Understanding Unit Cancellation in Equations for Confused Students

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The discussion focuses on understanding unit cancellation in the equation h = 3270 Pa / (1.0 g/cm^3 × 9.81 m/s^2) and how it results in a height measurement in centimeters. Participants emphasize the importance of correctly converting units, particularly noting that 1 g/cm^3 equates to 1000 kg/m^3, which is crucial for accurate calculations. There is a consensus that the original equation mixes units improperly, leading to confusion about the final result. The correct approach involves recognizing the relationship between different units of measurement and applying consistent conversions. Ultimately, the discussion highlights the need for clear communication of thought processes to facilitate learning and understanding in physics.
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Homework Statement



I've an equation here: h = 3270 Pa / 1,0 g/cm^3 × 9,81 m/s^2 = 33 cm

How did those units cancel out to make it cm? That's something I don't get.

Homework Equations



Given above.

The Attempt at a Solution



I've tried to cancel them out by looking at the same unit in the numerator and denominator such as, for example m / m , so m cancels out and so on but I don't get it. Does anyone know or can anyone explain why the answer is in centimeters? How do the units cancel out in the given equation? Thank you
 
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Forum rules require you to show your attempt so you need to actually show your attempt at cancelling out the units.

Technically you could put your answer in any length unit.
 
Orodruin said:
Forum rules require you to show your attempt so you need to actually show your attempt at cancelling out the units.

Technically you could put your answer in any length unit.

If we count it:

h = 3270 Pa / 1,0 g/cm^3 × 9,81 m/s^2 = 333 m

It should be 333 meters. How that becomes centimeters? As for your statement, there really isn't much to show. I can't / don't know how to cancel them out but above I explained the idea of canceling out, so I've proved that I know it.
 
JesseK said:
If we count it:

h = 3270 Pa / 1,0 g/cm^3 × 9,81 m/s^2 = 333 m

It should be 333 meters. How that becomes centimeters? As for your statement, there really isn't much to show. I can't / don't know how to cancel them out but above I explained the idea of canceling out, so I've proved that I know it.
You cannot put in different units and just assume that the answer will be in SI units. If you did the cancellation correctly, there would be a lot to show. Note that your density is in g/cm^3, not in kg/m^3!

As a hint: How many grams are there in 1 kg? What is (1 kg)/(1 g)?
 
Orodruin said:
You cannot put in different units and just assume that the answer will be in SI units. If you did the cancellation correctly, there would be a lot to show. Note that your density is in g/cm^3, not in kg/m^3!

As a hint: How many grams are there in 1 kg? What is (1 kg)/(1 g)?

There are 1000 grams in 1 kg. I still don't get it.
 
JesseK said:
There are 1000 grams in 1 kg. I still don't get it.
So how many kg/m^3 are there in 1 g/cm^3?
 
Orodruin said:
So how many kg/m^3 are there in 1 g/cm^3?

Well, 1 kg/m^3 = 0,001 g/cm^3 --> did you mean it vice versa? If so, then 1 g/cm^3 is 1000 kg/m^3
 
JesseK said:
Well, 1 kg/m^3 = 0,001 g/cm^3
No, this is wrong. 1 m^3 is not the same as 1 cm^3.
 
Orodruin said:
No, this is wrong. 1 m^3 is not the same as 1 cm^3.

That's right but 1 m^3 can be converted into cm^3, for example. I'd just like to let you know that I still haven't got the idea.
 
  • #10
JesseK said:
That's right but 1 m^3 can be converted into cm^3, for example.
Yes it can, just like 1 g can be converted into kilograms, the point is that you didn't do that. Hence your answer actually has the units cm^3 kg / (m^2 g).
 
  • #11
Orodruin said:
Yes it can, just like 1 g can be converted into kilograms, the point is that you didn't do that. Hence your answer actually has the units cm^3 kg / (m^2 g).

How can I solve this problem?
 
  • #12
How did you solve the problem of figuring out how many grams are in a kilogram? How would you solve the problem of figuring out what 30 km/h would be in m/s?
 
  • #13
Orodruin said:
How did you solve the problem of figuring out how many grams are in a kilogram? How would you solve the problem of figuring out what 30 km/h would be in m/s?

1 kg = 1000 g

kg hg dag g , so
1 0 0 0

30 km/h can be converted into m/s by dividing it by 3,6 , so it'd be 8,333333... m/s

However, I don't still get the idea in this specific instance.
 
  • #14
You are missing the point entirely.
JesseK said:
30 km/h can be converted into m/s by dividing it by 3,6
The point was not what is the actual factor, the point is how you obtain that factor.
 
  • #15
Orodruin said:
You are missing the point entirely.

The point was not what is the actual factor, the point is how you obtain that factor.

Well 1 km = 1000 meters and 1 hour is 3600 seconds. That gives me the factor.
 
  • #16
JesseK said:
Well 1 km = 1000 meters and 1 hour is 3600 seconds. That gives me the factor.
So what is cm^3 kg / (m^2 g) in cm? The procedure is exactly the same.
 
  • #17
Orodruin said:
So what is cm^3 kg / (m^2 g) in cm? The procedure is exactly the same.

I've got no idea.
 
  • #18
Yes you do, you did it when you converted km/h to m/s.
 
  • #19
Orodruin said:
So what is cm^3 kg / (m^2 g) in cm? The procedure is exactly the same.

cm^2 * kg / m^2 * g

m^2 / m = m

kg / g = 1000 g => cm * 1000 ?
 
  • #20
JesseK said:
m^2 / m = m
Where did this come from? The length units you have are cm^3/m^2 = cm (cm/m)^2.
 
  • #21
Orodruin said:
Where did this come from? The length units you have are cm^3/m^2 = cm (cm/m)^2.

I don't know how to do it, so could you possibly show me?
 
  • #22
JesseK said:
I don't know how to do it, so could you possibly show me?
The classical way to do it is algebraicly, multiplying by a series of "1"s that are chosen so that the units cancel.
For instance, you can write 1 as ##\frac{1\ hr}{3600\ sec}##
$$30 \frac{km}{hr} \cdot \frac{1\ hr}{3600\ sec} = \frac{1}{120} \frac{km}{sec}$$
$$\frac{1}{120} \frac{km}{sec} \cdot \frac{1000\ m}{1\ km} = \frac{1000}{120} \frac{m}{sec}$$
 
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  • #23
jbriggs444 said:
The classical way to do it is algebraicly, multiplying by a series of "1"s that are chosen so that the units cancel.

I call that the Cleverly-Multiply-by-1 method.

I prefer (and strongly advocate to my students) the Substitution method.
So, we have
$$30 \frac{\rm\ km}{\rm\ hr}=30\frac{\rm\ km}{(60\rm\ min)}=30\frac{\rm\ km}{(60\rm (60\rm \sec))}=\frac{30}{(60)(60)} \frac{\rm\ km}{\rm\ sec}$$
$$\frac{1}{120} \frac{\rm\ km}{\rm\ sec}=\frac{1}{120} \frac{\rm\ (1000\ m)}{\rm\ sec} = \frac{1000}{120} \frac{\rm\ m}{\rm\ sec}
$$
and the classics
$$1\rm\ m^3=1\rm (100\rm\ cm)^3=10^6\rm\ cm^3$$
$$1\rm\ cm^3=1\rm (10^{-2}\rm\ m)^3=10^{-6}\rm\ m^3$$
and
$$(1\rm\ y)=(1\rm\ (365\ days))
=(1\rm\ (365\ (24\ h)))
=(1\rm\ (365\ (24 (60\ m))))
=(1\rm\ (365\ (24 (60 (60\ s))))
=(365)(24)(60)(60)\ s
$$
 
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  • #24
JesseK said:
If so, then 1 g/cm^3 is 1000 kg/m^3

Right. Use the latter instead of the former in your original equation and you will then see how to get the 33 cm answer.

Note that in that original equation it's not just the units that don't work out right, it's also the numbers! I'm not sure where that original equation came from, but whoever did it needs to understand that you can't just mix units together like that and expect to get an answer that has any meaning.
 
  • #25
JesseK said:
I don't know how to do it, so could you possibly show me?
If you want spoon fed answers you are in the wrong place. We encourage students to think for themselves and offer guidance. Saying ”I have no idea, show me” is counter productive and will not help you in the long run.
 
  • #26
Orodruin said:
If you want spoon fed answers you are in the wrong place. We encourage students to think for themselves and offer guidance. Saying ”I have no idea, show me” is counter productive and will not help you in the long run.

I have tried to think about it myself as you have probably noticed. Saying that I don't get the idea and asking for help should be completely fine. However, I thank you for giving me this tip.
 
  • #27
JesseK said:
I have tried to think about it myself as you have probably noticed. Saying that I don't get the idea and asking for help should be completely fine. However, I thank you for giving me this tip.
If you have really tried to think about it you have unfortunately done a bad job in communicating it. Your posts have been very brief with no argumentation. To show that you have thought about it, you need to describe what you are doing and why you are doing it. This is the only way that we can follow and correct your thought process.

I am not saying this to be mean or to belittle your work. I am trying to give you an honest advice on how to benefit more from and getting the appropriate guidance.
 
  • #28
JesseK said:
I have tried to think about it myself as you have probably noticed.

Yes, but explaining your thinking is the only path to success. If all you can do is try, and you are not able to explain what you've tried, you will never be able to interact with others in a way that advances your knowledge and promotes your success.

By the way, did you read Post #24?
 
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  • #29
Mister T said:
Note that in that original equation it's not just the units that don't work out right, it's also the numbers! I'm not sure where that original equation came from, but whoever did it needs to understand that you can't just mix units together like that and expect to get an answer that has any meaning.
I think the numbers work out quite well to the given answer:
https://www.wolframalpha.com/input/?i=(3270+Pa)+/+(1.0+g/cm^3+*+9.81+m/s^2)
The natural interpretation is the height difference required for a pressure difference of 3270 Pa in a medium with density 1 g/cm^3 and a gravitational field of 9.81 m/s^2. The pressure differential given is about 0.03 bar and the density that of water. This result makes perfect sense and is compatible with the fact that (as any scuba diver knows) the pressure at a depth of 10 m is roughly 2 bar (1 bar from the atmosphere and 1 bar from the 10 m water column, meaning 0.1 bar/m).
 
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