Mass vs. Weight

Date: 05/27/99 at 14:56:08
From: ryan
Subject: Physics (F = mg)

If you calibrated a scale to read lb. (mass) at g = 32.192 ft/s^2, and 
you took this scale to the moon where g = 5.32 ft/s^2, placed a moon 
rock onto the scale and the scale read 13.37, what would be the mass 
of the rock on the moon?  What would be the weight of the rock on the 
moon?

My question concerns the calibration of the scale at one g, then going 
to the moon and taking masses at another g. Is this a proportionality 
problem?

Date: 05/28/99 at 08:47:32
From: Doctor Rick
Subject: Re: Physics (F = mg)

Hi, Ryan, thanks for your question!

There is a very important piece of information that you have left out. 
What kind of scale is this? It makes a huge difference whether it is a 
balance or a spring scale. I assume it is a spring scale, but before I 
get to that, let me copy an answer I sent to a teacher who asked:

>We have read your response about the difference between mass and 
>weight, and now have a follow question:
>
>We have a triple beam balance that claims to measure mass. It 
>measures in gram units, so we are led to believe it is measuring 
>mass. However, it appears to be like any other scale that truly 
>measures the force that the earth's gravity is putting on an object. 
>(Newtons) I realize mass and weight are used interchangeably, and 
>hope you can provide me with a concise explanation that I can share 
>with my students. 

Hi, Chris. Good question! As a matter of fact, triple beam balances do
measure mass, not gravitational force. The same is true for the simple
two-pan balances that come with the set of "weights." On the other 
hand, spring scales like those we have in our bathrooms measure 
gravitational force.

You see, a balance works by comparing the force of gravity on the 
sample you are weighing with the force of gravity on standard masses. 
In the case of the two-pan balance, you just use the set of weights to 
make up a mass equal to that of your sample. In the case of the 
triple-beam balance, you have fixed weights but you adjust the lever 
arms to get balancing torques.

If you were to take the balance to the moon where the gravitational 
force is less, the force pulling on your sample will be less; but the 
force pulling on the standard masses will be reduced in the same 
proportion. The result is that you will need exactly the same masses 
(or the same lever arms) to balance your sample.

A spring scale, on the other hand, works by comparing the force caused 
by gravity with the force exerted by a spring that is compressed a 
certain amount. The heavier the weight (and this time it really is 
weight), the more the spring has to be compressed to get a matching 
spring force.

If you stood on a spring scale on the moon, your body would exert less 
force on the scale. The spring still has the same spring constant that 
it had on the earth, so it would not have to compress as far to give 
the lower force; and the scale would say you weigh less.

In conclusion, your triple beam balance uses gravity as sort of an
intermediary in its measurement, but it compares mass against mass. It
wouldn't work in space where there is no gravitational force (in the
reference frame of the free-falling balance). But in any place where 
there is a uniform gravitational field, it should give the same 
measurement of mass. Spring balances, though, compare force against 
force, so they measure weight, not mass.

------

Now back to you, Ryan. That may help you understand the background of 
this problem. Did you catch the part about spring scales measuring 
force? When the scale reads m pounds, it is actually detecting a force 
of mg pound-feet per second squared, where g is the acceleration of 
gravity on EARTH since that's where it was calibrated. 

Knowing a reading m (whether on earth or on the moon), you can 
calculate the force detected by the scale. In the problem, this force 
is the gravitational force exerted by the moon on the mass. From this 
and the acceleration of gravity on the moon, you can calculate the 
actual mass.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/