Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have a buoy that can hold 2943.02 cubic millimeters of non-pressurized air, I need and equation to work out the maximum weight that the buoy can float.

share|improve this question
    
Well, it depends how heavy the buoy is. –  TonyK May 13 '11 at 12:20
    
The buoy is 0.8kg but I was hoping that, that could be incorporated into the weight floating. –  Alexander 'Bambi' Sadler May 13 '11 at 12:29
    
The Buoy is 0.8kg –  Alexander 'Bambi' Sadler May 13 '11 at 12:30
    
Buoyancy is related to the weight of the volume of material displaced. In your case this means the max weight would be equal to $3943mm^3$ of water, minus the 0.8kg weight of the buoy.. –  qftme May 13 '11 at 12:50
    
@qftme That volume of "pressurized air" is meaningless. One needs the Volume of the "buoy" (total volume!) –  Georg May 13 '11 at 13:11
show 5 more comments

closed as off topic by Georg, David Z May 13 '11 at 20:51

Questions on Physics Stack Exchange are expected to relate to physics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

3 Answers

The buoyancy is the weight of water displaced. Usually for problems like this we assume 1 g/cm^3 (which is fresh water at 4C). That is the weight it can support, but you have to count its own weight as part of what is supported.

share|improve this answer
add comment

Seems like a very tiny buoy! Also, the amount of air is given to surprising precision.

As a first (and excellent) approximation, you can assume that air is weightless. Water weighs roughly $1$ gram per cc, you have about $3$ cc of air, so the air can support about $3$ grams.

Not an awful lot, compared to your estimated weight of the buoy.

An important consideration would be the density of the material that the buoy is made of. If it is substantially greater than the density of water, the buoy will sink and support nothing. But if the buoy is bulky and made of material of low density, the solid material of the buoy could contribute a lot more to the buoyancy than the air. Imagine for example a buoy made of very lightweight wood with a tiny $3$ cc bubble of air in the middle. This bubble will contribute virtually nothing to the buoyancy. The "solid" material of the buoy will make a much greater contribution.

Please do check your units. As mentioned above, $2943.02$ cubic millimetres is really really tiny. Do you mean $2943.02$ cc? That would be more realistic, though still smallish. And the density of the "solid" part of the buoy is relevant.

share|improve this answer
add comment

2943 mm^3 equals 2.943 cc. This is the equivalent force of submerging a 0.7 inch diameter ball in water. It will give about 0.002943 kg * (9.8N/kg) of bouyant force in fresh water. You say it weighs 800 grams = 0.800 kg * (9.8N/kg). 0.800 kg is a lot more than 0.002943 kg. Unless the enclosure is built of something less dense than water, this bouy is going to drop like a rock without any load at all.

Perhaps you meant 2943 cc? That is a volume of just under 3 liters, and equates to a sphere of exactly 7 inches diameter. In that case, assuming negligible added volume of the construction material, then your net bouyant force in fresh water will be about (2943 grams - 800 grams)(1kg/1000g)(9.8N/kg) = 2.143 kg * (9.8N/kg) = 21.00N = 4.71 lbs.

To explictly answer your exact question; the equation is:

Net_Bouyant_Force = ((Total_Volume_of_bouy) * (Density_of_displaced_liquid)) - Weight_of_bouy

Make sure you keep your units straight. Density is often given as mass/volume. In the above equation you want to convert to weight/volume.

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.