There is no special name for the unit of momentum in any of the systems of units. The unit may be denoted at length as lb. by ft. per sec., kilogramme by metre per sec., gm. by cm. per sec. ART. 139.-Impulse. By an impulse is meant the cause which produces a change of momentum in a body, whether the change takes place in the magnitude or in the direction, or in both the magnitude and the direction. The second law of motion states that the impulse is measured by the change of momentum produced. Let denote the systematic unit of impulse, then ART. 140.-Momentum per Volume and Current. Liquid bodies, such as water and mercury, when at a uniform temperature throughout, are also uniformly dense. Suppose that the density of such a liquid is p M per L3, and that its velocity is v L per T, then by multiplying together these rates we get pv M by L per T = L3; that is, pv units of momentum per unit of volume of the liquid. Consider a cross-section perpendicular to the direction of the velocity. The density can be expressed as P M per L2 cross-section per L normal, and the velocity is v L normal per T; hence, by eliminating the common unit, pv M per L2 cross-section per T. This is the idea of current per unit cross-section, and it is ultimately equivalent to the idea of momentum per unit of volume. Let the cross-section be a L2, then apv M per T. This is the idea of current. EXAMPLE. Ex. A shell of 60 lbs. weight is moving before explosion at the rate of 400 feet per second. In consequence of the explosion, a piece weighing 14 lbs. is projected forward in the direction of motion with an additional velocity of 300 feet per second. What will now be the velocity of the remainder ? The mass of the piece is 14 lbs., and its additional velocity is 300 feet forward per second, therefore its additional momentum is By the third law of motion the additional momentum of the remainder of the shell has the same magnitude but the opposite direction, therefore it is 14 x 300 lb. by foot backward per second; but the mass of the remainder is 60 - 14 lbs.; its additional velocity is Hence the new velocity of the remainder is 400-91.3 feet forward per second, 1. Of two bodies moving with constant velocities, one describes 36 miles in 1 h. 20 m., the other 55 feet in 1 sec.; the former weighs 50 lb., the latter 72 lb. Compare their momenta. 2. How far would a cannon-ball weighing 10 lb. travel in one minute, supposing it to possess the same momentum as a rifle bullet of 2 oz. moving with the velocity of 1,000 ft. per sec. 3. A man whose weight is 12 stones falls freely from a height of 64 feet. Calculate, neglecting the resistance of the air, the velocity and momentum acquired on reaching the ground. 4. A mass of snow, 28 lbs. in weight, falls from the roof of a house to the ground, a distance of 40 feet. Calculate the momentum. 5. A ball weighing 10 lbs. is projected vertically upwards with an initial velocity of 1,660 feet per second. Find its velocity and its momentum after 30 seconds and after 60 seconds. 6. Find the momentum per cubic foot in the case of a stream flowing at 2 miles an hour. Express in terms of the F.P.S. unit. 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 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 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; 1 poundal per lb. foot per second per second, 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, 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. |