Help with Physics question

[Ludicrous Depth]

About to take my divemaster exam and not sure of the answer to one of the review questions. Any of you physics guru's out there who can chime in, I would really appreciate the help.

An object weights 566 pounds. It displaces 6.6 cubic feet of water. How much sea water do you need to dislpace to give the object 60 pounds of positive buoyancy?

Here are my two lines of reason, which one is correct?

6.6 X 64 = 422.4, so 566 - 422.4 = 143.6 * 80 = 223.6 / 64 = 3.49 or 3.5 cubic feet.
or
566 + 60 = 646 / 64 = 10.09 cubic feet.

 

[Jim Lapenta]

I'm not sure where your figure of 80 is coming from. Do you mean 60?

 

[Ludicrous Depth]

Yes, sorry it should be 60. I have changed it on the post

 

[TSandM]

Your first line of reasoning is correct. Since the object displaces water, it will experience a buoyant force equal to the weight of the water it displaces, making its apparent "weight" underwater less than on dry land. You need to do the calculations with the correct numbers, though.

 

[Crush]

D = the density of sea water is 1020 to 1029 kg/m3, depending on the temperature and salinity
D = 64 lbs / ft3 (assume 1020 kg/m3, convert to 63.7 lbs / ft3)

Bi = inherent buoyancy of the object (how much it wants to float, not to be confused with whether or not it floats)
= volume x density of water displaced
= 6.6 ft3 x 63.7 lbs / ft3
= 422.4 lbs

Bactual = inherent buoyancy - mass
= Bi - 566
= -144 lbs (rounding the answer) Note: negative buoyancy = it sinks

Breq = buoyancy required = 60 lbs

Badded = buoyancy you have to add (to satisfy your requirement that it is 60 lbs positively buoyant)
= Breq - Bactual
= 60 - (-144)
= 60 + 144
= 204 lbs

Badded = volume x density of water displaced, so
volume = Badded / density of water displaced
= 204 lbs / 63.7 lbs / ft3
= 3.2 ft3

 

[Jim Lapenta]

Ok, thought so. The displacement of the object itself provides 422.4lbs of lift as you stated. Leaving you with 143.6 lbs to make it neutral as you correctly found. In order to give it that extra 60lbs of positive buoyancy you will need to displace 203.6 lbs, Dividing 203.6 by 64 gives you a figure of 3.18+ cu ft you need to displace

 

[Ludicrous Depth]

So with correct numbers it should be;

6.6 X 64 = 422.4
566 - 422.4 = 143.6
143.6 + 60 = 203.6
203.6 / 64 = 3.18 which rounds up to 3.2 cubic ft.

 

[Ludicrous Depth]

Thanks to all, you are awesome

 

[Ludicrous Depth]

Ok, here is the last one

Object weighs 125 pounds
Displaces 2 cubic feet
How much weight do you need to make it 50 pounds negatively buoyant in sea water.

2 X 64 = 128
128-125 = 3
3 + 50 = 53 pounds needed

 

[bleeb]

An object weights 566 pounds. It displaces 6.6 cubic feet of water. How much sea water do you need to dislpace to give the object 60 pounds of positive buoyancy?

No arguments here about the math people have presented. My question is more what exactly is the exam question asking, total displacement (10.1 cu ft) or additional displacement above that of the object (3.5 + 6.6 cu ft)? It seems a literal interpretation could go either way, which is maybe part of what the OP is asking about. Is there some detail of terminology in the question that the exam creators normally use that help make this less ambiguous?