Can't Do This Scuba Math....A test for you

[scubajoh44]

I can not figure out how to do this math. Someone help. I've looked everywhere and my Encyclopedia of Rec Diving has similar problems but not the ones like on the DM test.

1. A 125 lb object is submersed in 126 ft of sea water. It displaces 2 cu. ft. Disregarding the minimal positive bouyancy, how much lead weight do you need to make the object 50lbs. negatively bouyant?
2. A 324 lb object is in 98 ft of sea water and displaced 3 cu. ft. If the object was taken to 139 ft., how much air would be needed to raise it to the surface?


Things I know are... 1 cu.ft of displaced sea water= 64 lbs.
1 ata = 33 ft of seawater

Tricky, tricky..... I just can't figure them out. I've asked two Physics teachers and they don't know either. You scuba geniuses figure them out. (or do yall remember either?)

 

[captain]

53 pounds of lead

8.53 cubic feet at 1 atmosphere

 

[The Kraken]

1: 53#
The object displaces 128 pounds of water but weighs only 125, that makes it 3# positively buoyant. In order to make it 50# negative, one must first make it neutrally buoyant, that means adding 3# more. Then to get it 50# negative, one must add an additional 50#.

2: 2 answers; 2.0625 cf and 10.746 cf (you did not indicate if the object was neutrally buoyant at 98')

*** Given that both objects are solids and incompressible.

Plus whatever minisucle amount of gas is required to offset the negative buoyancy of the lifting device.

the K

 

[redacted]

#1. Ignoring the volume of the lead, how about 53 lb.

#2. Assuming the object is not compressable and ignoring the weight of the air, 10.75 cu ft of surface air (2 & 1/16 cu ft at depth) should make it neutral.

Shame on those physics teachers. I guess they are trained to ask questions rather than answer them.

 

[Walter]

1. A 125 lb object is submersed in 126 ft of sea water. It displaces 2 cu. ft. Disregarding the minimal positive bouyancy, how much lead weight do you need to make the object 50lbs. negatively bouyant?

The item weighs 125 lbs and displaces 2 cu ft of sea water which weighs 128 lbs. The object is 3 lbs positively buoyant. To have it 50 lbs negative, you'll need 50 lbs of lead plus 3 lbs to overcome the positive buoyancy, so the correct answer is 53 lbs.

2. A 324 lb object is in 98 ft of sea water and displaced 3 cu. ft. If the object was taken to 139 ft., how much air would be needed to raise it to the surface?

The object displaces 3 cu ft of sea water weighing 192 lbs. The object weighs 324 lbs, so it is 132 lbs negatively buoyant. It needs 132 lbs of lift. 132/64 = 2.0625 cu ft of air at 139 feet to raise it to the surface. Since air volume is measured at 1 atm, 139/33 +1 = 5.21 atm at 139 ft. 5.21*2.0625 = 10.75 cu ft of air to raise it to the surface.

 

[mddolson]

Need to decide what info is important
1) One has to make an assumption, the information is incomplete.
If the object is solid, there is no effect taking it deeper.
However if this is a lift bag that weights 125lbs, Big difference!
At surface
Bouyant force =64 x2 (cuft) =-128 lbs
Object weight =+125 lbs
so object is 3 lbs positive(buoyant) (125-128 =-3)
Desired weight is 50 lbs (50=-3 +53)
Lead required is 53 lbs

If this object is compressible then the 2 cuft vol will decrease,
So from Boyle's Law
at constant temp PV is a contant so P1V1 = P2V2
P1 is 1 atm (at surface) V1 is 2 cuft
P2 is pressure at 126 fsw (=126/33 +1 at surf) =3.9 atm
V2 we calulate (P1 xV1)/P2 =(1x2)/4.8= 2/4.8= 0.42 cuft
The buoyant force is 0.42 cuft x 64 lbs/cuft =26.9 or 27 lbs
If we want the weight to be 50 but it weighs 125 -27 = 98 lbs.
We must infact remove weight to the tune of 98-50= 48 lbs

 

[kensuguro]

here's my take:
1.
anything that displaces 2cu.ft of water will be pushed up by 64lbs*2cu.ft=128 lbs of buoyant force. The object is pulled down 125lb by gravity, so 128lbs UP - 125lbs DOWN = 3lbs UP. You want to end up with 50lbs DOWN so you'd need 53lbs of weight.

The equasion would be:
objectWeight - objectVolume*seaWaterWeight = objectBuoyancy
in other words:
forceActingDown - forceActingUp = totalForce

2.
Again, I figure out the totalForce acting on the object:
324lbs DOWN - 64lbs*3cu.ft UP= 132lbs DOWN

Air would give you UP however much volume you add, but you also need to take the 139ft depth into account since the depth affects volume. You need to figure out how much volume of air you need to make the object neutrally buoyant at 139ft, and then figure out how much volume it actually is on land. (because air is compacted at 139ft depth)

The totalForce right now is 132lbs DOWN so we want 132lbs UP. (numbers are wierd.. so I'm not sure if this is right)
132lbs / 64lbs = 2.0625 cu.ft of air at 139ft.
the pressure at 139ft would be 139ft/33ft=4.2121ata + 1ata for pressure on land.
2.0625 * 5.2121=10.75cu.ft of air on land.

So, the answer is 10.75cu.ft of air on land. But I'm not too sure since the numbers didn't divide out evenly..

 

[The Kraken]

Yeah, I thought question #2 was a bit ambiguous.
At the depth of the object, 2.062 cu. ft. is required to make the object neutrally buoyant allowing it to be raised to the surface.

But if the question was related to how much air would be required from the diver's tank to effect the lift, then the answer would be 10.75 cu. ft.

the K

 

[mddolson]

2)Assuming object is not compressible moving from 98 fsw to 139 has no effect on it's vol
weight in water = dry weight -bouyant force
wt in water = 324 -(3 x 64) =132 lbs.
So we need 132 lbs of lift to raise the object from 139 fsw.
we need 132/64=2.06 cuft (at 139ft)
Pressure1 at 139 ft = 139/33 +1 =5.2 atm
Vol1 at depth =2.06 cuft
Pressure2 at surface is 1 atm

from boyle's law P1V1 = P2V2
V2= P1V1/P2
V2=2.06 x5.2 /1.0 =10.7 cu ft at surface

 

[scubajoh44]

WOW.....what scuba gurus. That is just what I needed. I figured that #1 was 53lbs, but thought that was too simple. (you know, one of those trick questions....) Thanks all. I might be asking some of you for more help later. (Physics test coming up)