128 



THE POPULAR EDUCATOR. 



velocity in falling, except so far as they are impeded by other 

 causes. 



A balloon, if we could make it strong enough not to burst, 

 would in a vacuum fall in exactly the same time as a ball of 

 lead. 



If we take a number of balls made of different substances 

 and arrange them side by side in a box, the bottom of which 

 turns on a hinge, and allow it to fly open, the balls will travel 

 in a straight line and all reach the ground together. A little 

 consideration shows that it is very natural that it should be so. 

 If we have a number of equal balls, made, for instance, of lead, 

 each will fall in the same time. Now let two or more be rolled 

 into one, and the large one will fall in the same time that the 

 small ones composing it did, though it is heavier, for there is 

 obviously no reason why the mere change of shape should alter 

 the speed. 



We want to know now what is the actual velocity with which 

 a body falls ; and this is often a useful thing to know, for by it 

 we can ascertain the height of a tower or the depth of a well. 

 We have only to drop a stone from the top, and notice how long 

 it takes to reach the bottom, and from this we can calculate the 

 height. 



A falling body is acted on by the attractiqn of the earth. 

 Now after any given time say, for instance, one second it has 

 acquired a certain velocity with which it would continue to 

 move if the attraction ceased. It does not cease, however, and 

 hence the body must fall with a constantly increasing velocity. 



This we can calculate by means of Atwood's machine. We 

 can, by diminishing the weight of the bar, decrease the velocity 

 in any proportion we like, and thus are able to measure the 

 space passed over. 



If the bar weighs as much as the weights do, then the moving 

 force is one-half of the mass moved, and the velocity with which 

 it descends is one-half of the velocity it would have were it free 

 to fall from its own weight alone. But to make the speed more 

 easily measurable, let us further diminish the weight of the bar 

 as compared with the weights. If we make w and w' to weigh 

 7J oz. each, and the bar, F, | oz., we shall have as convenient a 

 proportion as we well can. The total mass moved will in this 

 case be 1 pound, and the moving force | oz., or i of the mass ; 

 the velocity with which w falls will therefore, be Jj of that of 

 a falling body. 



Now raise w with the bar on it to the catch, and allowing it 

 to start at one tick of the pendulum, note how far it falls before 

 the next. The easiest way of doing this is to fix the ring a 

 little way under w, and, by shifting it up and down, ascertain 

 the place at which the second tick of the pendulum occurs at 

 exactly the same time as the sound of the bar striking on the ring. 

 This distance will be found to be 3 inches. Of course, you must 

 measure from the height of the under side of the bar, for that is 

 the part which strikes the ring. This, then, is the space passed 

 over in the first second, and if we multiply this by 64, we find 

 that 16 feet is the space a body, left free, will by its own weight 

 fall through in the first second. More exact experiments show 

 that the amount is 16^ feet, but we may take 16 as near enough 

 for most practical purposes. We have thus found the distance 

 w passes in one second ; but we want to know what momentum 

 it has acquired, that is, what space it would, from the velocity 

 it has received, pass over in the next second, supposing gravity 

 were to cease to act altogether. As it falls with an accelerating 

 velocity, it must be moving more quickly at the end 'of the 

 second than at the beginning, and thus its velocity at the end 

 must be greater than 3 inches. To ascertain this we leave the 

 ring as before, 3 inches under the bar. Now when w passes 

 through the ring, the bar rests on it, being too long to pass, 

 and therefore w falls from its own momentum alone. If, then. 

 we fix the shelf, E, at such a distance under D that the weight 

 strikes upon it at the third tick, the distance between D and a 

 will be that which w passes over from its momentum, and this 

 space we shall find to be 6 inches, or just double that passed 

 over in the first second. 



Now if the ring had been removed, and the bar left on during 

 this second, it would, by the second law of motion, have caused 

 w to fall through an additional 3 inches. It ought then to fall 

 through 6 inches from its own momentum, and 3 inches from 

 the force of gravity, making in all 9 inches ; and if we place the 

 stage 12 inches below the catch, we shall find that such is the 

 case. Thus it passes 3 inches in the first second, and 3 times 



3 inches, or 9 inches, in the second. By again arranging the 

 shelf and ring, we shall find that the momentum acquired after 

 two seconds is double that acquired after one, for it will carry 

 W through 12 inches in the third second. 



Similarly in this second it will move 12 inches from momen- 

 tum, and 3 inches from gravity, making in all 15 inches, or 5 

 times 3, and its momentum at the end will be 18 inches. 



Now if we arrange these results in a tabular form, we shall 

 find some simple laws which regulate them. Instead, however, 

 of putting down 3, 6, 9, etc., we will use 1, 2, 3. The pro- 

 portion is just the same, and if we had made the bar i instead 

 of g'j of the mass, these are the distances in feet which would 

 have been moved over. 



We see, thus, that the velocity increases in the exact proportion 

 of the number of seconds the body has been falling ; that the 

 spaces passed over in succes- . , 

 sive seconds are proportional 

 to the successive odd num- 

 bers ; and that the total space 

 fallen through in any number 

 of seconds is proportional to 

 the square of that number. 



Now we saw that the space 

 any body falls through in the 

 first second is 16 feet. Hence 

 in the second it is 48, or 16 x 3, 

 in the third 80, or 16 x 5. 

 Generally, then, if we multi- 

 ply the numbers in the above 



1. 



2. 

 3. 



4. 



** 

 5. 



6. 



table by 16, we shall have those 

 applicable to the case of fall- 

 ing bodies. This may be more clearly represented by Fig. 98. 

 In this diagram vertical height represents the time in seconds ; 

 breadth, the velocity; and area the total space passed over. 

 At the end of the first second it shows the space passed over 

 to be 16, and the velocity is represented by the horizontal line 

 drawn through 1, which we call 32. 



We can thus see at a glance the space passed over in any 

 given second, or the velocity at any given time. In the same 

 way, if we draw a line midway between 4 and 5 to represent 

 4 seconds, we can find the space passed over and the velocity 

 acquired in that time. The figure, in fact, gives us a good idea 

 of the action of a uniform accelerating force. 



Suppose, for instance, we drop a stone down a well and find it 

 is 4} seconds before we hear the splash, we know the depth is 

 324 feet. The stone falls 16, 48, 80, and 112 feet in the first, 

 second, third, and fourth seconds, and 68 in the last half second ; 

 these together make 324 ; or we may take the square of 4, 

 and multiply it by 16, and thus get the same result. 



Again, suppose we want to know the velocity any body 

 acquires in falling for 6 seconds, we have only to multiply 32 

 by 6, and we find it to be 192 feet per second. 



ANSWERS TO EXAMPLES IN LESSON XX. 



1. 60 x 10 x 70 = 42,000 units. 



2240 x 20 x 400 



2. The work done per minute is 



60 



or 298,666 



298 666 

 units. Hence the H.P. required is ' , which is a little 



over 9. 33 ' 000 



3. It would raise it from a depth of 663 feet. 



4. About 47 Ibs. per hour. 



5. About 5 days by means of a windlass, or 3| days by 

 ascending a ladder and allowing his own weight to raise 

 it. 



