« AnteriorContinuar »
38. DEF. The mass-velocity (or, the momentum) of a particle is that which varies as the mass and also as the velocity of the particle.
We shall choose as our unit mass-velocity that of a particle of 1 lb. moving with i velo.
We shall call this unit a pound-velo.
It follows that the number of pound-velos which a particle has, is the product of the number of lbs. in it by the number of velos with which it is moving.
39. DEF. The mass-acceleration of a particle is that which varies as the mass and also as the acceleration of the particle.
We shall choose as our unit mass-acceleration that of a particle of 1 lb. moving with i celo.
We shall call this unit a pound-celo.
It follows that the number of pound-celos which a particle has, is the product of the number of lbs. in it by the number of celos with which it is moving.
It should be noticed that the measure of the mass-acceleration of a particle is the rate of increase of its mass-velocity.
Example i. Find the mass-velocity of a train of 120 tons going 30 miles an hour.
I20 tons=120 X 20 X 11 2 lbs.
30 miles an hour=44 velos. Therefore, the mass-velocity of the train is 120 X 20 X 112 x 44 pound-velos, that is 11,827,200 pound-velos.
Example ii. Find the mass-acceleration of a man and bicycle of 20 stone while attaining a speed of 15 miles an hour in going 121 ft. from rest. [1 stone=14 lbs.] 15 miles an hour=22 velos.
(Ex. ii. p. 3.] We have
2as = va,
20 x 121=(22)”; therefore, a=2. Also
20 stone=20 X 14 lbs. Therefore, the required mass-acceleration is 560 pound-celos.
EXAMPLES. XII. Find the mass-velocity (or, momentum) of each of the following: 1. A mass of 3 lbs. having 3 velos.
3 2. A mass of i cwt. having 10 velos. 3. A train of 100 tons going at the rate of 30 miles an hour. 4. A bullet of 1 ounce going 1000 ft. per second. 5. A cannon ball of 1 cwt. going 1000 yds. per second. 6. A man of 12 stone running 12 miles an hour. [a stone=14 lbs.] Find the mass-acceleration of each of the following: 7. A mass of 10 lbs. having 8 celos. 8. A mass of 2 cwt. having 32 celos.
9. A train of 100 tons which has gained from rest a velocity of 30 miles per hour in i minute.
10. A bullet of i oz. which in traversing 36 inches from rest has gained a velocity of 1000 ft. per second. [16 oz.=1 lb.]
11. A cannon ball of 56 lbs. which in traversing the tube of a cannon io ft. long, from rest, has gained a velocity of 500 yards per second.
12. A skater of 12 stone, who starting from rest has acquired a velocity of 20 miles an hour in going 50 yds.
It follows from Art. 31 that the number of poundals in an external force acting on a mass, is equal to the number of pound-celos which it produces in that mass.
Example i. A force 32 poundals is applied to a ton; find/ (i) the acceleration produced, (ii) how long this force will take to produce 6 velos in the ton.
Let the number of celos produced by the 32 poundals be a; then, the mass acceleration produced is 22400 pound-celos. By Art. 40 the external force which produces 2240a pound-celos is
22402 poundals. But, this force is
32 poundals. Therefore, 22402 poundals=32 poundals ; that is,
22400=32; or, a=7o. Thus, the acceleration produced is * of a celo.
(i) W celo in i second produces + velo,
to celo in 6 x 70 seconds produces 6 velos. Therefore, if the force be continuously applied to the ton it will produce 6 velos in 6 minutes.
Example ii. A cannon ball of 250 lbs. starting from rest, leaves the mouth of a cannon with a velocity of 1000 ft. per second ; find the resultant force which acted upon it when in the cannon, supposing this force to be uniform and that the ball passed over 20 ft. while in the cannon.
[NOTE. By resultant force is meant, the force over and above that necessary to overcome the friction of the tube ; the pressure of the gases on the base of the shot is necessarily greater than this resultant force, which is only that part of this pressure which produces the acceleration of the shot.]
We have to find the number of celos which the ball had in the
s= 20 ft., and
The mass-acceleration is ma;
The number of poundals required to produce 6250000 pound celos is equal to the number of pound-celos produced, that is
EXAMPLES. XIII. 1. Find the number of poundals required to produce the motion in each of the examples 7 to 12 of Examples XII.
2. Find the number of poundals which by acting for 20 seconds will produce the mass-velocities in Questions 1 to 6 of Examples XII, respectively.
3. A locomotive engine can exert upon a train of 100 tons a force of 10000 poundals more than is necessary to counteract the friction ; how long would this engine take in working the train on a level line up to a speed of 30 miles an hour ?
4. A rifle bullet of i oz. leaves the barrel of a rifle which is 2 ft. 6 in. long with a velocity of 1200 feet per second ; find the resultant force, supposed uniform, with which it was acted on in the barrel.
5. A mass of 10 lbs. moves over 16 ft. from rest in 2 seconds; supposing its motion to be that of uniform acceleration, what force is acting on it?
6. It is observed that i ton is moving in a straight line with a velocity of 10 miles per hour; 3 minutes later it is observed to be moving in the same straight line with a velocity of 12 miles per hour; supposing a uniform external force to have acted on the mass during those 3 minutes, how many poundals did it contain ? L. D.
41. DEF. The weight of a mass is the force with which it is attracted by the earth.
When we support a mass in our hand so as to keep it at rest, we can only do so by the application of an upward push or force; this upward force is equal and opposite to the weight of the mass.
42. Since the weight of a mass is a force, and since our unit of force is a poundal, in order to measure the weight of a given mass we must express its weight in poundals.
The number of poundals in any force is equal to the number of pound-celos which it produces; accordingly we have to ascertain how many celos the weight of any particular mass produces in that mass.
43. The following experiment is important.
A chamber, such as the receiver of an air-pump, is exhausted of air. In the chamber are placed several masses of different kinds and of different sizes; for instance, a leaden bullet, an iron nail, a bronze penny, a gold sovereign, a glass ball, a piece of cork, a feather; these are all placed on a ledge so arranged that they can be simultaneously let fall.
When a mass is ‘let fall' in a vacuum, the only force acting upon the mass is its own weight; accordingly in this experiment when the masses are let fall, the weight of each mass produces in its own mass motion, in a vertical direction downwards, which is that of uniformly increasing velocity.
Now it is observed that, having been simultaneously let fall from the same height, these different masses all reach the bottom of the chamber simultaneously.
But, when a number of points, starting from rest and each moving in the same direction with uniformly increasing velocity, pass over equal distances in the same interval of time, they must all be moving with the same acceleration. Hence,
44. The weight of a mass is a force, which produces in its own mass a certain definite acceleration; which acceleration is the same for all masses of whatever magnitude and of whatever kind of material. 45
This definite acceleration, which is called the acceleration due to gravity, is different at different places on the earth's surface.
It is greatest near the Poles, (where it is about 32'255...celos) and it gradually diminishes to about 32.091...celos at the equator.
It is the same at places having the same latitude.
It has been ascertained by observation that at Greenwich the number of celos produced in any mass when falling freely towards the earth's surface (that is, when moving under the action only of its own weight) is 32'19....
46. It is usual to use the letter g to indicate the number of units of acceleration produced in a mass by the action of its own weight.
Hence, at Greenwich g=32:19....
The acceleration produced in a mass m lbs. by its own weight is s celos; hence the amount of mass-acceleration produced in mass m lbs. by the weight of m pounds is mg pound-celos; the force required to produce mg poundcelos is mg poundals. Therefore
the weight of m lbs. is the force mg poundals.
the weight of 1 lb.
Example ii. How many celos will a force equal to the weight of 3 lbs. when applied to a mass 12 lbs. produce in it?
The force is 3 x 32, or, 96 poundals and 96 poundals produces in 12 lbs., 19 or 8 celos.