# Thread: buoyancy

1. How many newtons in force would be required to sink a buoyant object (with positive buoyancy) 1 x 1 feet in surface area to a depth of 500 feet. Is it possible to convert this force into watts required to sink this object to a depth of 500 feet. Would appreciate even a rough estimation. Water is fresh water.

2.

3. Originally Posted by Adnkir
How many newtons in force would be required to sink a buoyant object (with positive buoyancy) 1 x 1 feet in surface area to a depth of 500 feet. Is it possible to convert this force into watts required to sink this object to a depth of 500 feet. Would appreciate even a rough estimation. Water is fresh water.
Can't answer with the info you have provided. What is needed is the volume displaced by the object when fully submerged (not surface area) and its mass.

From this you can get the net upthrust on it and hence the work (not power) that needs to be done on it to push it down to a chosen depth.

4. Originally Posted by exchemist
Originally Posted by Adnkir
How many newtons in force would be required to sink a buoyant object (with positive buoyancy) 1 x 1 feet in surface area to a depth of 500 feet. Is it possible to convert this force into watts required to sink this object to a depth of 500 feet. Would appreciate even a rough estimation. Water is fresh water.
Can't answer with the info you have provided. What is needed is the volume displaced by the object when fully submerged (not surface area) and its mass.

From this you can get the net upthrust on it and hence the work (not power) that needs to be done on it to push it down to a chosen depth.
You would need to know it own weight as well. For then you could work out the weight needed to submerge it. That times the depth will be the work required to send it to the bottom.

5. Originally Posted by Robittybob1
Originally Posted by exchemist
Originally Posted by Adnkir
How many newtons in force would be required to sink a buoyant object (with positive buoyancy) 1 x 1 feet in surface area to a depth of 500 feet. Is it possible to convert this force into watts required to sink this object to a depth of 500 feet. Would appreciate even a rough estimation. Water is fresh water.
Can't answer with the info you have provided. What is needed is the volume displaced by the object when fully submerged (not surface area) and its mass.

From this you can get the net upthrust on it and hence the work (not power) that needs to be done on it to push it down to a chosen depth.
You would need to know it own weight as well. For then you could work out the weight needed to submerge it. That times the depth will be the work required to send it to the bottom.
That's why I asked for its mass.

6. Originally Posted by exchemist
That's why I asked for its mass.
So you did. I must have read that sentence incorrectly sorry.

7. Originally Posted by Robittybob1
Originally Posted by exchemist
Originally Posted by Adnkir
How many newtons in force would be required to sink a buoyant object (with positive buoyancy) 1 x 1 feet in surface area to a depth of 500 feet. Is it possible to convert this force into watts required to sink this object to a depth of 500 feet. Would appreciate even a rough estimation. Water is fresh water.
Can't answer with the info you have provided. What is needed is the volume displaced by the object when fully submerged (not surface area) and its mass.

From this you can get the net upthrust on it and hence the work (not power) that needs to be done on it to push it down to a chosen depth.
You would need to know it own weight as well. For then you could work out the weight needed to submerge it. That times the depth will be the work required to send it to the bottom.
Thanks fellows. Sorry I did not provide more detailed information. The weight of the object is 5 kg and its dimensions are 1 foot x 1 foot and its height is 3 inches. It takes about 6 kg of weight to fully submerge it and to sink it to a depth of 1 meter. Can we convert the work required to sink it to a certain depth to kilowatts/hour required in electricity terms.

8. Originally Posted by Adnkir
Originally Posted by Robittybob1
Originally Posted by exchemist
Originally Posted by Adnkir
How many newtons in force would be required to sink a buoyant object (with positive buoyancy) 1 x 1 feet in surface area to a depth of 500 feet. Is it possible to convert this force into watts required to sink this object to a depth of 500 feet. Would appreciate even a rough estimation. Water is fresh water.
Can't answer with the info you have provided. What is needed is the volume displaced by the object when fully submerged (not surface area) and its mass.

From this you can get the net upthrust on it and hence the work (not power) that needs to be done on it to push it down to a chosen depth.
You would need to know it own weight as well. For then you could work out the weight needed to submerge it. That times the depth will be the work required to send it to the bottom.
Thanks fellows. Sorry I did not provide more detailed information. The weight of the object is 5 kg and its dimensions are 1 foot x 1 foot and its height is 3 inches. It takes about 6 kg of weight to fully submerge it and to sink it to a depth of 1 meter. Can we convert the work required to sink it to a certain depth to kilowatts/hour required in electricity terms.
I may have got this wrong but something odd here. If I convert the dimensions of the object to metric units, say 30 cm x 30 cm by 8 cm, it should have a displacement when fully submerged of 7.2 litres, creating an upthrust of 7.2 kg weight. If it weighs 5kg it should take only a further 2.2kg weight to make it sink, rather than 6kg weight.

But assuming you have measured the weight required to make it sin and it really is 6kg, then the force corresponding to that is 9.8 x 6 = 59 N. One Joule is one Newton applied through 1 metre, so if you apply 59N through 1 m the work done is 59J.

1 Watt is 1 Joule/sec. So 1 kWh is 1000 x 3600 J = 3.6 x 10⁶ J, or 3.6. megajoules, MJ. So to sink it once you need 59/3.6 x 10⁶, = 1.6 x 10-⁵ kWh.

That is, if I've placed my decimal points correctly, not always easy to get right.

9. Originally Posted by Adnkir
Thanks fellows. Sorry I did not provide more detailed information. The weight of the object is 5 kg and its dimensions are 1 foot x 1 foot and its height is 3 inches. It takes about 6 kg of weight to fully submerge it and to sink it to a depth of 1 meter. Can we convert the work required to sink it to a certain depth to kilowatts/hour required in electricity terms.
That isn't right. The volume of your object is 1/4 of a cubic foot. A cubic foot of water at 32 F weighs 62.42 lb, so the object is displacing 62.42/4=15.6 lb. The weight of water displaced is the buoyant force. A weight of 15.6 lb is equal in mass to 15.6/2.2= 7.1 kg. so the net force is the difference between the buoyant force and the weight, which is 7.1-5 = 2.1 kilograms-force, not 6 kilograms. One newton is .1 kilograms force, so 2.1/.1=20.6 Newtons. Applying this force over a distance of 1 meter gives you 20.6 newton-meters of work. If this were converted without loss to electricity, Google calculator says 20.6 newton meters = 0.00572222222 watt hours.

10. Originally Posted by Harold14370
Originally Posted by Adnkir
Thanks fellows. Sorry I did not provide more detailed information. The weight of the object is 5 kg and its dimensions are 1 foot x 1 foot and its height is 3 inches. It takes about 6 kg of weight to fully submerge it and to sink it to a depth of 1 meter. Can we convert the work required to sink it to a certain depth to kilowatts/hour required in electricity terms.
That isn't right. The volume of your object is 1/4 of a cubic foot. A cubic foot of water at 32 F weighs 62.42 lb, so the object is displacing 62.42/4=15.6 lb. The weight of water displaced is the buoyant force. A weight of 15.6 lb is equal in mass to 15.6/2.2= 7.1 kg. so the net force is the difference between the buoyant force and the weight, which is 7.1-5 = 2.1 kilograms-force, not 6 kilograms. One newton is .1 kilograms force, so 2.1/.1=20.6 Newtons. Applying this force over a distance of 1 meter gives you 20.6 newton-meters of work. If this were converted without loss to electricity, Google calculator says 20.6 newton meters = 0.00572222222 watt hours.
Snap, Harold. THe displacement doesn't agree with the stated weight required to submerge the object.

11. Originally Posted by exchemist
Snap, Harold. THe displacement doesn't agree with the stated weight required to submerge the object.
Yes, I agree. Either the 5 kg is wrong, or the 12 by 12 by 3 inches is wrong.

12. Originally Posted by Harold14370
Originally Posted by exchemist
Snap, Harold. THe displacement doesn't agree with the stated weight required to submerge the object.
Yes, I agree. Either the 5 kg is wrong, or the 12 by 12 by 3 inches is wrong.
Why measure it in inches and express the weight in kilos!

13. Originally Posted by Robittybob1
Originally Posted by Harold14370
Originally Posted by exchemist
Snap, Harold. THe displacement doesn't agree with the stated weight required to submerge the object.
Yes, I agree. Either the 5 kg is wrong, or the 12 by 12 by 3 inches is wrong.
Why measure it in inches and express the weight in kilos!
If you are in India or Pakistan the shadow of Imperial units still looms large. And even in Britain, if you buy a length of garden hose, you buy it by the metre, but they ask whether you want 1/2 inch or 3/4 inch diameter. Really!

14. [QUOTE=exchemist;551388]
Originally Posted by Robittybob1
Originally Posted by Harold14370
Originally Posted by exchemist
Snap, Harold. THe displacement doesn't agree with the stated weight required to submerge the object.
Yes, I agree. Either the 5 kg is wrong, or the 12 by 12 by 3 inches is wrong.
Why measure it in inches and express the weight in kilos!
If you are in India or Pakistan the shadow of Imperial units still looms large. And even in Britain, if you buy a length of garden hose, you buy it by the metre, but they ask whether you want 1/2 inch or 3/4 inch diameter. Really![/QUOTE

15. hahahaha

16. Originally Posted by exchemist
And even in Britain, if you buy a length of garden hose, you buy it by the metre, but they ask whether you want 1/2 inch or 3/4 inch diameter. Really!
It's always a problem buying 'ose.
https://www.youtube.com/watch?featur...;v=qu9MptWyCB8

 Bookmarks
##### Bookmarks
 Posting Permissions
 You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement