It should be remarked, that the minus sign of the quantity obtained, after taking the square root for the value of z, is taken to get that value; because we worked from the bottom in the last case (z). SECTION III. 1. "What must be the length of the stroke of the piston of an engine whose area is 1000 square inches, that, making 20 single strokes per minute under a mean effective pressure of 15 lbs. per square inch, the engine may yield 10 H.P.?" 10 x 33000 20 = the units of work done per stroke. 4 1000 X 15 X = effective units of work done through one foot. stroke. 5 10 x 33000 × 5 20 x 1000 X 15 X 4 = 1.375 feet, length of 2. "State concisely the method by which the number of units of work done per stroke upon each square inch of the piston of the engine may be determined when the steam is working expansively." Let the stroke from the point at which the steam is cut off be considered as divided into any number of equal parts; find the pressure of the steam upon one inch of the piston at these points by "Marriotte's law;" and let the units of such pressure be transferred to length by ordinates drawn upon a straight base, at a distance from each other equal to the length of the portions of the stroke; join the other extremities of the ordinates, and a figure will be formed whose area represents the work performed upon one inch of the piston during the expansive working of the steam; which work may be found by the application of "Thomas Simpson's rule." Add to this the work done upon one inch before the steam is cut off; the result will be the work done upon one square inch of the piston during one stroke when the steam is worked expansively. See Appendix i., Mechanics, Ques. 3. 3. "Investigate an expression for the work accumulated in a body of a given weight moving with a given velocity." Let W the given weight; and V velocity. the given In order to find the accumulated work of the weight W, when it has acquired the velocity V, we may consider it as having fallen freely from the requisite height to produce that velocity. This height, or space in feet when found, multiplied by W will obviously give the accumulated work; in the same manner as 16 units are accumulated by the falling of a weight of one pound 16 feet, at the end of which the acquired velocity is 32. From Ques. 2. Section iv. we obtain, By multiplying this space by W in lbs. the accumulated units of work are obtained. U = 321. V2 X W 29 in which g is the constant quantity 4. A train which weighs 400 tons is travelling at the rate of 20 miles an hour; what friction must be put upon it by the breaks, in addition to the friction of the rail, that it may be brought to rest within the space of 200 yards, the steam being thrown off?" the 3 882 X 400 × 2240 9 × 2 × 32 =11860895'7 Now the work absorbed by the usual friction over space of 200 yards will be 400 X 7 X 200 × 3 = 168000 By taking this from the previous result, the whole work due to additional friction will be obtained. This work of additional friction is distributed over the whole train of 400 tons. If the friction per ton be required, the operation is obvious. SECTION IV. 1. "State consisely the statical principle of equality of moments, and describe a method of proving it by experiment." When two forces acting in opposition to each other, are such in combined magnitude and direction as reciprocally to neutralize their effects, there is said to be an equality of moments. Thus, in a lever of the first kind, let F represent the position of the fulcrum, A and B, the extremity of the long and the short arm respectively, W the weight and P the power in the same units of weight as W. Then the statical principle of the equality of moments is expressed by the equation PX AF W x B F: or, in words, a force multiplied by the distance from its centre of action is equal to an opposing force multiplied by its distance from the same point, when equilibrium takes place; i.e., the moments on each side are equal. The term moments is a modification of momentum; and the effects which these terms indicate, when closely compared, bear a striking analogy to each other:-—the momentum of a body is its weight, or quantity of matter, multiplied by its quantity of motion. Now the arms of the lever, in length, are to each other as the arcs described by their extremity; therefore, if C and e be substituted for the lengths of those respectives arcs, the above equation will become P x C W c. The resemblance between these moments and the momentum of a body will readily suggest itself. A method of proving the principle by experiment will be found in "Tate's Mechanics," Page 64. But the principle could be verified by the lever, by taking into consideration the weight of the arms and thus producing a compound power and weight; the former composed of P acting at the distance of the extremity of the long arm plus the weight of that arm acting at the distance of half its length, and the latter, of W acting at its previous distance plus the weight of the short arm acting at the distance of half its length. NOTE. It will be understood that we suppose the lever to be of uniform weight throughout its length. The circular, uniform board, which Mr. Tate recommends for the illustration of the principle is preferable; for, the centre of gravity of the board is at the fulcrum, hence, P and W only need be taken into consideration. 2. "Investigate an expression for the velocity acquired by a body falling by gravity freely through a given space." When, at the latitude of Greenwich, and near the earth's surface, a body is acted upon directly by the force of gravitation, all opposing forces being excluded, it falls, during the first second of its descent, through a space of 16.0954 feet. This quantity being nearly equal to 16 ft. 1 in., is usually expressed in practice as 16 ft. Now, the velocity of the falling body during the second must have increased from nothing to such a quantity that the space passed over is 16 ft. Hence, the acquired velocity at the end of one second will be 2 × 16 ft., which is termed the force of gravity, for which we shall substitute g. Now, as the force of gravity is uniform and constant, it necessarily follows that an equal increase of velocity will be acquired by the falling body in equal times; thus, at the end of 1, 2, 3, n sec., the velocity will be g, 2 g, 3 g, ng, or V = tx g .... .... (1) The space passed over in 1, 2, 3, .... n sec., will be During 1st second 16, for which put f 16+2 × 16 3 f 2nd Hence the space passed over in n seconds is f+3ƒ + 5ƒ+ &c. to (2n-1)ƒ .. (a) |