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Thread: buoyancy

  1. #1 buoyancy 
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    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.


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  3. #2  
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    Quote Originally Posted by Adnkir View Post
    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.


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  4. #3  
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by Adnkir View Post
    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.
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  5. #4  
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    Quote Originally Posted by Robittybob1 View Post
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by Adnkir View Post
    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.
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  6. #5  
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    Quote Originally Posted by exchemist View Post
    That's why I asked for its mass.
    So you did. I must have read that sentence incorrectly sorry.
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  7. #6 Thanks fellows 
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    Quote Originally Posted by Robittybob1 View Post
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by Adnkir View Post
    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.
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  8. #7  
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    Quote Originally Posted by Adnkir View Post
    Quote Originally Posted by Robittybob1 View Post
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by Adnkir View Post
    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.
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    Quote Originally Posted by Adnkir View Post
    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.
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  10. #9  
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    Quote Originally Posted by Harold14370 View Post
    Quote Originally Posted by Adnkir View Post
    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.
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    Quote Originally Posted by exchemist View Post
    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.
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  12. #11  
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    Quote Originally Posted by Harold14370 View Post
    Quote Originally Posted by exchemist View Post
    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!
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    Quote Originally Posted by Robittybob1 View Post
    Quote Originally Posted by Harold14370 View Post
    Quote Originally Posted by exchemist View Post
    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!
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    [QUOTE=exchemist;551388]
    Quote Originally Posted by Robittybob1 View Post
    Quote Originally Posted by Harold14370 View Post
    Quote Originally Posted by exchemist View Post
    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
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    hahahaha
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    Quote Originally Posted by exchemist View Post
    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
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