Example 8. Let x=the required velocity; then 32: 100 00::20x, and x=6250=the answer.

Example 9. As there are 5280 feet in one mile, 195000 × 5280 1.029. 600,000 which divided by 7000, the number of grains in one pound avoirdupois=1147085.7142 pounds =65 tons+.

Example 10. The earth's diameter being about 8000 miles, and spheres being as the cubes of their diameters, the mass of the earth would have to the mass of the ball, the ratio of 80003-512,000,000,000,000, to 1; consequently, if the tenth part of a mile were divided into 512 millions of millions of equal parts, one of these parts would be pretty nearly the space through which the earth would move tow. ards the falling body. In the tenth part of a mile there are somewhat less than 6400 inches: if this were divided into 512 millions of millions, each part would be the eighty millionth part of an inch. To perform the operation algebraically :

Let x=the space which the earth would move through to meet the ball, then-x=the space which the ball would move through to meet the earth: therefore, 80003:(1)3:: 1 -xx. By multiplying extremes and means, and clearing of fractions, 512,000,000,000,000x 6,336,000, and the answer.

10

x=

6,336,000

8 0

5 12,00 0,00 0,00 0,00 0 8 0,808,175,5 05

Example 11. As action and reaction are equal, it is immaterial to which end of the rope the force is applied. If a man stands in a boat and pulls upon a rope which is fastened to a post on shore, the force of the man is expended on the post in one direction, namely, towards the shore. Call the man A, and let another man B, take the place of the post. If B pulls with a force just equal to that of A he will do nothing more than what the post did before, and therefore the two men together will bring the boat ashore no sooner than A would have done alone in the former case. If A pulls with more force than B, he pulls B towards him; and the reaction, or the force which carries the boat ashore, is the same as before, namely, the force of B. If B were to pull with more force than A, he would pull A out of the boat, were not A attached firmly to the boat, in which case

the velocity of the boat would be accelerated.* The sum of the motions of any two bodies in any one line of direction, towards the same part, cannot be changed, by any action of the bodies upon each other; whatever forces these actions are carried by, or the bodies exert among themselves.†

EXAMPLES FOR PRACTICE.

Example 1. By Art. 53, S-mT2, and by substituting numbers, S=12a × 16,, =144 × 16,2=2316 feet.

Example 2. By Art. 24, V=m÷2=15950÷55-290. S=V2÷4m=(290) 643-1307.2 feet. Again, T=V 2m 290321-9 seconds.

÷

Example 3. T=√

conds.

V=2√m S=2√16, X250=126.8 feet.

12

Example 4. S=mT2=16,, x36-579-the depth of the first well, and S=16,,×100=16081=the depth of the second; and 16081-579 = 10291 = the difference in the depths of the wells.

Note. An ingenious solution of a problem similar to the 4th, may be found in the second volume of the Mathematical Recreations, London, 8vo.

Example 5. S=V2÷4m=1100÷643=155.44 feet.

12

Example 7. S=mT2=16,1, × 12.25=197.02, V=2mT= 323112.6, and S-TXV=24×112.6=281.5, which added to 197.02-478.52 feet.

Example 8. S=m× 2T—1=16,, 61.76=993.3 feet.

S

Example 9. T=

m

330

=4.53. By Art. 55,

S=mx2nT-n2 16,1,18.18=292.3 feet.

Example 10. S-TXV+mT2=(17×5)+(16,2×25)= 4872.

Example 11. S=T× V—mT2 = (55×2) — (16,', × 4) = 453.

Example 12. S=T× V+mT2, and by transposing, TX VS-mT2, and V=(S—mT2)÷T=350-(16,1,×16)÷4= 23 feet in a second.

Example 13. By example 3d, (Art. 57.), let a=150+40 =190, and let b=150, and x=ba÷4a=22500÷760=29.6 =the number of feet from the top of the steeple.

Example 14. By example 4th (Art. 57.), let a=27, and b=45: But x=b2÷4a=2025÷108-18.75 the distance from the window to the point where the bullet overtakes the stone; and 45+18.75=63.75 the number of feet from the top of the tower, and 200-63.75-136.25 the answer.

COMPOSITION AND RESOLUTION OF MOTION.

Example 1. 33° 45'+64° 12.25' 97° 57.25', and 180° -97° 57.25'=83° 2.15′; therefore, sin 64° 12.25′ : sin 83° 2.25';:20: the side subtending the angle 83° 2.25'. By the tables, 9.954496:9.996782::1.301030 22 = the length of the line described by the ship. Again, sin 64° 12.25': sin 33° 45'::5; velocity of the current; and, by the tables, 9.954496 9.744739: 0.698970; 0.489213=3.085 = the number of miles with which the current moves per hour.

Example 2. 78° 45'-22° 30'=56° 15'. Let the course; then, log. 6: log. 2.5: :sin 56° 15' sin x=20° 16.' *22° 30′+56° 15'+20° 16'99° 1, and 180°—99° 1′=80° 59' the course. Again, 56° 15'+20° 16' 76° 31', and 180°— 76° 31′=103° 29′: hence, sin 103° 29′;66::sin 56° 15' the side subtending 56° 15', which=56.43, and this divided by 6=9.40 hours 9 hours and 24 minutes.

Example 3. 1000-766=234. Let x=the course: then, by Trig. 139, log. 60 R::234 tan x; and, by the tables, 1.7781513 10.0000000 :: 2.3692159: 10.5910646′ = N. 75° 37.' E.

Example 4. By the 3d example, tbe earth at the equator moves 1000 miles per hour, and while in lat. 40° it moves 766 miles: Therefore, if a man were taken up in latitude 40°, and at the same instant set down on the equator, he would be 1000-766-234 miles farther west than when in latitude 40°.

Example 5. As the boat moves at the rate of one mile per hour, and as the current runs 3 miles per hour, the boat will move three fourths of a mile while the current runs two miles and one fourth. Let x=the required angle; then, 13::R tan x=' 71° 34'.

Example 6. Let x=the required ratio; then, R: log..75:: sec 71° 34′; x; and, by the tables, 10.000000 : 1.87506 13::10.5000367: 3750980=2.372, which multiplied by 4 and the product divided by 3, the quotient becomes 3.162= the answer.

Example 7. By Art. 68, the curve will be a parabola. By Bridges' conic sections (Barnards edition) Art. 19, abscissa ordinate::ordinate: parameter. The abscissa represents the space passed over in the eagle's fall, and the or dinate=60=the number of feet from the place from whence the eagle was shot: but (Art. 53.) space=mT2=16,11⁄2 × 42 =16,1,×16=2573; that is, (Alg. 530) P×257}=60×60 =3600; therefere, the parameter=13.99=the answer.

Note. If we regard accuracy, we may say of this solution, as has been remarked of a labored solution of the problem for measuring the depth of a well by the time elapsed between the commencement of the fall of a stone from its mouth, and the noise of its dash in the water, namely, "that it is rather curious than useful."

QUESTIONS FOR PRACTICE.

Example 1. Bisect the longest side of a triangle, and from the point of direction draw a straight line to the opposite angle; then by Art. 100, this line is the center of

gravity of the triangle. In this line take c at one third the distance from the point of bisection. Call the triangle ABC, and the center of gravity c, and let A, B, C, represent the three equal bodies. The center of gravity of the bisected line AC, is the point of bisection which call G. The pressure upon G=A+C. But Bc=2Gc, and A+C =2B; therefore, Bc Ge::A+C: B; hence (Art. 96.) c is the center of gravity of A+B and C, and of the three bodies, A, B, C.

Example 2. Let O be the given point, and the four bodies be severally represented by A, B, C, D, placed in the following order.

Also, let G represent their center of gravity: by Art. 104, 3X1+5×2+7x4+9x7_104

OG=

1+2+4+7

= =7 feet.

14

B

Example 3. Let the center of gravity be represented by G, and assume A: B::BG: AG: then by composition (substituting the weight of the several bodies,) 5:3::5:3; hence G is the center of gravity of A, B. By Euclid I. 47, CG2=32+42=25, CG=5. Again, by Trig. 135, log. 5: R::log. 4

sin AGC=53° 8'. Assume Ce Gc G A+B C; or C: A+B::Gc: Cc; then C+A+B¦A+B::Gc+Cc: Cc; and, by substituting numbers, 20:8::5 :40-20-2 feet.

A

Example 4. See Fig. 53. GF: BF::R; cos GFB ; and substituting numbers, log. 4 log. 2.::1: cos GFB =60°.

COLLISION OF BODIES.

Example 1. a:b::2÷3:1; therefore a=26÷3, and b=3a÷2. A=2, and B=1. By Art. 121, common velo

city=(Aa—Bb)÷(A+B)=?

2-3a-2

a÷6=the answer.