Cor. 4. If one of the forces, as C, be a weight, which is sustained by two strings drawing in the directions DA, DB; then the force or tension of the string AD is to the weight C, or tension of the string DC, as DE to DC; and the force or tension of the string BD is to the weight C, or tension of CD, as CE to CD. Cor. 5. Let fand fi be two forces acting si multaneously in directions making an angle ø; then in the triangle DEC we have DE=ƒ; EC=fi angle DEC=- ቀ፡ hence by the principles of trigonometry, we have DC2 DE2 + EC2—2DE.EC cos DEC; = and therefore the magnitude of the resultant R is found from the equation R= √ƒa +ƒ2—2ff, cos (π-)=√ƒ2+ƒ‚2+2ƒfƒ, cos p. PROP. IV. 15. To find the resultant of several forces concurring in a point, ana situated in the same plane. Let P1, P2, P3, P4, be any four forces acting on the point P, throught which draw the axes of co-ordinates PX, PY at right angles to each other. Let Pp. represent the magnitude and direction of the force p1, and draw p, B, p, A parallel to the axes XX' and YY. Then putting angle p, PX= a1, we have the two rectangular forces PA, PB, equivalent to the given force p1; but by trigonometry PA= Pp1 cos APp1 = P1 cos a, and PB = p1 sin a1. In like manner, if a, a, a, be the angles which the direction of the forces P2, P3, P4, make with PX, we shall have each of the proposed forces resolved into two others acting in the directions of the two axes, and therefore the sum, X, of all the component forces in direction PX, gives P Y D P Y P3 P (1) X=p1 cos a1 + P2 cos a2 + P3 cos a + p1 cos a and the sum, Y, of all the other component forces in direction PY, gives Y=p, sin a, + P2 sin a2 + P3 sin a, + p1 sin a, ..... ....(2) Hence the single force X, in direction PX, and the single force Y, in direction PY, may be substituted for the four given forces, and the resultant of the two forces X, Y, will be the resultant of the four forces P1, P2, P3, P1. But X and Y are two forces acting at right angles to each other, and their result. ant, R, is the diagonal of the rectangle XY; hence we have R = √/X2+Y2 ... (3) Let R make an angle with the axis of X; then we have and any one of these three equations will give the position of the resultant. In precisely the same manner may the magnitude and direction of any number of forces in the same plane be found. Cor. I. By means of a series of parallelograms the resultant of any number of forces may be found geometrically. For the diagonal of a parallelogram whose sides represent the first two forces will be their resultant, and this diagonal may be made the side of another parallelogram, having the third force for the other side, and so on. Or describe a polygon, whose sides beginning from the point, are successively equal and parallel to the given forces, and in the same direction; then the straight line which joins the point and the extremity of the last side completes the polygon, and represents the magnitude and direction of the resultant of the proposed forces. Cor. 2. If three forces act on the same point in different places, and if the parallelopiped, whose adjacent edges represent these forces, be completed, its diagonal will represent their resultant both in magnitude and direction. Cor. 3. Let P1, P2, P3, P4,.... be any forces, and let each of these forces be resolved into three other forces in reference to three rectangular axes; then, collecting into one sum the component forces which act in the same axis, we can find the resultant of the three components thus obtained in the following manner : B1, 7, be the angles which p, makes with the three axes Then each of the given forces may be resolved into three others; viz. p into the three forces p2 cos a, p2 cos B2, p2 cos 1⁄2, and so on; hence X = p, cos a1 + p2 cos a2+ p ̧ cos a ̧+p, cos a, +. Y =p1 cos B1+ p2 cos B2+ P3 cos B2+p1 cos B1 +. z=p, cos y1 + p2 cos y2+ P, cos y1+ p1 cos 1 +. hence RX2+Y2+Z2 = magnitude of resultant. And if a, B, y, be the angles which R makes with each axis, we have H 16. Ex. 1. Let the four forces p1, P2, P3, P4, concurring in a point P, and situated in the same plane, be respectively denoted by the numbers 4, 6, 12, 10, and let the angles included by their directions be P1 Pp2 = 15°, p2 Pp1 = 30°, p、 Pp1 = 60°; required the magnitude and direction of the resultant of these forces. We might assume any two rectangular axes whatever, PX, PY; but the solution will be simplified by taking one of the axes in the direction of one of the given forces; let, therefore, the axis of X coincide with the direction o Hence X = p1+ P2 cos 15°+p3 cos 45°+p, cos 105° = 4 + 3 √6+ 3 √2+6√/2− 5 √6+ 5 √2 = 4 +10√2−√6. 2 2 Y = p, sin 15°+p, sin 45°+p, sin 105° = 3√6 − 3 √2+6√2+ 35 √6+ 5 √2 = 7√2+4√6. 2 2 •. R = √/X2+ Y2 = {(4+10√2−√/6)2 + (7√/2+4√/6)°} angle RPX = 51° 27′ 22′′ = angle included by force p, and the resultant. Ex. 2. Two forces, represented by 7 and 5, act at an angle of 60°; find their resultant, and the angle it makes with the less force. Ans. R 10-4403065, and 4 = 35° 30′. E. 3. The resultant of two forces is 24, and the angles it makes with them are 30° and 45°; find the component forces. Ex. 4. Resolve a given force into two others, such that (1.) Their sum shall be given, and act at a given angle. (2.) Their difference shall be given, and act at a given angle. Ex. 5. If a stream flows at the rate of two miles an hour, find the course which a boat, rowed at the rate of four miles an hour, must pursue, that it may pass directly across the stream. Ex. 6. Two chords AB, AC of a circle, represent two forces; one of them, AB, is given; find the position of the other, when the resultant is a maximum. Ex. 7. Three forces represented by 13, 14, 15, acting at a point, keep each other in equilibrium; find the angles which their directions make with each other. Ans. 112° 38', 120° 30′, and 126° 52'. Ex. 8. Three forces P1, P2, P3, act upon a given point and keep it at rest; given the magnitude and direction of p1, the magnitude of p1, and the direction of p2, to find the magnitude of p, and the direction of p3. Ex. 9. A string 15 inches in length is attached at its extremities to two tacks, in the same horizontal line, at the distance of 10 inches from each other; a weight of 12lbs. is suspended between the tacks, by means of a string attached to the first, at the distance of 7 inches from one of its extremities; find the strain upon each tack. Ex. 10. A cord PABQ passes over two small pulleys A, B, whose distance AB is 6 feet, and two weights of 4 and 3lbs., suspended at the extremities P and Q respectively, support a third weight W of 5lbs. ; find the position of the point C to which the weight W is attached, when AB is inclined to the horizon at an angle of 30°. Ex. 11. P and Q are two equal and given weights suspended by a string passing over three fixed points, A, B, C, given in position; find the actual pressure, and also the horizontal and vertical pressures on each of the three points A, B, C. Also, compare the pressures on A, B, C, when the angles at A, B, C are 150° 90°, 120° respectively. ON THE MECHANICAL POWERS. 17. WEIGHT and Power, when opposed to each other, signify the body to be moved, and the body that moves it; or the patient and agent. The power is the agent, which moves, or endeavours to move, the patient or weight. 18. A Machine, or Engine, is any mechanical instrument contrived to more bodies; and it is composed of the mechanical powers. 19. Mechanical Powers are certain simple machines, which are commonly employed for raising greater weights, or overcoming greater resistances, than could be effected by the natural strength without them. These are usually accounted six in number; namely, the Lever, the Pulley, the Wheel and Axle, the Wedge, the Inclined Plane, and the Screw. 20. Centre of Motion, is the fixed point about which a body moves. And the Axis of Motion, is the fixed line about which it moves. 21. Centre of Gravity, is a certain point, upon which a body being freely suspended, it will rest in any position. OF THE LEVER. 22. A Lever is any inflexible rod, bar, or beam which serves to raise weights, while it is supported at a point by a fulcrum or prop, which is the centre of motion. The lever is supposed to be void of gravity or weight, to render the demonstrations easier and simpler. There are three kinds of levers. 26. A Fourth Kind is sometimes added, called the Bended Lever. As a hammer drawing a nail. W 27. In all these machines, the power may be represented by a weight, which is its most natural measure, acting downwards; but having its direction changed, when necessary, by means of a fixed pulley. PROP. V. 23. When the Weight and Power keep the Lever in equilibrio, they are to each other reciprocally as the Distances of their Lines of Direction from the Prop. That is, P: W:: CD: CE; where CD and CE are perpendicular to WO and AO, which are the Directions of the two Weights, or the Weight and Power W and P A D E For, draw CF parallel to AO, and CB parallel to WO: Also, join CO, which will be the direction of the pressure on the prop C; for there cannot be an equili- Pô brium unless the directions of the three forces all meet in, or tend to, the same point as O. Then, because these three forces keep each other in equilibrio, they are proportional to the sides of the triangle CBO or CFO, which are drawn in the direction of those forces; therefore, P: W:: CF: FO or CB. But, because of the parallels, the two triangles CDF, CEB are equiangular, therefore Hence, by equality, CD: CE:: CF : CB. P: W:: CD: CE. B F That is, each force is reciprocally proportional to the distance of its direction from the fulcrum. And it will be found that this demonstration will serve for all the other kinds of levers, by drawing the lines as directed. Corollary. 1. When the two forces act perpendicularly on the lever, as two weights, &c.; then, in case of an equilibrium, D coincides with W, and E with P; consequently then the above proportion becomes P: W:: CW: CP, or the distances of the two forces from the fulcrum, taken on the lever, are reciprocally proportional to those forces. Corollary. 2. If any force P be applied to a lever at A; its effect on the lever, to turn it about the centre of motion C, is as the length of the lever CA, and the sine of the angle of direction CAE. For the perp. CE is as CA X sine of angle at A. Corollary. 3. Because the product of the extremes is equal to the product of the means, therefore the product of the power by the distance of its direction, is equal to the product of the weight by the distance of its direction. That is, P X CE W x CD. Corollary. 4. If the lever, with the weight and power fixed to it, be made to move about the centre C; the momentum of the power will be equal to tho momentum of the weight; and their velocities will be in reciprocal proportion |