (author)2009-03-29

Is type A. So what formula or force applied here?

zoohunter (author)2009-03-29

The actual diagram

frollard (author)2009-03-29

Okay, The counterweight has a different weight when its in the air, or submerged. Does it stay underwater all the time?

If so, you want something like lead (sealed to prevent contaminating the water) or some other metal...or concrete/stone.

Say your weight is 1meterx1meterx1meter, (one cubic meter, or 1000 litres), and weighs 2000 kilograms out in the air. 1000 litres of water that it will displace will 'float' it that much, so it will be 2000-1000 kilograms = 1000 kilograms underwater.

For this reason, a water-weight wont work when its submerged, because the water weighs the same as the water it displaces.

zoohunter (author)2009-03-30

Yes both bulk suppose to be inside the water all the time. So 1 of the bulk will move down in order to lift the other bulk up above the water. So I need to show some calculation for this problem only. Let said, the bulk 40tonne attached with single wire. When both bulk at same level, the bulk height from ground will be 15m and to be lifted up above water. Water level at 20m height.

frollard (author)2009-03-30

As soon as the 2nd weight comes out of the water, its apparent weight changes because the buoyancy no longer acts on it - meaning you need extra counterweight. Essentially what I'm saying is the pully still has to do all the work (with extra force being added)

frollard (author)2009-03-30

right, as soon as your 'load' comes out of the water, it will become significantly heavier, because the water will no longer have it's volume of buoyancy. As soon as either comes out of the water, it will need 'its volume in water' extra weight to lift any higher.

alchemistdagger (author)2009-03-29

I am not clear on exactly what you are trying to lift out of the water but I can offer a general explination. For the most part, whey you are trying to lift something out of the water, the force needed above water will be more than what is required to lift it in the water, due to buoyancy. So just looking at the force needed to lift a mass above water. If you are using ropes and pulleys, if you have one pulley with one rope, you would need a mass equal to what you are trying to lift. The more pulleys you use the less mass (and less force) to lift the mass you are trying to lift. Now when you are actually pulling the mass out of the water into the air, the mass of the object will increase for a short time, because some water will be attached to the mass so you will be lifting the mass and some amount of water, until the water drains off. hope this helps

zoohunter (author)2009-03-29

Thanks alchemist, "Now when you are actually pulling the mass out of the water into the air, the mass of the object will increase for a short time, because some water will be attached to the mass so you will be lifting the mass and some amount of water, until the water drains off." What do I need to consider here? How do i calculate the water or resistance force?

alchemistdagger (author)2009-03-29

I think a big factor in the force needed to pull an object out of the water (transitioning from water to air) is the speed you want the object to be lifted. If you can move it very slowly, there should be very little change in the required force compared to when it is completely out of the water. If however, you pull it out of the water very quickly, then there may be a layer of water above the object that will be lifted out of the water as well as a vacuum beneath the object created by the water that is trying to pull the object back in the water. These are just suggestions off the top of my head. It really depends on what the object is, and how its surface interacts with the water, is it a complex shape that can hold water, will it soak up water. And how fast you want to remove it from the water. I don't know if there is an equation that can take account of all the variables, is seems more like a trial and error type of experiment or maybe you can make a scale test model. I guess going back to your original question, I would just calculate the force by mass times acceleration (gravity) and just double or tripple it to be on the safe side and just pull it out of the water slowly. I really can't say more without having a better understanding of what the setup is.