Hi determine the height of the mountain (using atmospheric pressure)

• Krokodrile
In summary: Suppose I defined 1 kgf as the weight of a 1 kg mass. How many Newton’s would that be?9.8 N. This is because 1 kgf = 1 kg * 9.8 m/s^2 = 9.8 N.
Krokodrile
Homework Statement
determine the height of the mountain from the data shown. Consider the average air density of 0.074 lbm and g= 32 ft/s^2
Relevant Equations
p = p*g*h
i have the first convertions of the atmospheric pressure:
1872 lbf/ft^2 = 89,632.6 Pa
2016 lbm/ft^2 = 96,325.4 Pa
g= 9.72 m/s^2

But, i don't have idea how pass the air density of "lbm" to international units ;,(. And other cuestion: its fine pass lbm/ft^2 and lbf/ft^2 to Pa simirlarly?

The units are all over the place here. Is this the actual problem statement word for word and is the image from the problem?
• Density should have units of mass/length^3, but lbm is a unit of mass only.
• The units in the picture show lbf/s^2 and lbm/s^2 for the pressures. These are wrong as the dimensions of pressure are force/area while lbf/s^2 is force/time^2 and lbm/s^2 is mass/time^2.
I'd really recommend getting the input units correct before doing anything else.

docnet
Orodruin said:
The units are all over the place here. Is this the actual problem statement word for word and is the image from the problem?
• Density should have units of mass/length^3, but lbm is a unit of mass only.
• The units in the picture show lbf/s^2 and lbm/s^2 for the pressures. These are wrong as the dimensions of pressure are force/area while lbf/s^2 is force/time^2 and lbm/s^2 is mass/time^2.
I'd really recommend getting the input units correct before doing anything else.
yes, i wrong in the densty, its lbm/ft^3. yes, the units of pressures its lbf/ft^2 (top) and lbm/ft^2(bottom) (my mistake, sorry). But, i too have the question about these units: why lbf and lbm?

Ok, the lbm/ft^2 is probably a mistake on the problem writer's part. It should likely be lbf/ft^2, which is the correct units for pressure (force/area).

Regarding your question on lbm, 1 lbm is defined as the mass for which the gravitational force in standard Earth gravity is 1 lbf. Since F = mg, 1 lbf = (1 lbm)g or in other words: 1 lbm = (1 lbf)/g. If you convert the 1 lbf to N and insert the appropriate value for g you will get the numerical value for 1 lbm expressed in kg. In order to get the SI units of kg/m^3 for the density, you will of course also have to convert the ft^3 to m^3.

docnet
Orodruin said:
Ok, the lbm/ft^2 is probably a mistake on the problem writer's part. It should likely be lbf/ft^2, which is the correct units for pressure (force/area).

Regarding your question on lbm, 1 lbm is defined as the mass for which the gravitational force in standard Earth gravity is 1 lbf. Since F = mg, 1 lbf = (1 lbm)g or in other words: 1 lbm = (1 lbf)/g. If you convert the 1 lbf to N and insert the appropriate value for g you will get the numerical value for 1 lbm expressed in kg. In order to get the SI units of kg/m^3 for the density, you will of course also have to convert the ft^3 to m^3.
I have all the units in international system, but i can't eliminate the s^2 of gravity and obtain m:

The unit Pa is not kg/m^2 [mass/area], it is N/m^2 [force/area], which equates to (kg m/s^2)/m^2 = kg/(m s^2).

docnet
Orodruin said:
The unit Pa is not kg/m^2 [mass/area], it is N/m^2 [force/area], which equates to (kg m/s^2)/m^2 = kg/(m s^2).
Oohh, thanks for that, now i can eliminate the units, i don't know if i was bad because i got 1/m in the result:

I would do this differently. 2016 - 1872 = 144 psf

144/0.084= 1714 ft

Chestermiller said:
I would do this differently. 2016 - 1872 = 144 psf

144/0.084= 1714 ft
in this case why the gravity was omited? and why 0.084 and not 0.074?

Krokodrile said:
in this case why the gravity was omited? and why 0.084 and not 0.074?
I misread the 0.074 as 0.084.
In imperial units, 1 lbm weighs 1 lbf. In these units, F = ma/g

Last edited:
Chestermiller said:
I misread the 0.074 as 0.084.
In imperial units, 1 lbm weighs 1 lbf.
I have difference of 11.92 m with my answer, i must check it again. The gravity was omited because the mountain its not afected by her? i don't understand why in 144/0.074 the gravity doesnt appears :(

Krokodrile said:
I have difference of 11.92 m with my answer, i must check it again. The gravity was omited because the mountain its not afected by her? i don't understand why in 144/0.074 the gravity doesnt appears :(
lbm is not the proper unit to use with such units . The correct unit to use is the slug, which is lbm/32.2. In this way, the 32.2’s cancel out.

Suppose I defined 1 kgf as the weight of a 1 kg mass. How many Newton’s would that be?

Chestermiller said:
Suppose I defined 1 kgf as the weight of a 1 kg mass. How many Newton’s would that be?
9.80? as 1 kgf = 9.80N

Chestermiller said:
lbm is not the proper unit to use with such units . The correct unit to use is the slug, which is lbm/32.2. In this way, the 32.2’s cancel out.
if the slug 32.2s, its fine to cancel it with 32.0?

Krokodrile said:
if the slug 32.2s, its fine to cancel it with 32.0?
That was just rounded off to 32.

Chestermiller said:
That was just rounded off to 32.
then it would be something like this?:

Ah...The Imperial System...What was the name of that doomed satellite?

Gordianus said:
Ah...The Imperial System...What was the name of that doomed satellite?
Mars Global Surveyor satellite?

Krokodrile said:
then it would be something like this?:
View attachment 280215
It should be $$\frac{144\ lb_f/ft^2}{0.074 \frac{lb_m}{ft^3}(32.2\ \frac{ft}{s^2})\left(\frac{1}{32.2\ \frac{lb_mft}{lb_fs^2}}\right)}$$

Chestermiller said:
It should be $$\frac{144\ lb_f/ft^2}{0.074 \frac{lb_m}{ft^3}(32.2\ \frac{ft}{s^2})\left(\frac{1}{32.2\ \frac{lb_mft}{lb_fs^2}}\right)}$$
thank you so much. I understand all the process, except for the 1/32: where does it come lbm/lb?

Krokodrile said:
thank you so much. I understand all the process, except for the 1/32: where does it come lbm/lb?
It is the factor that converts lbm to slugs. Those units are the same as lbm/slug

Chestermiller said:
It is the factor that converts lbm to slugs. Those units are the same as lb/slug
so, the ft/s^2 its the slug units and lbm/lb is the lfbm to slug factor?

Krokodrile said:
so, the ft/s^2 its the slug units and lbm/lb is the lfbm to slug factor?
$$32.2 lb_m=1\ slug= 1 \frac{lb_f s^2}{ft}$$

Chestermiller said:
$$32.2 lb_m=1\ slug= 1 \frac{lb_f s^2}{ft}$$
oooohhh, and 1/32 its because we can eliminate with the other 32. Ok, i get it. Thank you so much again for your time, Sir.

Krokodrile said:
Oohh, thanks for that, now i can eliminate the units, i don't know if i was bad because i got 1/m in the result:
View attachment 280209
You just did the cancellation of the length units wrong. Nominator has one negative power of length, denominator has three negative and one positive power of length so two negative powers overall, thus the power of length in the final result is one positive power of length, i.e., meters.

1. How does atmospheric pressure help determine the height of a mountain?

Atmospheric pressure decreases as altitude increases. By measuring the atmospheric pressure at the base and summit of a mountain, the difference can be used to calculate the mountain's height.

2. What instruments are used to measure atmospheric pressure?

Barometers are typically used to measure atmospheric pressure. Digital barometers are more accurate and commonly used in modern mountain height calculations.

3. How accurate is using atmospheric pressure to determine the height of a mountain?

Using atmospheric pressure to determine the height of a mountain can provide an accurate estimate, but it may not be exact. Other factors such as weather conditions and topography can affect the accuracy of the measurement.

4. Can atmospheric pressure be affected by other factors besides altitude?

Yes, atmospheric pressure can also be affected by temperature, humidity, and weather patterns. These factors should be taken into consideration when using atmospheric pressure to determine the height of a mountain.

5. Are there any limitations to using atmospheric pressure to determine the height of a mountain?

Yes, there are some limitations to this method. It may not be accurate for very tall or steep mountains, and it may not work in areas with extreme weather conditions. Additionally, it may not be possible to measure the atmospheric pressure at the exact base and summit of a mountain, leading to a less precise calculation.

• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
15
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
12
Views
608
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
950
• Introductory Physics Homework Help
Replies
9
Views
1K
• Introductory Physics Homework Help
Replies
9
Views
2K
• Mechanical Engineering
Replies
3
Views
1K