Physics question from DM exam

[dumpsterDiver]

the answer is 50 lbs.. Sorry

 

[Steve_Dives]

That isn't exactly what the question was asking.

The answer to the question in the opening post is 50 pounds. However, that question doesn't require knowing the density of sea water or Archimedes Principle in order to answer it.

So the question would probably ask how to make the object 50 pounds negative. The student figures out whether the object is positive or negative to begin with and then add the appropriate amount of lead.

 

[kadleck]

I'm clearly missing something here; but if the object is 3 lbs positive, how is it "lying in 56 feet of depth"? Surely it would be floating?

 

[LowVizWiz]

Hey, maybe next time just spoon feed him the answer so he doesn't learn anything...oh wait.

I was typing while you posted and after I saw your post I realized I probably should have let him think about it.

Thanks for pointing out that I did jump the gun
I'm sorry if I ruined your teaching session

Wiz

 

[ChrisA]

What amount of lead (disregarding the water displaced by the mass of the lead) would be required to add 20 kg/50 lbs of negative buoyancy to the object?

The answer ti "50 pounds". You don't need to know anything about the object. Any object gains 50 pounds when you add 50 pounds to it.

I think maybe you did not remember the question very well.

 

[LowVizWiz]

I'm clearly missing something here; but if the object is 3 lbs positive, how is it "lying in 56 feet of depth"? Surely it would be floating?

It's a hypothetical question, the depth as I said is there to "throw you off" just as in real life you could not disregard the displacement of the lead added.

SparticleBarne, you can still teach something here as it seems that even giving the answer does not clarify the issue.

 

[SparticleBrane]

"What do you want me to do? LEAVE? Then they'll keep being wrong!"



Alright, let's look over the question again, shall we?

An object weighing 50 kg/125 lbs displaces 2 cubic feet of sea water and is lying in 56 feet of depth. What amount of lead (disregarding the water displaced by the mass of the lead) would be required to add 20 kg/50 lbs of negative buoyancy to the object?

Now, let's look at Archimedes' Principle (already posted once by me )...

Any body fully or partially submerged in a fluid is buoyed up by a force equal to the weight of the fluid displaced.

Facts:

• Object "displaces 2 cubic feet of sea water"
• Object weighs 125lbs
• We are "disregarding the water displaced by the mass of the lead" -- eg, pretend the lead weights used to make the object negative aren't buoyed up by the water for ease of calculation.

Seawater weighs, on average, 64lbs per cubic foot. If the object displaces 2 cubic feet, then it is displacing 128lbs. Thus, according to Archimedes' Principle, the object is being buoyed up by a force of 128lbs. It weighs 125lbs...meaning the object is 3lbs positive because there's 3lbs more force buoying it upwards than there is pulling it down by gravity.


Obviously, the way to make an object that is neutrally buoyant 50lbs negative is to add 50lbs of weight to it (again, disregarding the buoying force of water on the lead weights). In this particular scenario, since the object is 3lbs positive already, you have to add 53lbs -- 3lbs to overcome the fact that the object is already 3lbs positive, and then 50lbs to make it 50lbs negative.



The fact that the wrong answer has been propagated by an OW instructor (and many divers) makes me sad.

 

[LowVizWiz]

Well, first off, buoyancy is never negative. Secondly:

Am I missing something?
???

You can be positively, negatively or neutrally buoyant

 

[emttim]

Hey, maybe next time just spoon feed him the answer so he doesn't learn anything...oh wait.


This was a teaching moment, before the answer was given. If you help lead a person to the correct answer (without giving them the answers), making sure they understand the CONCEPTS that bring the answer about, they'll learn much more than if you just give them the answer or make them memorize a formula.

"What do you want me to do? LEAVE? Then they'll keep being wrong!"



Alright, let's look over the question again, shall we?




Now, let's look at Archimedes' Principle (already posted once by me )...



Facts:

• Object "displaces 2 cubic feet of sea water"
• Object weighs 125lbs
• We are "disregarding the water displaced by the mass of the lead" -- eg, pretend the lead weights used to make the object negative aren't buoyed up by the water for ease of calculation.

Seawater weighs, on average, 64lbs per cubic foot. If the object displaces 2 cubic feet, then it is displacing 128lbs. Thus, according to Archimedes' Principle, the object is being buoyed up by a force of 128lbs. It weighs 125lbs...meaning the object is 3lbs positive because there's 3lbs more force buoying it upwards than there is pulling it down by gravity.


Obviously, the way to make an object that is neutrally buoyant 50lbs negative is to add 50lbs of weight to it (again, disregarding the buoying force of water on the lead weights). In this particular scenario, since the object is 3lbs positive already, you have to add 53lbs -- 3lbs to overcome the fact that the object is already 3lbs positive, and then 50lbs to make it 50lbs negative.



The fact that the wrong answer has been propagated by an OW instructor (and many divers) makes me sad.

You seem to get pretty bent out of shape by the fact that I didn't respond to your post right away...sorry, I have other things to do in my spare time.

And for your general info, I did try to solve it on my own. Usually someone asks a question when they can't figure something out because they're looking for answers. Unfortunately, I was hoping that maybe all of the responses wouldn't be idiotic. Even if yours is correct, it still fits the aforementioned criteria. Sorry, I thought we could all be adults here.

 

[SparticleBrane]

You seem to get pretty bent out of shape by the fact that I didn't respond to your post right away...sorry, I have other things to do in my spare time.

Actually, I'm pretty bent out of shape that several certified divers (who are supposed to be able to answer questions like this) and an OW instructor are all handing out incorrect information.

Tim I hope you got the answer you were looking for -- I'm actually not irked at you at all.