THE PULLEY. A pulley is a small grooved wheel, movable about a pivot, the pivot itself being at the same time either fixed or movable. A single pulley that only turns on its axis, and does not move out of its place, serves only to change the direction of the power, but gives no mechanical advantage, since it is evident that the power and weight are equal. It is, however, very convenient in accommodating the direction of the power to that of the resistance. Thus, by pulling downwards, we are able to draw a weight upwards. The advantage gained is always as twice the number of movable pulleys, without taking any notice of the fixed pulleys necessary to compose the system of pulleys. The principle upon which the weight is sustained by means of a pulley or system of pulleys is very simple, and may be readily understood from the following rules: CASE I. To find the Weight that may be raised by a known power and a given number of fixed and movable or stationary and rising pulleys. Rule.-Multiply the power by twice the number of movable pulleys, and the product is the weight the power is equal to. Example.-Required the weight that a power of 180 lbs. will raise by a block and tackle, the bottom or movable block consisting of four sheaves or pulleys. 180 × 8=1440 lbs. weight. When only one Cord or Rope is used. Rule.-Divide the weight to be raised by the number of parts of the rope engaged in supporting the lever or movable block. Example.-Required the power that will raise a weight of 3300 lbs. with a single movable pulley, as shown in the margin. 3300 3 1100. Ans. The single movable pulley may be so constructed, as shown in the marginal figure, that the weight will be three times the power, the number placed at each rope expresses the part of the weight it sustains. Example.-What power will it require to raise 900 ibs. when the lower block contains six sheaves, and the end of the rope is fastened to the upper block, and what power when fastened to the lower block? 900 6x2=75 lbs. 1st Ans. 900 6×2+1=683 lbs. 2d Ans. W Or, if n, represents the number of parts of the rope which sustains the lower block, we shall have the following formula: W=nxP. CASE II. In the system represented in the adjoining figure, called the Spanish Burton, the tension of the rope, PB, is equal to the power; and this rope, being finally attached to the pulley which sustains the weight, supports a part of the weight equal to the power. The rope from C to B balances the united tensions of both parts of the rope, extending from B to the weight and power, and therefore its tension is twice the power; and being brought under the pulley which sustains the weight, and finally attached to the fixed point, it sustains a part of the weight equal to four times the power: thus the whole weight must be equal to five times the power. The power being taken as the unit, the number placed at each rope expresses the part of the weight it sustains, i. e. as 5 to 1. Rule. Divide the weight to be raised by the sum of the numbers placed at each rope, and the quotient will be the power. Example-Required the power necessary to raise 4500 lbs. with a single movable pulley, and with two ropes, as shown in the above figure. 4500-5 900 lbs. NOTE. The first cord, B P, is obviously not taken into the account. When more than one Rope is used. In a system of pulleys where the ends of one rope are fastened to the support and power, and the ends of the other to the lower and upper blocks, the weight is to the power as 4 to 1, and with any number of ropes, the ends being fastened to the support, W=2" × P. Example.-What weight will a power of 3 lbs. sustain in a system of 4 movable pulleys and 4 ropes 3×2×2×2×2=48 lbs. Ans. MECHANICAL CENTERS. These are the centers of gravity, oscillation, percussion, and gyration. CENTER OF GRAVITY. The center of gravity of a body, or any system of bodies connected, is a point about which, if suspended, all the parts are in equilibrio. Hence, if a body be suspended or supported by this point, the body will rest in any position into which it is put. We may, therefore, consider the whole weight of a body as centered in this point. The common center of gravity of two or more bodies, is the point about which they would equiponderate, or rest in any position. If the centers of gravity of two bodies, A and B, be connected by a right line, the distances, A C and B C, from C, the common center of A B gravity, are reciprocally as the weights or bodies A and B; that is, as AC BC: B: A. If a line be drawn from the center of gravity of a body, perpendicular to the horizon, it is called the line of direction, because it is the line the center of gravity would describe if the body fell freely. Thus, the adjoining figure represents a loaded wagon on the declivity of a hill. The line C F represents the horizon; D E the base of the wagon. If the wagon be loaded in such a D manner that the center of gravity be at B, the perpendicular, B D, will fall within the base, and the wagon will stand. But if the load be altered, so that the center of gravity be raised to A, the perpendicular, A e, will fall outside of the base, and the wagon will be overset. From this it follows that a wagon, or any carriage, will be most firmly supported when the center of gravity falls exactly between the wheels; and that is the case on a level road. The center of gravity, in the human body, is between the hips, and the base is the feet. The method of determining the center of gravity mechanically in various bodies of uniform density is as follows: OF SURFACES. To find the Center of Gravity in the Surface of any Parallelogram, Rhombus, Rhomboid, Regular Polygon, Ellipse, Circle, or Lune. Rule.-Draw diagonals, diameters, or radii, intersecting one another; the point of intersection is the center of gravity; i. e., the center of gravity is the geometrical center of these figures. To find the Center of Gravity of a Triangle. Rule. Draw a line from any angle to the middle of the opposite side, then of this line from the angle will be the position of the center of gravity. To find the Center of Gravity of a Trapezium. Rule.-Draw the two diagonals, and find the centers of gravity of each of the four triangles thus formed, then join each opposite pair of these centers of gravity, and the two joining lines will cut each other in the center of gravity of the figure. To find the Center of Gravity of a Circular Arc. Rule.-Radius of circle × chord of arc length of arc ter of gravity from the center of the circle. = distance of the cen To find the Center of Gravity of the Sector of a Circle. Rule.-2 x chord of are x radius of circle 3 x length of arc center of gravity from the center of the circle. = distance of the To find the Center of Gravity of a Semicircle. Rule.-4 × radius of circle 3 x 3.14159 =3 distance from the center To find the Center of Gravity of a Segment of a Circle. Rule.-Chord of the segment cubed 12 x area of segment = distance from the center. To find the Center of Gravity of a Parabola. The distance of the center of gravity from the vertex is Rule. of the axis. To find the Center of Gravity of a Cylinder Cone, Frustrum of a Cone, Pyramid, or Frustrum of a Pyramid. Rule. The center of gravity is at the same distance from the base as that of the parallelogram, triangle, or trapezoid, which is a right section of either of the above figures. OF SOLIDS. To find the Center of Gravity of a Sphere, Spherical Segment, or Zone. Rule. At the middle of their height, is their center of gravity. To find the Center of Gravity of a given Pyramid or Cone. Rule.-Draw a straight line from the vertex to the center of gravity of the base, and divide it in the ratio of 3 to 1, the greatest segment being next the vertex. The point so found is the center of gravity of the pyramid. This construction applies to a pyramid of any number of sides, and therefore also to a cone. Or, of the line joining the vertex and center of gravity of the base. To find the Center of Gravity a Frustrum of a Cone or Pyramid. (R+r)2+2 R2 Rule.-hx (R+r)2 – R r where h the height, and R and r the radii, the greater and lesser ends in a cone and the sides of a pyramid. To find the Center of Gravity of a Paraboloid. Rule. The distance from the vertex is of the axis or abscissa. To find the Center of Gravity of the Frustrum of a Paraboloid. Rule.-hx: 2R2+r2 R2+2 where h the height, and R and r the radii of the greater and lesser ends. To find the Center of Gravity of a Spherical Segment. Rule. The distance from the center=3·1416 v2(r-v)2 where v is S the versed sine, s the solidity of the segment, and r the radius of the sphere. To find the Center of Gravity of a Spherical Sector. Rule.—The distance from the center=4(r−2). To find the Center of Gravity of any System of Bodies. Rule. The distance of the common center of gravity from a B D+B D+B" D'+, &c. given plane = B+B+B+, &c. ; BBB" being the masses or solid contents of the bodies, and D. D' D' the distances of their respective centers of gravity, from the given plane. FRAUDULENT BALANCES. Commercial balances are frequently misconstructed for fraudulent purposes; but the end to be obtained by the use of such a balance, nay be readily defeated by the following Rule.-First establish an equilibrium between the substance to be weighed and the weights; then transpose them, and the weight will preponderate, if the article weighed be lighter than the weight, and contrawise. Then, To ascertain the True Weight, Find the weight that will produce equilibrium after transposition, and let this and the former counterpoise be reduced to the same denomination of weight; then multiply together the two weights thus expressed, and extract the square root of their product; the result will be the true weight. Example. If one counterpoise weigh 7 lbs., and the other 94, the product is 64, the square root of which is 8. Hence, 8 lbs. is the true weight; and in this way the most precious metals may be accurately weighed on the most fraudulent balances, which may be proved by the following formula: Let a and b be the arms of the two balances, A B the two counterpoises, and x the true weight. |