Put AG GE = 21 = a, cc = 6 = b, AC = √ AG2+Gc3 = 61 = c, AD = FE = FH = x the radius of the ball. The two triangles ACG and DCF are equiangular; theref. AG: AC: DF: FC, that is, a : c :: x: CX CX - a = FC; hence CF = GC — FC=b ——, and GH =GF + FH = b + x = C1, GF a a the height of the segment immersed in the water. Then (by rule 1 for the spherical segment, p. 427 vol. 1.), the content of the said immersed segment will be (6DF 2GH) X GH3 ×·5236 = (2x (2x − b + — ) × (x + b − CI)2 × 1·047%, a a which must be a maximum by the equation; the fluxion of this made 0, and divided by 2x and the common factors, gives 2a+c c-a a this reduced gives x = (c-a) x (c+2a) of the ball. Consequently its diameter is 4 quired. a 211, the radius inches, as re PRACTICAL EXERCISES CONCERNING FORCES WITH THE RELATION BETWEEN THEM AND THE TIME, VELOCITY, AND SPACE DESCRIBED. BEFORE entering on the following problems, it will be convenient here, to lay down a synopsis of the theorems which express the several relations between any forces, and their corresponding times, velocities, and spaces, described; which are all comprehended in the following 12 theorems, as collected from the principles in the foregoing parts of this work. LET f, F, be any two constant accelerative forces, acting on any body, during the respective times t, T, at the end of which are generated the velocities v, ▼, and described the spaces, s, Then, because the spaces are as the times and velocities conjointly, and the velocities as the forces and times; we shall have. S. And if one of the forces, as F, be the force of gravity at the surface of the earth, and be called 1, and its time r be= 1"; then it is known by the experiment that the corresponding space s is 16 feet, and consequently its velocity v=2s= 324, which call 2g, namely, g 16 feet, or 193 inches. Then the above four theorems, in this case, become as here below : = And from these are deduced the following four theorems, for variable forces, viz. In these last four theorems, the force f, though variable, is supposed to be constant for the indefinitely small time ¿, and they are to be used in all cases of variable forces, as the former ones in constant forces; namely from the circumstances of the problem under consideration, an expression is deduced for the value of the force f, which being substituted in one of these theorems, that may be proper to the case in hand; the equation thence resulting will determine the corresponding values of the other quantities, required in the problem. When a motive force happens to be concerned in the question, it may be proper to observe, that the motive force m, of a body equal to fq, the product of the accelerative force, and the quantity of matter in it 9; and the relation between these three quantities being universally expressed by this equation m = qf, it follows that, by means of it any one of the three may be expelled out of the calculation, or else brought into it. Also, the momentum, or quantity of motion in a moving body, is qu, the product of the velocity and matter. It is also to be observed, that the theorems equally hold good for the destruction of motion and velocity, by means of retarding forces, as for the generation of the same, by means of accelerating forces. And to the following problems, which are all resolved by the application of these theorems, it has been thought proper to subjoin their solutions, for the better information and convenience of the student. PROBLEM I. To determine the time and velocity of a body descending, by the force of gravity, down an inclined plane; the length of the plane being 20 feet, and its height 1 foot. : Here, by Mechanics, the force of gravity being to the force down the plane, as the length of the plane is to its height, therefore as 20:1:1 (the force of gravity) = ƒf the force on the plane. Therefore, by theor. 6, v or 20=√4× 16=2×4 per second. And, By theor. 7, tor ✔ or 8 4gfs is 4 × 16 × 1 × feet nearly, the last velocity seconds, the time of descending. PROBLEM II. If a cannon ball be fired with a velocity of 1000 feet per second up a smooth inclined plane, which rises foot in 20: it is proposed to assign the length which it will ascend up the plane, before it stops and begins to return down again, and the time of its ascent. 4jg 4X 16 X 60000000 193 -31088016 feet, or nearly 59 miles, the distance moved. If a ball be projected up a smooth inclined plane, which rises 1 foot in 10, and ascend 100 feet before it stop; required the time of ascent, and the velocity of projection. FIRST, by theor. 6, v4gfs=4X16, XX100= 810-25-36408 feet per second, the velocity. 100 gf 16 X ✓10 7.88516 seconds, the time in motion. PROBLEM IV. If a ball be observed to ascend up a smooth inclined plane 100 feet in 10 seconds, before it stop to return back again: required the velocity of projection, and the angle of the plane's inclination. 2s 200 FIRST, by theor. 6, v=-= =20 feet per second, the velocity. t 10 gt 16X100 193 length of the plane is to its height, as 193 to 12. Therefore 193; 12: 100: 6-2176 the height of the plane, or the sine of elevation to radius 100, which answers to 3° 34', the angle of elevation of the plane. PROBLEM PROBLEM V. By a mean of several experiments, I have found, that a cast iron ball, of 2 inches diameter, fired perpendicularly into the face or end of a block of elin wood, or in the direction of the fibres, with a velocity of 1500 feet per second, penetrated 15 inches deep into its substance. It is proposed then to determine the time of the penetration, and the resisting force of the wood, as compared to the force of gravity, supposing that force to be a constant quantity. 2s 2X13 FIRST. by theor 7, t=== cond, the time in penetrating. = 4gs 1500' = = 81000000 13 x 193 And, by theor. 8, f: =32284. That is, the resisting force of the wood, is to the force of gravity, as 32284 to 1. But this number will be different, according to the diameter of the ball, and its density or specific gravity. For, since fis as--by theor. 4, the density and size of the ball remaining the same; if the density, or specific gravity, n, vary, and all the rest be constant, it is evident that ƒ will be as n; and therefore fas when the size of the ball only is constant. nva 8 But when only the diameter d varies, all the rest being constant, the force of the blow will vary as d3 or as the magnitude of the ball; and the resisting surface, or force of resistance, strength or firmness of the substance penetrated, and is here supposed to be the same, for all balls and velocities, in the same substance, which is either accurately or nearly so. See page 581, &c. vol. 1, of my Tracts. Hence, taking the numbers in the problem, it is dnva 1 × 7×1500* f= 1 = 44 X 15002 =2538 162 the value of ƒ for elm wood. Where the specific gravity of the |