Physics question from DM exam

[amascuba]

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?

SB is right. It's Archimedes Principle. There are a few things that you need to know when dealing with Archimedes Principle:

* How much water an object displaces
* How much the object weighs
* How much water weighs

We have the answer to those questions in the question being asked:

1. The object displaces 2 cubic feet of water (In this question we are using cubic feet as an measurement. If I'm not mistaken the DM exam also uses gallons/liters as a form of measurement as well, so you will have to take that into account during those questions)
2. The object weighs 125 pounds
3. There are couple things to look at here to answer this question.
3a. We are talking about sea-water and not freshwater. That will make a slight difference in the weight of water and have to take that into account.
3b. The displacement is measured into cubic feet. We have to account for this when calculating our measurement of water.
3c. Sea-water weighs roughly 64 pounds per cubic feet.

Since the object displaces two cubic feet of water we have to multiply 64 lbs/cf by 2

64 x 2 = 128 lbs/cf

Since the object weighs 125 pounds we need to find out how negative/positive the object is. We do this by subtracting the weight of the displaced water by the weight of the object.

128 - 125 = 3 lbs positive.

The question is how much lead would make the object 50 lbs negative. So first you have to account for the object being 3 lbs positive. You would have to add 3 pounds to make the object neutral. Once the object is neutral you can add another 50 pounds to make it 50 lbs negative.

So the answer is 53 lbs of lead.

As far as depth is concerned, it's a useless statement because the DM exams assume that the object is solid and doesn't have any air space that condenses to throw off the buoyancy swing. In this case, the object is going to be 3 lbs positive and is going to displace 2 cf of water no matter where it is in the water column, until the 53 lbs of lead is added.

 

[ChristyV]

The object displaces 2 cubic feet of water. Each cubic foot of sea water weighs 62.4 lbs. You multiply the 62.4 by the 2 cubic feet. That is the Archimedes principle at work. It is being buoyed up by 124.8 lbs. You subtract that amount from the weight of the object (125 lbs) to get the weight you actually need to lift. When you subtract it, you find that the object is barely negatively buoyant as it is. To make it negatively buoyant by 50 lbs, you would subtract the .2 left and ...to be precise add 49.8 lbs. I was so blessed here on Maui to have an instructor, Laura Scott out of Lahaina Divers, who can really teach physics. She taught remedial science (she's a Stanford honors science grad) I had to sit (voluntarily) through the physics lecture 4 times and get individual help, but if she could teach this stuff to me with all my math phobias, she could teach it to anyone. If your friend finds it too difficult he should just fly out here to Maui and look her up at the Lahaina Dive Shop. She's the best! Christy

 

[ChristyV]

OOOPS! As soon as I hit send, I realized I gave the weight of fresh water...62.4 lbs and not salt water which is heavier at 64 lbs. The problem can be reworked the same way though multiplying 2 cubic feet by 64 and then deducting that from the weight of the object. I hope Laura doesn't read that!!! Christy

 

[kadleck]

The question is...what is six times seven

Well I botched that question. It's actually "What is seven multiplied by nine?"

I think in the Hitchhiker's Guide to the Galaxy, the ultimate question (giving the answer 42) was actually "What is six multiplied by nine?"

I must go and check. But I seem to remember that 6x9 is 42 (in base 13).:shocked2:

 

[NudeDiver]

SB is right. It's Archimedes Principle. There are a few things that you need to know when dealing with Archimedes Principle:

* How much water an object displaces
* How much the object weighs
* How much water weighs

We have the answer to those questions in the question being asked:

1. The object displaces 2 cubic feet of water (In this question we are using cubic feet as an measurement. If I'm not mistaken the DM exam also uses gallons/liters as a form of measurement as well, so you will have to take that into account during those questions)
2. The object weighs 125 pounds
3. There are couple things to look at here to answer this question.
3a. We are talking about sea-water and not freshwater. That will make a slight difference in the weight of water and have to take that into account.
3b. The displacement is measured into cubic feet. We have to account for this when calculating our measurement of water.
3c. Sea-water weighs roughly 64 pounds per cubic feet.

Since the object displaces two cubic feet of water we have to multiply 64 lbs/cf by 2

64 x 2 = 128 lbs/cf

Since the object weighs 125 pounds we need to find out how negative/positive the object is. We do this by subtracting the weight of the displaced water by the weight of the object.

128 - 125 = 3 lbs positive.

The question is how much lead would make the object 50 lbs negative. So first you have to account for the object being 3 lbs positive. You would have to add 3 pounds to make the object neutral. Once the object is neutral you can add another 50 pounds to make it 50 lbs negative.

So the answer is 53 lbs of lead.

As far as depth is concerned, it's a useless statement because the DM exams assume that the object is solid and doesn't have any air space that condenses to throw off the buoyancy swing. In this case, the object is going to be 3 lbs positive and is going to displace 2 cf of water no matter where it is in the water column, until the 53 lbs of lead is added.

None of this crap matters!!! It has nothing to do with diving or depth or displacement or anything else. The question asks "What amount of lead (disregarding the water displaced by the mass of the lead) would be required to add 50 lbs of negative buoyancy to the object?" and the answer is 50 pounds.

IF the question were asking how much weight do you need to add to give the object a net buoyancy of -50 pounds, only then would the answer be 53 pounds.

 

[boulderjohn]

Let me first try to step in from the perspective of a quarter century of teaching the English Language in countless classes. As the question is worded, the answer is 50 pounds, for exactly the reason several people have already stated. The wording is not ambiguous. It does not say you have to make the object 50 pounds negative; it says you have to add 50 pounds.

Now let me step in as someone who has taken the test, although it has been several years. The question has obviously been misremembered. There is no way a question with that wording appeared on the test. The most likely wording would have asked you to make the object 50 pounds negative, in which case 53 pounds is the answer. That is probably the way it was worded, but we won't know without seeing the actual question.

 

[amascuba]

None of this crap matters!!! It has nothing to do with diving or depth or displacement or anything else. The question asks "What amount of lead (disregarding the water displaced by the mass of the lead) would be required to add 50 lbs of negative buoyancy to the object?" and the answer is 50 pounds.

IF the question were asking how much weight do you need to add to give the object a net buoyancy of -50 pounds, only then would the answer be 53 pounds.

Re-read the question.

disregarding the water displaced by the mass of the lead

That's a little misleading. They aren't telling you to diregard the displacement of the object. They are telliing you to disregard the displacement of the lead that you would ADD to the object to make it Negative.

What amount of lead would be required to add 50 lbs of negative buoyancy to the object?

Meaning that they want the object to be 50 lbs NEGATIVE. If the object is starting off 3 lbs positive to begin with, then you need to add 53 lbs of lead to make the object 50 lbs negative.

 

[amascuba]

The object displaces 2 cubic feet of water. Each cubic foot of sea water weighs 62.4 lbs. You multiply the 62.4 by the 2 cubic feet. That is the Archimedes principle at work. It is being buoyed up by 124.8 lbs. You subtract that amount from the weight of the object (125 lbs) to get the weight you actually need to lift. When you subtract it, you find that the object is barely negatively buoyant as it is. To make it negatively buoyant by 50 lbs, you would subtract the .2 left and ...to be precise add 49.8 lbs. I was so blessed here on Maui to have an instructor, Laura Scott out of Lahaina Divers, who can really teach physics. She taught remedial science (she's a Stanford honors science grad) I had to sit (voluntarily) through the physics lecture 4 times and get individual help, but if she could teach this stuff to me with all my math phobias, she could teach it to anyone. If your friend finds it too difficult he should just fly out here to Maui and look her up at the Lahaina Dive Shop. She's the best! Christy

Fresh water weighs approximately 62.4 lbs
Salt water weighs approximately 64 lbs.

Salt water is more dense because of the salt, hense the heavier weight.

oops.. you corrected yourself.

 

[TSandM]

What amount of lead would be required to add 50 lbs of negative buoyancy to the object?

I find it fascinating that some very intelligent people are construing this sentence to mean how much weight is required to make the object 50 pounds negative. That is not what the sentence says at all. Whether it is what the original question said, or whether it is what the original question MEANT, we don't know. But it is not what THIS question says. It says we are ADDING 50 lbs of negative buoyancy, which requires 50 lbs of lead.

It almost certainly was miswritten or misremembered in the original question. But the answer is simple, and Sparticle, I'm surprised at you! Parse the sentence.

 

[Charlie99]

I can assure you, I can do basic math. :cool2:

But can you READ?

You are answering the question as you believe it to be on the exam rather than answering the question the OP actually posted.

There's a difference between "Make something 50 pounds negatively buoyant" and "add 50 pounds of negative buoyancy".