Units of acceleration [f/s^2] vs [g]

[jahmookcity]

Homework Statement

Force, mass, acceleration

Homework Equations

F=ma

The Attempt at a Solution

Hi

I am having a hard time understanding the units of acceleration. Normally I am used to distance/time^2 being the standard unit [ m/s^2 , ft/s^2 ]

Now I am seeing this gravity unit [g] being thrown around. Here are 2 examples of simple calculations that use this unit (note highlighted regions of text)
http://imgur.com/S14dnYS,oB56WXS
http://imgur.com/S14dnYS,oB56WXS#1

**The problem I have is:**
If g represents gravity, shouldn't 1.5g = 1.5*32.2[ft/s^2] or 1.5*9.8[m/s^2]. I like doing all my calculations using consistent units and this [g] unit is throwing me off.

For example:
If you look at yellow highlighted text in the 2nd link above, you will see some person claiming that from F=ma:
F = 600lbm x 1.5g = 900lbf

...I feel like this person is incorrect and the F=ma equation should be:

F = 600lbm x 1.5 x 32.2 ft/s2 = 28,980lbf.

 

[Chestermiller]

Hi Jamookcity. Welcome to Physics Forums.

The problem you are encountering with the calculations you refer to is that they are using "English units." A unit of mass in this system is the lb_m, and the unit of force is lb_f. The lb_m is defined such that it has has a weight of 1 lb_f at the surface of the earth, where the acceleration of gravity is 32.2 ft/sec². So you can see that, in these units, F is not equal to ma. In these units F = ma/32.2. The dilemma of using lb_m is overcome in one of two ways. One of these is to properly define the mass such that F does equal to ma. This is done by defining the unit of mass as the Slug M, which is equal to the mass in lb_m divided by 32.2: M = m/32.2. So, F = Ma. The other way of overcoming this is to include a constant in the F = ma equation, called g_c. g_c = 32.2 (lb_fft)/(lb_msec²). Then F = ma/g_c.

I know that all this sounds crazy to someone who has never worked with it. But it all goes back to the historical development in the English system of units. Back in the old days, when the metric system was not used as extensively in the US, we were taught to work with the English system, and we eventually got used to it.

Chet

 

[Bystander]

**The problem I have is:**
If g represents gravity, shouldn't 1.5g = 1.5*32.2[ft/s^2] or 1.5*9.8[m/s^2]. I like doing all my calculations using consistent units and this [g] unit is throwing me off.

Good here.

For example:
If you look at yellow highlighted text in the 2nd link above, you will see some person claiming that from F=ma:
F = 600lbm x 1.5g = 900lbf

"Pound mass," and "pound force." Remember back in high school, or whenever, that for English units, a mass of one slug at Earth's surface exerted a force of 32 pounds? Therefore, the "pound mass" is 1/32 of a slug.

...I feel like this person is incorrect and the F=ma equation should be:

F = (600lbm/32.2) x 1.5 x 32.2 ft/s2 = 28,980lbf.

... and Chet beat me to it.

 

[jahmookcity]

Hi Jamookcity. Welcome to Physics Forums.

The problem you are encountering with the calculations you refer to is that they are using "English units." A unit of mass in this system is the lb_m, and the unit of force is lb_f. The lb_m is defined such that it has has a weight of 1 lb_f at the surface of the earth, where the acceleration of gravity is 32.2 ft/sec². So you can see that, in these units, F is not equal to ma. In these units F = ma/32.2. The dilemma of using lb_m is overcome in one of two ways. One of these is to properly define the mass such that F does equal to ma. This is done by defining the unit of mass as the Slug M, which is equal to the mass in lb_m divided by 32.2: M = m/32.2. So, F = Ma. The other way of overcoming this is to include a constant in the F = ma equation, called g_c. g_c = 32.2 (lb_fft)/(lb_msec²). Then F = ma/g_c.

I know that all this sounds crazy to someone who has never worked with it. But it all goes back to the historical development in the English system of units. Back in the old days, when the metric system was not used as extensively in the US, we were taught to work with the English system, and we eventually got used to it.

Chet

Good here.
"Pound mass," and "pound force." Remember back in high school, or whenever, that for English units, a mass of one slug at Earth's surface exerted a force of 32 pounds? Therefore, the "pound mass" is 1/32 of a slug.
... and Chet beat me to it.

Ok, thank you both for your responses. I understand this now. Also, I agree with both of you that SI is easier to understand and work with.

 

[Orodruin]

F is not equal to ma. In these units F = ma/32.2

I never liked arguments like this. I think it is more physical to always have F = ma, which is a relation for dimensionful quantities and thus includes the units. In the end you will be left with a new dimensionful quantity with units depending on what units you put in. This may not be the units you desired and in that case you have to do unit conversion, which is just a matter of changing units, not fundamentally changing the physical relationship.

 

[Chestermiller]

I never liked arguments like this. I think it is more physical to always have F = ma, which is a relation for dimensionful quantities and thus includes the units. In the end you will be left with a new dimensionful quantity with units depending on what units you put in. This may not be the units you desired and in that case you have to do unit conversion, which is just a matter of changing units, not fundamentally changing the physical relationship.

Later in the post I explained the g_c (i.e., the 32.2) has units of $\frac{lb_m}{lb_f}\frac{ft}{sce^2}$. It all goes back historically to the use of the lb_m unit in the English system. In the extensive scientific and engineering literature in which the English system is used, g_c appears abundantly, and it is important for the Starter to be aware of this if he encounters it. His unawareness of it was what led him to get the wrong answer.

Chet

 

[Orodruin]

The only thing I am saying is that I do not want the OP to think that the physics depends on the units used to describe it. Thus, to be picky, the 32.2 comes with a unit (lb_m ft / lb_f s²) with which it essentially evaluates to 1, just as when converting from m to cm, you would multiply by 1 = 100 cm/m.

 

[lep11]

I personally prefer the metric system over imperial units. inches, feet, gallons, miles and pounds are just stupid.

 

[jtbell]

Just as an aside, this sort of question is fine in the discussion forums e.g. General Physics, even if it arises from school coursework.

The homework forums are intended for help with working through textbook-type exercises, whether they are part of schoolwork or not.

 

[Chestermiller]

I personally prefer the metric system over imperial units. inches, feet, gallons, miles and pounds are just stupid.

I personally prefer computers and calculators over slide rules and abacus. Slide rules and abacus are just stupid. (lol)

But seriously, there is still tons of equipment out there that was designed using the imperial system. In the US, flow rates and production rates are still often expressed in lb_m/hr. Equipment sizes and product sizes are still in imperial units. Economically, it's very expensive to make the conversion to metric, particularly in such a huge economy.

Chet