What is the significance of g_c in different unit systems?

[georg gill]

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?

 

[Andrew Mason]

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

 

[georg gill]

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

 

[D H]

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:

$$g_c=32.1740\frac{0.453593 kg \cdot 0.30408 m}{4.482216 N}=1$$
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
$$g_c =<br /> 32.1740486 \frac {0.45359237\ \text{kg} \cdot 0.30408\ \text{m}/\text{s}^2}<br /> {4.44822162\ \text{N}}=1$$
Now that is unitless and it is indeed one.

What about English customary units? Here we have
$$g_c =<br /> 32.1740486 \frac {1\ \text{lbm} \cdot 1\ \text{ft}/\text{s}^2} {1\ \text{lbf}}$$
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.

 

[OldEngr63]

g_c 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.

 

[gmax137]

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.

 

[OldEngr63]

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.