What conversions are needed for solving a density equation?

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The discussion revolves around solving a density equation involving conversions between units. Participants emphasize the importance of ensuring consistent units when adding or subtracting quantities, specifically converting Newtons to grams or kilograms as needed. The expression provided contains terms that can cancel out, leading to confusion about whether a negative density result is valid. Clarifications are sought regarding the interpretation of the equation and the definitions of variables involved. Ultimately, the complexity of the problem suggests that additional context is necessary for accurate resolution.
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Homework Statement



I need to solve the problem listed below. I do not need the answer. I just need to know if there's any conversions that are needed. Do I turn Newtons into kg and kg into g or...? Any kind of leads to help solve this problem would be greatly appreciated. Thank you.

Homework Equations



(3.70 N)(1.00 g/cm3) / (3.70 N) – (2.35 g/cm3)

The Attempt at a Solution



I am not very good with conversions.
 
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Makali said:
(3.70 N)(1.00 g/cm3) / (3.70 N) – (2.35 g/cm3)
First check to see if any of the units cancel (Numerator/Denominator) in your expression.

Whether or not you need to do conversions will depend upon how you want your density expressed. Do you want a result in g/cm3 or kg/cm3, or perhaps kg/m3?
 
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g/cm^3. The N will cancel in the numerator and denominator. 1 in the numerator can just be "thrown out". Since the numbers in the denominator are not multiplication, can we separate them to cancel with the numerator? so the answer that is left is -2.35 g/cm3 but can you have a negative density? Not sure if I am mathing correctly or not.
 
Makali said:

Homework Statement



I need to solve the problem listed below. I do not need the answer. I just need to know if there's any conversions that are needed. Do I turn Newtons into kg and kg into g or...? Any kind of leads to help solve this problem would be greatly appreciated. Thank you.

Homework Equations



(3.70 N)(1.00 g/cm3) / (3.70 N) – (2.35 g/cm3)

The Attempt at a Solution



I am not very good with conversions.

If you add or subtract quantities they must have the same units. So, you need to convert (N *g/cm^3 *1/N) to g/cm^3. You can look up the conversion factors (if any) in your textbook or on-line.

Note added in edit: the other responses did not appear on my screen until after I posted my reply. That type of thing happens to me a lot in this forum.
 
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Thank you!
 
Am I correct with my method stated above your comment, Mr Vickson? :)
 
Makali said:
Since the numbers in the denominator are not multiplication, can we separate them to cancel with the numerator?
Hang on, what are you taking to be the denominator? If I apply the standard parsing rules for mathematical expressions, there are two separate terms:

(3.70 N)(1.00 g/cm3) / (3.70 N)

and

(2.35 g/cm3)

only one of which has a denominator. If that is not the case, then you should use parentheses to group terms to force the correct interpretation.
 
Ray Vickson said:
Note added in edit: the other responses did not appear on my screen until after I posted my reply. That type of thing happens to me a lot in this forum.
I can see the responses, but I cannot see the problem.
 
3.70 * 1 in the numerator is 3.70.. then the denominator is (3.70 N) - (2.35 g/cm3). The 3.70 N would both cancel due to division, correct? So the answer would be 2.35 g/cm3 but it's subtraction so would the answer then be -2.35 g/cm3? I am not sure if density can be negative or not. I'd assume it can't be.
 
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Makali said:
g/cm^3. The N will cancel in the numerator and denominator. 1 in the numerator can just be "thrown out". Since the numbers in the denominator are not multiplication, can we separate them to cancel with the numerator? so the answer that is left is -2.35 g/cm3 but can you have a negative density? Not sure if I am mathing correctly or not.
The first term evaluates to 1 g/cm3, so you should throw it out entirely. You will still get a negative result for your density. That suggests that there is something wrong in the statement of the problem. Maybe if you explain how the expression (3.70 N)(1.00 g/cm3) / (3.70 N) – (2.35 g/cm3) came about, we can help you figure out what went wrong.
 
  • #11
tnich said:
The first term evaluates to 1 g/cm3, so you should not throw it out entirely. You will still get a negative result for your density. That suggests that there is something wrong in the statement of the problem. Maybe if you explain how the expression (3.70 N)(1.00 g/cm3) / (3.70 N) – (2.35 g/cm3) came about, we can help you figure out what went wrong.
(Edited to add "not".)
 
  • #12
The question is
Calculate the density ρ1 of the aluminum block using equation (3) in the lab manual:
The equation in the manual is p0 = W0 pf / Fb = W0pf / w0- T

Al = 3.70 N but is 2.35 N submerged in water
 
  • #13
Makali said:
The question is
Calculate the density ρ1 of the aluminum block using equation (3) in the lab manual:
The equation in the manual is p0 = W0 pf / Fb = W0pf / w0- T

Al = 3.70 N but is 2.35 N submerged in water
That helps, but the problem statement is still incomplete. There is clearly more to it.
What aluminum block is the problem referring to?
We still need definitions of the variables in the equation. My guess is that p0 is the density of aluminum, pf is the density of water, W0 is the weight of the aluminum block and Fb is the weight of water displaced. I'm not sure if W0 and w0 are meant to be the same variable. I don't know what T is either.
 
  • #14
tnich said:
That helps, but the problem statement is still incomplete. There is clearly more to it.
What aluminum block is the problem referring to?
We still need definitions of the variables in the equation. My guess is that p0 is the density of aluminum, pf is the density of water, W0 is the weight of the aluminum block and Fb is the weight of water displaced. I'm not sure if W0 and w0 are meant to be the same variable. I don't know what T is either.
Also, it is kind of difficult to estimate density of aluminum by its displacement of water since it is more dense than water and will sink to the bottom of the bucket. So unless you can tell us what the problem is you are trying to solve, we really can't do much to help you.
 
  • #15
I ended up figuring it out. Thanks.
 
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