Physics question from DM exam

[emttim]

Ok, so I already did the exam so it's not for me, but I was trying to help out a DM today because he was studying for his physics exam and one of the questions on the exam I simply cannot figure out after the fact, maybe someone could shed some light on the issue.

The question as best I can remember is as follows:

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?

I really don't see any practical application to this problem to begin with, but I'm just curious how you'd figure it out.

 

[SparticleBrane]

Think -- you have all the information needed:

• Weight of object
• Amount of water object displaces
• Amount negative that you want the object to be

Depth is, for all intents and purposes, irrelevant.


Practical application?
Good general information -- also helps you figure out exactly how much weight to move around when switching from seawater to fresh and vice versa.

 

[LowVizWiz]

The object displaces 128# of sea water making it 3# positive (1 cubic foot of sea water = 64#)
so my guess would be to add 53 pounds of lead.

The depth is just there to throw you off.

 

[SparticleBrane]

The object displaces 128# of sea water making it 3# positive (1 cubic foot of sea water = 64#)
so my guess would be to add 53 pounds of lead.

The depth is just there to throw you off.

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.

 

[MurkyRockDiver]

Sounds like the old... Doc I have a friend who has a rash, what do you think he should do so I can tell him...

 

[b1gcountry]

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

Am I missing something? to make the object 50 lbs heavier, disregarding the buoyancy of the lead, you add 50lbs of lead.

???

 

[SparticleBrane]

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

Am I missing something? to make the object 50 lbs heavier, disregarding the buoyancy of the lead, you add 50lbs of lead.

???

Archimedes' Principle, anyone...?

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

 

[Steve_Dives]

Am I missing something? to make the object 50 lbs heavier, disregarding the buoyancy of the lead, you add 50lbs of lead.

Yeah, same thing ran through my head. If you want to make it 50 lbs heavier, add 50 lbs. If you want to make it weigh 50 lbs when submerged, add 53 pounds. They're different questions.

 

[tfsails]

Soooo...if a cubic foot of lead weighs, say, 500 lbs., then it's only 446 lbs. "negative" at the bottom of the ocean...right??

 

[SparticleBrane]

Yeah, same thing ran through my head. If you want to make it 50 lbs heavier, add 50 lbs. If you want to make it weigh 50 lbs when submerged, add 53 pounds. They're different questions.

That isn't exactly what the question was asking.

First, the question stated the SPECIFICALLY DISREGARD the water displaced by the mass of the lead...

Soooo...if a cubic foot of lead weighs, say, 500 lbs., then it's only 446 lbs. "negative" at the bottom of the ocean...right??

No.