Let db and qb represent the magnitudes and directions of two forces that act to move the body b. By completing the parallelogram there will be obtained a diagonal force fb, whose magnitude and direction are equal to the effect produced by d band qb. fb is called the resultant of db and q b. If three or more forces act in different directions to move a body b, find the resultant of any two of them, and consider it as a single force. Between this and the next force find a second resultant. Thus, pb, qb, and rb are magnitudes and directions of the forces. pb + qb + rb ≤ gb+rb fb, the magnitude and direc tion of the three forces, pb, qb, and rb. A SINGLE MOVABLE PULLEY. F: W = 1:2, or F = W. If the force F be applied at a and act upwards, the result will be the same. A DOUBLE MOVABLE PULLEY. F: W1: 4, or F = W. Let u= number of parts of rope, not counting the free end. FW ÷ U. v: v 1: u. 2= QUADRUPLE MOVABLE PULLEY. F = { W. F: W = 1: 8. Let u = number of parts of rope, not counting the free end; then, F= W ÷ u. v: v = 1: u. V= velocity of W; COMPOUND PULLEY. u = number of movable pulleys. v = velocity of F. Hydrostatics treats of liquids at rest under the action of forces. If a liquid is acted on by a pressure, the pressure per unit of area exerted anywhere on the mass of liquid is transmitted undiminished in all directions, and acts with the same force on all surfaces, in a direction at right angles to those surfaces. General Law for the Downward Pressure on the Bottom of Any Vessel. The pressure on the bottom of a vessel containing a liquid is independent of the shape of the vessel, and is equal to the weight of a prism of the liquid whose base is the same as the bottom of the vessel, and whose altitude is the distance between the bottom and the upper surface of the liquid, plus the pressure per unit of area upon the upper surface of the liquid multiplied by the area of the bottom of the vessel. General Law for Upward Pressure.-The upward pressure on any submerged horizontal surface equals the weight of a prism of the liquid whose base has an area equal to the area of the submerged surface, and whose altitude is the distance between the submerged surface and the upper surface of the liquid, plus the pressure per unit of area on the upper surface of the liquid multiplied by the area of the submerged surface. General Law for Lateral Pressure.-The pressure on any vertical surface due to the weight of the liquid is equal to the weight of a prism of the liquid whose base has the same area as the vertical surface, and whose altitude is the depth of the center of gravity of the vertical surface below the level of the liquid. Any additional pressure is to be added. Pressure on Oblique Surfaces.-The pressure exerted by a liquid in any direction on a plane surface is equal to the weight of a prism of the liquid whose base is the projection of the surface at right angles to the given direction, and whose height is the depth of the center of gravity of the surface below the level of the liquid. If a cylinder is filled with water, and a pressure applied, the total pressure on any half section of the cylinder is equal to the projected area of the half cylinder (or the diameter multiplied by the length of the cylinder) multiplied by the depth of the center of gravity of the half cylinder, multiplied by the weight of a cubic inch of water, plus the diameter of the shell, multiplied by the pressure per square inch, multiplied by the length of the cylinder. If d = the diameter, and — the length of the cylinder, the pressure due to the weight of the water when the cylinder is vertical upon the half cylinder 12 dxlxx the weight of a cubic inch of water = d XX the weight of a cubic inch of water; d and I are to be measured in inches. The pressure in pounds per square inch due to a head of water is equal to the head in feet multiplied by .434. The head equals the pressure in pounds per square inch multiplied by 2.304. EXAMPLE. (a) What is the pressure per square inch corresponding to a head of water of 175 ft.? (b) If the pressure had been 90 lb. per sq. in., what would the head have been? SOLUTION.-(a) 175 X .434 75.95 lb. per sq. in. 207.36 ft. (b) 90 x 2.304 HYDROKINETICS. Hydrokinetics, also called hydrodynamics and hydraulics, treats of water in motion. When water flows in a pipe, conduit, or channel of any kind, the velocity is not the same at all points of the flow, unless all cross-sections of the pipe or channel are equal. That velocity which, being multiplied by the area of the cross-section of the stream, will equal the total quantity discharged, is called the mean velocity. Let Q = quantity that passes any section in 1 second; A = area of the section; v = mean velocity in feet per second. The vertical distance between the level surface of the water and the center of the aperture through which it flows, is called the head. Let V = mean velocity of efflux through a small aperture; h head in feet at the center of the aperture; w = Then, V = weight of water flowing through the aperture per second. V2gh; that is, the velocity of efflux is the same as if the water had fallen through a height equal to the head. Let Q = theoretical number of cubic feet discharged per second; Vm = mean velocity through orifice in feet per second; = area of orifice; A h Qa = theoretical head necessary to give a mean velocity Vmi actual quantity discharged in cubic feet per second. Then, for an orifice in a thin plate, or a square-edged orifice (the hole itself may be of any shape, triangular, square, circular, etc., but the edges must not be rounded), the actual quantity discharged is = .615 AVm. The weir is a device used for measuring the discharge of water. It is a retangular orifice through which the water flows. = If d the depth of the opening in feet, and b its breadth in feet, the area of the opening is A = dxb, and the theoretical discharge is Q db XV2gd, the head for this case being taken as d. = The actual discharge when the top of the weir lies at the surface of the water is Qa = = .615 Q = .615 × db × √2 g d = .615 Xb√2gd3: = 3.288b v ds. |