On the whole, this will be less than the real result, since during each tenth of a second we have supposed the body to retain the velocity it had at the beginning of the time. Suppose now we assume that during each tenth of a second the body moves with the velocity it had at the end of that interval, then the space passed over will be as follows

That is to say, the whole space passed over will now be represented by the united area of the outer series of steps in Fig 5.

It will readily be seen that this arrangement gives us a result as much above the truth as the other is below it, and 4'410 + 5'390

that the mean of the two, or

the true result.

2

= 4'90 will be

Indeed we see from the figure that if we had sub-divided the second into a very great number of parts instead of into ten, the two results, of which one is above and the other below the truth, would differ from one another and from the truth only by an exceedingly small quantity, and in fact that the area of the triangle A B C represents graphically the space moved over by a falling body under gravity during the first second; and generally if AB represent the whole time

of fall, and BC the velocity at the end of the time, then

AB X BC will represent the space passed over.

2

which we have seen is the space moved over during the first second.

2

The space which the body will pass over during the next second is easily found, for the body will commence with the velocity of 9.8 metres, and end with that of 196 metres per second, having during the interval a mean velocity of 9.8+ 19.6 = 147 metres per second; this, therefore, is the space which it will describe. Thus during the first second it described 49 metres, and during the next second 14.7 metres : in all, at the end of the second second, it will have fallen 4'9 +14'7 19.6 metres. It may be shown in like manner that at the end of the third second it will have fallen 44'1 metres from its point of rest. If we refer to the graphical representation, the rule connecting space and time becomes obvious; for in Fig. 5, if the base AB be taken to represent the whole time t from the commencement of the fall in seconds, and the vertical height BC the velocity v, at the end of this time, then the area of the triangle (= vt) will, as we have seen, represent the whole space passed over. But ข = 98 t (Art. 20); hence, substituting this value for v in the above expression, we have s, or whole space passed over = & vt = 1 × (9·8 t) t = 4′9 t2.

=

Hence if t = 1 (second) s (space passed over)

This proof may be illustrated in the following manner. Let us suppose that the moveable chamber of Art. 19, being

49 metres in height, contains within it a man, who holds a stone loosely in his hand at the level of the inside roof of the chamber. Now let the whole arrangement be allowed to fall freely down the shaft of a mine under the influence of gravity, and at the moment when the fall begins, suppose that the man lets the stone go. Where will the stone be at the end of the first second? A little reflection will convince us that, although free, the stone will still be at the top of the carriage, for the whole arrangement is falling as fast as it can, and there is therefore no reason why the stone should gain upon the other parts.

At the end of one second, therefore, the top of the carriage will have fallen through 49 metres, and the stone will still be at the top.

Now let us imagine that at the end of the first second external friction is made to operate on the carriage sufficiently to stop the tendency of gravity to increase the speed. It will thus cancel, as it were, the further effect of gravity upon the carriage, leaving it to continue its descent as if there were no gravity, in which case it would, in accordance with the first law of motion, continue to move with the velocity it has already attained, that is to say, with the velocity of 9.8 metres a second. During the second second, the top of the carriage will therefore have fallen through 98 metres. Altogether therefore, in the two seconds it will have fallen through 49 +9.8 = 14'7 metres. But in the meantime what will have happened to the stone? Inasmuch as the stone was free within the carriage, it would not be acted upon by the friction arrangement from without, and would therefore, according to the second law of motion, obey gravity in the moving carriage, just as truly as if the carriage were at rest. During the second second, the stone will therefore have fallen, as regards the carriage, through 4′9 metres, that is to say from the top to the bottom, and it will strike the bottom precisely at the end of the second second.

We thus perceive, that while the carriage has fallen through 14'7 metres, the stone, which has all the time been perfectly free, has fallen in addition, through 49 metres, this being

the distance from the top to the bottom of the carriage. The stone has therefore fallen in two seconds through 147 +49 196 metres, while in one second it only fell through 4'9 metres.

=

We thus realize how the distance fallen through varies as the square of the time.

22. Oblique Motion under Gravity.-We may now easily

pass to the case of oblique motion under gravity. Suppose, for instance, a projectile, such as a bombshell, to be fired off at A with a velocity which would bring it, were there no gravity, to A' at the end of the first second, to A" at the end of the second second, to A"" at the end of the third second, and so on, describing equal spaces in equal times by the first law of motion. What will be its actual path

under the action of gravity? Were the shell at rest, gravity would cause it to be at the end of the first second 4'9 metres below the place it would have occupied had there been no gravity, and the same will happen whatever be its velocity. Now, at the end of the first second it would have been at A' had there been no gravity, and hence its real position at that time will be at B′, 4'9 metres below A'. In like manner at the end of the second second it would have been at A" had there been no gravity, and hence its true position will be at B", 196 metres below a". Also at the end of the third second it would be at A"" without gravity, and hence it will be at B'"', 441 metres below A"", through the action of gravity. The true position of the projectile will therefore be a curve, bending further and further from the original line of impulse,

such as we have shown in the above figure; and finally the projectile will reach the ground again at C. The curve A B C, denoting the path of a projectile, may be shown to be a parabola.

LESSON IV.-SECOND LAW: ACTION OF TWO OR MORE FORCES.

We have hitherto supposed only one force to act, and have found that its action is unaffected by the state of rest or motion of the body to which it is applied, and it is very easy to step from this to the case where two forces act at the same moment on a body.

23. Two or more Forces in the same Direction.-Let us, in the first place, suppose that the forces act in the same direction. For instance, imagine a piece of iron to fall by the action of gravity; at the end of one second it will have the velocity of 9.8 metres. Suppose, now, that at the same moment at which gravity began to act upon it, and to cause it to fall, it was acted upon by a magnet so placed as to give it in one second a downward velocity of 98 metres. Then, owing to the joint effect of gravity and of the magnet, it will at the end of one second, according to this law of motion, have acquired a velocity of 98 + 9.8 196 metres per second, and have passed over the space of 9.8 metres instead of 49 metres, which it would have passed over through gravity alone.

Hence we are able to extend our definition of force (Art. 14). We thus see that if a double force be applied to the same body, it will produce a double velocity in unit of time, a triple force, a triple velocity, and so on; in fact a force may be measured by the velocity which it generates when applied for one second to a given body. Thus, suppose our unit of force is that which when applied to unit of mass produces unit of velocity in unit of time, then a force represented by two will produce in the same mass a velocity equal to two in unit of time, and so on. So that in fact, regarding both mass and velocity, we see that the magnitude of a force is represented