Two more dive physics problems

[n03]

Since it looks like we have some people on here that like to solve physics problems I'm going to put two more out that I'm not sure about:

1. An object is five ft. square and five ft. high is floating in the ocean. Six inches of the object is above the waterline. What is the minimum line strength that would be needed to lift the object out of the water?
1) 7,500-lb test (based on some calculations below I think it's this answer, but I'm not sure)
2) 10,000-lb test
3) 6,000-lb test
4) 1,000-lb test

The calculations I did for number one are:
The volume of the submerged portion of the object is 5*5*4.5 = 112.5 cu.ft.
Using the formula Buoyancy = Volume * Water_density - Object_weight = 0
Object_weight = 112.5 cu.ft. * 64 lb/cu.ft = 7200 lb
Does this mean that we need 7,500 lb line strength? Why wouldn't it be just 7,200 lb in this case?

2. If a diver takes a sealed, rigid, cubical container measuring 10 inches per side to a depth of 20 ft. in the ocean, the total crushing force (assume the wall thickness is zero) on the container would be about:
1) 3,560 lb
2) 5,256 lb
3) 5,340 lb
4) 14,076 lb

 

[~ dry]

These are from the SSI Science of Diving final exam, yes?

Wow, that was a toughie.

 

[Akimbo]

It depends a bit on the jurisdiction. The typical minimum safety factor for non-man-rated static loads is 4:1. Dynamic loading, like on a rolling vessel, often increases it between 6 and 8:1. Man-rated load safety factors start at 10:1.

Edit: It also depends on how you define “Test”. Test is some places means a destructive test measuring the ultimate point of failure. Other tests are around 2x the expected working load. I have never known X Lbs or Kilograms to be the rated “safe working load”.

 

[n03]

It depends a bit on the jurisdiction. The typical minimum safety factor for non-man-rated static loads is 4:1. Dynamic loading, like on a rolling vessel, often increases it between 6 and 8:1. Man-rated load safety factors start at 10:1.

And what does this mean in practical terms for the problem at hand?

 

[Akimbo]

And what does this mean in practical terms for the problem at hand?

Because the question was:

… What is the minimum line strength that would be needed to lift the object out of the water?...

The question wasn’t “what is the weight of the load”. The strength of the line must include the safety margin unless you intend for the line to break when the object clears the water.

 

[rhwestfall]

likely means you need to solve your own exam problems.....

 

[n03]

Because the question was:



The question wasn’t “what is the weight of the load”. The strength of the line must include the safety margin unless you intend for the line to break when the object clears the water.

So which answer would you pick? How much margin do I need to add for line safety? Are there specific regulations on this?

---------- Post added August 14th, 2014 at 05:48 PM ----------

likely means you need to solve your own exam problems.....

If the book explained how to solve them, then I would have obviously solved them myself. The book doesn't even say what total crushing force is! I've done actual physics courses in university, and it was never this frustrating to figure out what rules we are playing by. SSI is simplifying physics, supposedly to make it more approachable, but it's often hard to tell what the simplifications actually mean. Not to mention the specifics of the dive industry regulations, which, for some reason, they don't always deem necessary to include in the printed material.

 

[Rich Keller]

Since it looks like we have some people on here that like to solve physics problems I'm going to put two more out that I'm not sure about:

1. An object is five ft. square and five ft. high is floating in the ocean. Six inches of the object is above the waterline. What is the minimum line strength that would be needed to lift the object out of the water?
1) 7,500-lb test (based on some calculations below I think it's this answer, but I'm not sure)
2) 10,000-lb test
3) 6,000-lb test
4) 1,000-lb test

The calculations I did for number one are:
The volume of the submerged portion of the object is 5*5*4.5 = 112.5 cu.ft.
Using the formula Buoyancy = Volume * Water_density - Object_weight = 0
Object_weight = 112.5 cu.ft. * 64 lb/cu.ft = 7200 lb
Does this mean that we need 7,500 lb line strength? Why wouldn't it be just 7,200 lb in this case?

2. If a diver takes a sealed, rigid, cubical container measuring 10 inches per side to a depth of 20 ft. in the ocean, the total crushing force (assume the wall thickness is zero) on the container would be about:
1) 3,560 lb
2) 5,256 lb
3) 5,340 lb
4) 14,076 lb

The weight of the amount of water a floating object displaces will give you the objects weight but I would still use a 10,000lb test line anyway assuming "test" means the safe working load of the line.

The pressure on the box would vary depending if the top, bottom or middle of the box were at the 20' mark. The pressure in salt water increases at .445 psi per foot of depth so in the case of a 10" tall object it will make a difference. I will assume the middle of the box is at the 20' mark and go with 5,340lbs.

PS: These types of questions are better asked before Happy Hour starts. Will our names also appear on your certification card if you pass? Getting at least honorable mention would be nice.

 

[rhwestfall]

problem #1 : Clue #1 - archimedes principle

essentially, what is the weight of the water it is displacing?

 

[n03]

The pressure on the box would vary depending if the top, bottom or middle of the box were at the 20' mark. The pressure in salt water increases at .445 psi per foot of depth so in the case of a 10" tall object it will make a difference. I will assume the middle of the box is at the 20' mark and go with 5,340lbs.

Thanks! Do you mind telling how you did the calculations?
I calculated the absolute pressure at 20' = .445 psi * 20 ft + 14.7 psi = 23.6 psi. But how do you apply this to the container? Based on my understanding I would have multiplied 23.6 psi by 100 sq.in of the top surface of the container that the water column pressure is acting on... but this clearly doesn't produce the right answer.

---------- Post added August 14th, 2014 at 06:24 PM ----------

problem #1 : Clue #1 - archimedes principle

essentially, what is the weight of the water it is displacing?

Is it 7200 lb *0.9 (% of the weight submerged) = 6480 lb? Do we care about the weight of the water though?