PlanetPhysics/Difference Between Mass and Weight

The beginner often finds it difficult to distinguish between the 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 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 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 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 motion, the ratio of their masses is given by the 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 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 of the same mass placed at the 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

$$ f_1 = -\frac{v_1^2}{r_1} = - \frac{4 \pi^2 r_1}{P^2} $$

and

$$ f_2 = -\frac{v_2^2}{r_2} = - \frac{4 \pi^2 r_2}{P^2} $$

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.