17. Where, in the above case, must the centre of gravity be if we are justified in putting the weight of the lever out of the question?

18. A uniform bar 12 feet long is supported by two men, one of them at an end; where must the other be in order that he may support three fifths of the whole weight?

19. Three men are to carry a log which is of uniform size and density, and has a length of 12 feet. If one of the three lifts at an end, and the other two lift by means of a lever, how ought the lever to be applied in order that each man may bear one third of the weight?

20. A rod AB weighing 10 lbs. is found to balance about a point 8 feet distant from A; a weight of 6 lbs. is then fastened to A; about what point will the rod now balance ?

21. A rod AB, whose length is 5 feet, and mass 10 lbs., is found to balance itself if supported on a fulcrum 3 feet from A; if this rod were placed horizontally on two points, one under A and the other under B, what pressure would it exert on each point?

22. A uniform bar 20 inches long and weighing 2 lbs. is used as a common steelyard, the fulcrum being 5 inches from one end. Find the greatest mass which can be weighed with a movable weight of 4 lbs.

23. Forces equal to the weights of 1 lb., 2 lbs., 3 lbs., 4 lbs. act along the sides of a square taken in order; find the magnitude and line of action of their resultant.

24. A uniform beam 18 feet long rests in equilibrium upon a fulcrum 2 feet from one end; having a weight of 5 lbs. at the farther and one of 110 lbs. at the nearer end to the fulcrum; find the weight of the beam.

SECTION XXXVIII.—GRAVITATION.

ART. 168.-Law of Gravitation. The law of gravitation, formulated by Newton, asserts that the attractive force between any two particles of matter is proportional to the mass of the attracted particle and to the mass of the attracting particle, and is inversely proportional to the square of the distance between them. Hence it is expressed by

k FM attracted by M attracting per (L distance)2.

The unit of force F might be defined by means of this law, by making k=1; but when it is defined as in Art. 141, then k is not 1, unless by chance.

ART. 169.-Intensity at Unit Distance. When the attracting mass considered is constant, say m M, we deduce

mk FM attracted per (L distance)2.

This is called the intensity of the attraction at unit distance. If, further, the distance considered is constant, say d L, we deduce

This is called the intensity of the attraction at a given distance.

ART. 170.-Intensity of a Field. Round any distribution of matter there is a field of force, and at each point of the field there is a value for

F per M attracted.

The intensity at the place is the resultant of the several intensities due to the different elements of matter in the distribution.

A body of spherical form and of uniform density produces an intensity of force at any place outside it the same as if the whole of its mass were concentrated at the centre.

ART. 171.-Potential. When a particle is moved from one position to another position in a field of force, the amount of work done is independent of the path taken, depending only on the starting point and the final point. This gives us the idea of change of potential, which means the amount of work done in moving unit mass from one position to another. It is measured in terms of

W per M moved.

Ex. 1. How far would a body fall towards the earth in one second from a point at a distance from the earth's surface equal to the earth's radius ?

The earth, being very approximately a sphere, attracts as if its mass were situated at its centre.

Hence,

*Newcomb's Popular Astronomy, p. 542.

Ex. 2. The moon's mass is 136 x 1021 lbs., the moon's radius 5.7 x 106 ft.; the mass of the earth is 11,920 x 1021 lbs.; the radius of the earth 21 x 106 ft. Find how far a stone at the moon's surface would fall in a second, the disturbance due to attraction of the earth being neglected.

32 poundal per lb. = 11,920 × 1021 lb. attracting per (21 × 106 ft.)2;

32 × 212 × 1012

X

=

X poundal per lb. lb. attg. per (ft. dist.), 11,920 × 1021

136 x 1021 lb. per (5.7 x 106 ft.);

32 × 212 × 1012 × 136 × 1021

X

poundal per lb.,

i.e.,

ft. per sec. per sec;

11,920 × (5·7)2

16 x 212 x 136

...

ft. = sec.2,

11,920 × (5·7)2

1 sec.2,

2.5 ft.

EXERCISE XXXVIII.

1. Suppose that the earth shrank until its diameter were 6,000 miles, what would be the effect on the weight of an inhabitant? The diameter of the earth is approximately 8,000 miles.

2. If we suppose the mass of the sun to be 300,000 times the mass of the earth, and its radius to be 100 times the radius of the earth, find the attraction at the surface of the sun of a mass which at the surface of the earth is attracted by the force one pound weight.

3. The diameter of Jupiter is 10 times that of the earth, and its mass 300 times that of the earth; by how much per cent. of his former.weight would the weight of a man be increased by being removed from the surface of the earth to the surface of Jupiter?

4. The intensity of gravity at the surface of the planet Jupiter being about 2'6 times as great as it is at the surface of the earth, find approximately the time which a heavy body would occupy in falling from a height of 167 feet to the surface of Jupiter.

5. Find the intensity of the earth's attraction at the distance of the moon, taking 32 feet per second per second as its value at the surface of the earth. The diameter of the moon's orbit is 480,000 miles.