What is the significance of g_c in different unit systems?

  • Thread starter Thread starter georg gill
  • Start date Start date
  • Tags Tags
    Units
Click For Summary
SUMMARY

The significance of g_c in different unit systems is that it represents a dimensionless quantity used to compare gravitational acceleration across systems. In the English system, g_c is defined as 32.1740 lb_m·ft/lb_f·s², while in SI units, it simplifies to 1 when expressed as g_c = 32.1740486 (0.45359237 kg·0.30408 m)/(4.44822162 N). This unitless nature allows for consistent application of Newton's second law, F = ma, across various measurement systems. Understanding g_c is crucial for engineers and pilots who need to relate forces experienced during maneuvers to gravitational acceleration.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Familiarity with unit conversion between SI and English customary units
  • Knowledge of gravitational acceleration (9.8 m/s²)
  • Basic grasp of dimensional analysis
NEXT STEPS
  • Research the implications of unit consistency in engineering calculations
  • Explore the differences between SI and English customary systems in practical applications
  • Learn about dimensional analysis techniques for unit conversion
  • Investigate the historical context of measurement systems and their evolution
USEFUL FOR

Engineers, physicists, and pilots who require a clear understanding of gravitational forces and their implications in different measurement systems will benefit from this discussion.

georg gill
Messages
151
Reaction score
6
What is g_c

in pound force. It is said that it is dimensionless

It says in my book that in america where they measure acceleration in feet per s^2 we have:

32.1740 \frac{lb_m\cdot ft}{lb_f\cdot s^2}

then they say in my book that in SI-units we get:

g_c=32.1740\frac{0.453593 kg \cdot 0.30408 m}{4.482216 N}=1
what does this mean?
 
Physics news on Phys.org
georg gill said:
What is g_c

in pound force. It is said that it is dimensionless

It says in my book that in america where they measure acceleration in feet per s^2 we have:

32.1740 \frac{lb_m\cdot ft}{lb_f\cdot s^2}
This is the gravitational acceleration of an object near the Earth surface measured in feet/second^2 . It is equal to about 9.8m/sec^2

then they say in my book that in SI-units we get:

g_c=32.1740\frac{0.453593 kg \cdot 0.30408 m}{4.482216 N}=1
what does this mean?
g_c is a unit of acceleration = 9.8 m/sec^2. It allows one to compare a force that one may experience (due to centripetal acceleration, say) to the force of gravity. This is often used by pilots when doing a manoeuvre. An acceleration of 4g is 4 x 9.8 m/sec^2 = 39.2 m/sec^2 or 128 ft/sec^2.

AM
 
Andrew Mason said:
This is the gravitational acceleration of an object near the Earth surface measured in feet/second^2 . It is equal to about 9.8m/sec^2

g_c is a unit of acceleration = 9.8 m/sec^2. It allows one to compare a force that one may experience (due to centripetal acceleration, say) to the force of gravity. This is often used by pilots when doing a manoeuvre. An acceleration of 4g is 4 x 9.8 m/sec^2 = 39.2 m/sec^2 or 128 ft/sec^2.

AM

but how come they say it is 1 when you use SI-units? I am disregarding your point here I guess
 
Last edited:
georg gill said:
but how come they say it is 1 when you use SI-units? I am disregarding your point here I guess
First off, what you wrote in the opening post isn't quite right. Here's what you wrote:
georg gill said:
g_c=32.1740\frac{0.453593 kg \cdot 0.30408 m}{4.482216 N}=1
what does this mean?
That isn't unitless (it has units of seconds squared) and you have some of the numbers wrong. What you should have written is
<br /> g_c =<br /> 32.1740486 \frac {0.45359237\ \text{kg} \cdot 0.30408\ \text{m}/\text{s}^2}<br /> {4.44822162\ \text{N}}=1<br />
Now that is unitless and it is indeed one.

What about English customary units? Here we have
<br /> g_c =<br /> 32.1740486 \frac {1\ \text{lbm} \cdot 1\ \text{ft}/\text{s}^2} {1\ \text{lbf}}<br />
This looks like it should have a value of 32.174086. It doesn't. The value is once again one. It has to be; it is a unitless quantity. Unitless quantities are the same regardless of how one represents quantities with units such as length, mass, and time.What I think your book is alluding to is the form of Newton's second law. Newton's second law does not say F=ma. It says that force is proportional to mass times acceleration: F\propto ma or F=kma, where k is some constant of proportionality that varies with the representation system.

In addition to the obvious, Newton's second law also tells us is that force, mass, and acceleration are not three independent quantities. There are only two independent quantities here. The route chosen by the developers of the metric system was to make that explicit: Choose the unit of force such that the acceleration of an object with a mass of one unit of mass unit subject to a force of one unit of force will be one unit of distance per unit of time squared. In other words, F=ma. The constant of proportionality is one.

The old English system had concepts of mass, force, distance, and time in place prior to Newton's time. This constant of proportionality is something other than one in English units. One pound force accelerates a one pound mass object by exactly (9.80665/0.3048) ft/s2. The English system constant of proportionality k thus has a numerical value of exactly 0.3048/9.80665, or approximately 1/32.1740486.
 
gc is a confusion factor that results from using mixed units in calculation.

Newton's Second Law reads simply F = m a with no need for any additional constant of proportionality if proper units are used. There are four quantities involved in this equation: Force, mass, length, and time, and one equation relating the four of them. Thus only three of them can be independent. Consider the following table:

....Force...Mass...Length...Time...Derived Unit
(1)...SI...(---)...kg.....m...sec...N ~ Newton force
(2)...cgs...(---)...gm....cm...sec...dy ~ dyne force
(3)...USC1...lb...(---)...ft...sec...slug = lb s2/ft
(4)...USC2...lb...(---)...in...sec...lb s2/in
(5).archaic..(---)...lbm...ft...sec...pdl = poundal = lbm s2/ft

In 40+ years of American engineering practice, I have used the US Customary System 2 (lb-in-s) almost exclusively. I have used SI in a few cases, and I think I used cgs once. I have never, ever used the archaic system based on (lbm-ft-sec) because the poundal is simply too awkward to work with.

What ever system you use, it is critical to be consistent and use the same system through out all of your work. If you are handed a set of field data marked "masses" you can rest assured that they are really weights, so simply cross out the heading (do not change the values on the data sheet), and write over it "weights." Then you will know how to properly incorporate this data into your calculations.
 
Last edited:
Whenever I begin to get confused by this I think: If I push on 1 pound-mass with a force of 1 pound-force, the acceleration will be 32.2 ft/sec^2.

Contrast that with : If I push on one kg with a force of one Newton, the acceleration will be 1 meter/sec^2.

This, I think, is what DH's post is getting at.

By the way, the convenience of the 'english' units is that (here on the surface of the earth) one pound-mass weighs one pound-force. Easy to remember, compared to one kg weighs 9.8 Newtons. Either system, you have to remember a number (either 9.8 or 32.17), so it's just pick your poison.
 
gmax137 said:
Whenever I begin to get confused by this I think: If I push on 1 pound-mass with a force of 1 pound-force, the acceleration will be 32.2 ft/sec^2.

I'm afraid you are correct; this is when you begin to get confused because you are mixing unit systems -- pounds-mass and pounds-force in the same equation. That always makes for difficulties because that when Newton's Law gets violated.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
12
Views
2K
Replies
59
Views
8K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
26
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K