7. Compare the amounts of momentum in a pillow of 20 lbs. which has fallen through one foot vertically, and an ounce-bullet moving at 200 feet per second. 8. Calculate the momentum of a hammer of 5 tons, let fall half a foot.

9. A ball of 56 lb. is projected with a velocity of 1,000 feet per sec. from a gun weighing with the carriage 8 tons. Find the maximum velocity of recoil of the gun.

10. A pipe with a diameter of 2 inches delivers water at the rate of 9.8 gallons per minute; what is the velocity of the water.

SECTION XXVII.-FORCE.

ART. 141.—General Unit. Force is rate of impulse with respect to time, and is expressed in terms of I per T.

Since

11=M by L per T,

1 I per T = (M by L per T) per T;

that is, unit of force is equivalent to unit of momentum per unit of time. The rate-unit I per T is conveniently denoted by one letter F.

As the mass of the body considered is supposed not to change as time goes on, the above unit (M by L per T) per T is equivalent to M by (L per T per T); and thus the unit of force may be considered as derived from the unit of acceleration by introducing the unit of mass as a factor.

Another equivalent mode of viewing the unit is (M by L) per T per T.

Hence the unit may be written

M by L per T per T,

and any one of these interpretations may be given to it.

ART. 142. Special Units. In the British system of scientific units the principal unit of force is the

pound by foot per second per second.

It is denominated the poundal, a term invented for the purpose by Professor James Thomson.

In the French system the unit principally used is the kilogramme by metre per second per second.

In the C.G.S. system the unit chosen is

gramme by centimetre per second per second;

it is denominated the dyne.

The founders of the C.G.S. system have adopted the prefixes mega and micro to denote respectively a multiple and a sub-multiple of one million. Thus

Units of force such as the above are called absolute units, because they are defined entirely in terms of the fundamental units of length, mass, and time.

ART. 143.-Intensity of a Force; Inertia. Since

1 FM by L per T per T,

1 F per ML per T per T,

and 1 F per (L per T per T) = M.

By F per M is expressed the intensity of a force; it is equivalent to the acceleration.

By F per (L per T per T) is expressed inertia; it is equivalent to the mass.

ART. 144.-Gravitation Measure of Force. In practice it is very convenient to measure a force by comparing it with the weights of known masses at the place. Let the acceleration produced by gravity at the place be

then, since

we have

32.2 foot per second per second;

1 poundal per lb. foot per second per second,
32-2 poundals per lb.

If the counterbalancing mass is m lbs., the force is

m 32-2 poundals.

When the mass is half an ounce m is, hence one poundal is nearly equivalent to the weight of a half ounce.

Similarly, if the intensity of gravity at the place be given as 981 cm. per second per second,

then, since

we have

1 dyne per gm. = cm. per second per second, 981 dynes per gm.

If the counterbalancing mass is m gm., the force is

m 981 dynes.

If the value of M is given, say m M, but not that of L per T per T, then all that we know is expressed by

m FL per T per T ;

so that the force is not given absolutely, but only relatively to the constant intensity of gravity at the place.

In Britain the standard intensity of gravity is the intensity at the latitude of London at the level of the sea, namely,

32.187 feet per second per second.

In France, it is the intensity at the latitude of Paris at the level of the sea, namely,

9.8087 metre per second per second.

EXAMPLES.

Ex. 1. A force acts on a mass of 8 oz. for 6.9125 mins., and produces a velocity of 10 feet per sec. Express the magnitude of the force in poundals and dynes.

.. by substitution,

453.6 x 30.48

gm. by cm. per sec. per sec.,

2 x 6 X 6.9125

i.e., 166 + dyne.

Ex. 2. What force, expressed in pounds weight, will in a minute give a mass of one ton a velocity of 10 miles per hour?

112 x 20 x 44

the change of momentum is

Now, this is given uniformly in 60 seconds,

lb. by ft. per sec.

3

If the intensity of gravity at the place is 32.2 feet per sec. per sec., then

Ex. 3. A body under the action of a constant force traverses in the tenth second from rest twice the distance it would have traversed in the fifth second from rest under the action of gravity. Compare the forces.

Suppose the intensity of the force to be a poundals per lb., then the acceleration it produces is x feet per sec. per sec, therefore x/2 feet per sec., therefore the distance traversed during the tenth second is (1029)x/2 feet, i.e., 19x/2 feet.

Suppose that the intensity of gravity at the place is 32-2 poundals per lb., then it may be shown in the same manner that the

distance traversed during the fifth second from rest is (5o – 4o)32·2/2 feet. Now it is given that

The intensity of the force is 30.5 poundals per lb.

Ex. 4. A mass of 2 lbs. is drawn along a smooth horizontal table by a mass of 1 lb. hanging vertically; required the space described in 4 seconds.

The force acting is 1 × 32.2 poundals, and the mass moved is 3 lb., therefore the intensity is

Ex. 5. A mass of 488 grammes is fastened to one end of a cord which passes over a smooth pulley. What mass must be attached to the other end in order that the 488 grammes may rise through a height of 200 centimetres in 10 seconds, the intensity of gravity at the place being 980 dynes per gramme ?

Suppose x gms.; then, as the intensity of gravity is 980 dynes per gm., the force acting is

i.e., (x-488)980 gm. by cm. per sec. per sec.

The mass moved is x + 488 gms., therefore the acceleration is