PlanetPhysics/Difference Between Mass and Weight

The beginner often finds it difficult to distinguish between the [[../Mass/|mass]] of a body and its weight. He is apt to ask such a question as this, "When I buy a pound of fruit what do I get, one pount-mass or one pound-weight?"\footnotemark\ The difficulty is due to the fact that the common methods for comparing the masses of bodies make use of their weights.

There are two general methods by which masses may be compared, both of which are based upon [[../Newtons1stLaw/|Newton's laws of motion]]. Let $F_{1}$ and $F_{2}$ be the resultant forces acting upon two bodies having masses $m_{1}$ and $m_{2}$ , and $f_{1}$ and $f_{2}$ be the [[../Acceleration/|accelerations]] produced. Then the force equation gives

$F_{1} = m_{1} f_{1}$ $F_{2} = m_{2} f_{2}$

and

$\frac{m_{1}}{m_{2}} = \frac{F_{1}}{F_{2}} \cdot \frac{f_{2}}{f_{1}}$

(1) If the forces are of such [[../AbsoluteMagnitude/|magnitudes]] that the accelerations are equal then the masses are proportional to the forces; for when $f_{1} = f_{2}$ , the last equation becomes

$\frac{m_{1}}{m_{2}} = \frac{F_{1}}{F_{2}}$

This gives us a method of comparing masses, of which the common method of weighing is the most important example. If $W_{1}$ and $W_{2}$ denote the weights of two bodies of masses $m_{1}$ and $m_{2}$ , then by the equation that gives the magnitude of the force

$f = \sqrt{{\ddot{x}}^{2} + {\ddot{y}}^{2}}$

then we obtain

$W_{1} = m_{1} g$ $W_{2} = m_{2} g$

and

$\frac{m_{1}}{m_{2}} = \frac{W_{1}}{W_{2}}$

where $g$ is the common acceleration due to gravitational attraction.

(2) If the forces acting upon the bodies are equal the masses are inversely proportional to the accelerations:

$\frac{m_{1}}{m_{2}} = \frac{f_{2}}{f_{1}}$

This gives us the second method by which masses may be compared. The following are more or less practicable applications of this method:

(a) Let $A$ and $B$ (Fig. 61) be two bodies connected with a long elastic string of negligible mass, placed on a perfectly smooth and horizontal table.

\includegraphics[scale=.8]{Fig61.eps}

Suppose the string to be stretched by pulling $A$ and $B$ away from each other. It is evident that when the bodies are released they will be accelerated with respect to the table and that the accelerating force, that is, the pull of the string, will be the same for both bodies. Therfore if $f_{1}$ and $f_{2}$ denote their accelerations at any instant of their [[../CosmologicalConstant2/|motion]], the ratio of their masses is given by the [[../Bijective/|relation]] $\frac{m_{1}}{m_{2}} = \frac{f_{2}}{f_{1}}$

(b) Suppose the bodies whose masses are to be compared to be fitted on a smooth horizontal rod (Fig. 62) so that they are free to slide along it.

\includegraphics[scale=.8]{Fig62.eps}

If the rod is rotated about a vertical axis the bodies fly away from the axis of rotation. If, however, the bodies are connected by a string of negligible mass they occupy [[../Position/|positions]] on the two sides of the axis, which depend upon the ratio of the masses. So far as the motion along the rod is concerned, each body is equivalent to a [[../Particle/|particle]] of the same mass placed at the [[../CenterOfGravity/|center of mass]] of the body.

Suppose, as it is assumed in Fig. 62, the horizontal rod to be hollow and tohave smooth inner wall; further suppose the centers of mass of the given bodies to lie on the axis of the rod. Then if at the center of mass of each body a particle of equla mass is placed and the two particles connected by means of a massles string of proper lengths, the positions of the particles will remain at the centers of mass of the given bodies even when the rod is set rotating about the vertical axis.

Now let $m_{1}$ and $m_{2}$ be the masses of the particles and $f_{1}$ and $f_{2}$ their accelerations due to the rotation of the tube about the vertical axis. Then since the tensile force in the string is the same at its two ends, the forces acting upon the particles are equal. Therefore we have

$F = m_{1} f_{1} = m_{2} f_{2}$

or

$\frac{m_{1}}{m_{2}} = \frac{f_{2}}{f_{1}}$

But if $r_{1}$ and $r_{2}$ denote the distances of the particles from the axis of rotation, and $P$ the period of revolution, then

$f_{1} = - \frac{v_{1}^{2}}{r_{1}} = - \frac{4 π^{2} r_{1}}{P^{2}}$

and

$f_{2} = - \frac{v_{2}^{2}}{r_{2}} = - \frac{4 π^{2} r_{2}}{P^{2}}$

Therefore

$\frac{m_{1}}{m_{2}} = \frac{r_{2}}{r_{1}}$

gives the ratio of the masses of the particles as well as those of the given bodies.

References

This article is a derivative of the public [[../Bijective/|domain]] [[../Work/|work]], "Analytical [[../Mechanics/|mechanics]]" by Haroutune M. Dadourian, 1913. Made available by the internet archive

\footnotetext{This question may be answered in the following manner. "The fruit which you get has a mass of 1 pd. (about 453.6 gm.) and which weighs 1 lb. (about $4.45 \times 1 0^{6}$ dynes). If the fruit could be shipped to the moon during the passage the weight would diminish down to nothing and then increase to about one-sixth of a pound. The zero weight would be reached at a point about nine-tenths of the way over. Up to that position the wight would be with respect to the earth, that is, the fruit would be attracted towards the earth; but from there on the weight would be with respect to the moon. The mass of the fruit, however, would be the same on the earth, during the passage, and on the moon. It would be the same with respect to the moon as it is with respect to the earth. Mass is an intrinsic property of matter, therefore it does not change ( Except for relativistic effects... Ben). Weight is the result of gravitational attraction; consequently it depends upon, (a) the body which is attracted, (b) the bodies which attract it, and (c) the position of the former relative to the latter. It is evident therefore that when a body is moved relative to the earth its weight changes." (Don't forget about the Equivalence principle... Ben)}

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PlanetPhysics/Difference Between Mass and Weight

References

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