exact values under any condition. The student is, therefore, advised to use the values given in the tables, except where careful experiments have been made that give reliable values for the particular case under consideration. To find the force, in pounds, necessary to overcome friction, the coefficient taken from the table is multiplied by the perpendicular pressure, in pounds, on the surface considered. If the force acts at an angle to the surface, the perpendicular force can be found by resolving the given force into two components, one perpendicular and the other parallel to the surface. .18 Wrought iron on cast iron. Slightly greasy .18 .15 93. In the case of a weight sliding along a horizontal plane surface, the pressure is equal to the weight. When the surface is inclined, the weight acts vertically downwards, and the pressure perpendicular to the surface can be found by the principle of the resolution of forces. many cases the pressure on the surfaces is due to the combined action of several forces that must be combined into one common resultant force. 94. The work that must be done in overcoming the resistance of friction depends on the distance through which the resistance is overcome. It may be calculated by the following rule: Rule 9.—Multiply the total pressure in pounds by the distance in feet and by the coefficient of friction. EXAMPLE.-The average perpendicular pressure on the guide of a steam engine due to the force impelling the piston is 2.500 pounds. The pressure due to the weight of the crosshead and connecting-rod is 400 pounds. The crosshead moves at the rate of 500 feet per minute; what horsepower is required to overcome the friction on the guides ? SOLUTION. The total perpendicular pressure is 2,500+ 400 = 2,900 pounds. Since the lubrication is usually intermittent, the coefficient of friction, for a brass slipper working on a cast-iron guide, may be taken as .08. The resistance being overcome through a distance of 500 feet each minute, the work done in overcoming friction is 2,900 × 500 × .08 116,000 foot-pounds per minute. Then, the horsepower is 116,000 33,000 3.52 H. P., nearly. Ans. = 95. In the case of a shaft rotating in a bearing, the distance through which friction is overcome each minute is found by multiplying the circumference of the journal by the number of revolutions per minute. For a shaft, or any other body rotating in a bearing, the force required to overcome friction, as calculated by multiplying the pressure by the coefficient of friction, is the force that must be applied at the surface of the journal. 96. Allowable Pressures.-It has been found by experience that when the pressure per unit of area exceeds a certain amount, the lubricant will be forced out and the bodies rubbing on each other will heat and, finally, seize. The pressures that can safely be allowed on the bearings. of steam engines, on guides, thrust bearings, crankpins, crosshead pins, etc. vary considerably, being dependent to a large extent on the character of the workmanship, the degree of finish, the variation of the pressure, the character of the lubrication, and the quality of the lubricant. With fair workmanship, the following pressures per square inch. represent average practice in steam-engine work: TABLE III. PRESSURES PER SQUARE INCH ALLOWABLE IN 97. For crankpins, wristpins, and guides, the allowable pressures given represent the pressures corresponding to the maximum load, which in the case of a wristpin and crankpin occurs when the crank, connecting-rod, and piston rod are in a straight line, and in the case of guides, when the connecting-rod and crank are at right angles to each other. In the case of pins and journals, the area to be considered in calculating the pressure on the bearing is the projected area, which is found by multiplying the length of the journal by its diameter. EXAMPLES FOR PRACTICE. 1. A body weighs 90 pounds; what is its mass? Ans. 2.799, nearly. 2. What force will be required to accelerate a body at the rate of 2 feet per second, the body weighing 450 pounds, and the frictional resistances being equal to 10 per cent. of the weight of the body? Ans. 72.99 lb., nearly. 3. What work is done in raising 950 pounds 17 feet? Ans. 16,150 ft.-lb. 4. If an engine does 205,000 foot-pounds of work per minute, what is its horsepower? Ans. 6.21 H. P., nearly. 5. What is the kinetic energy of a shell fired from a cannon with a velocity of 1,800 feet per second, the shell weighing 1,000 pounds? Ans. 50,373,135 ft.-lb., nearly. 6. Taking the coefficient of friction at .15, what horsepower will be required to pull 100 pounds at a uniform speed of 5 feet per second along a level surface? Ans. .136 H. P. CENTER OF GRAVITY. 98. The center of gravity of a body is that point at which the body may be balanced, or it is the point at which the whole weight of a body may be considered as concentrated. This point is not always in the body; in the case of a horseshoe or a ring it lies outside of the substance of, but within the space enclosed by, the body. In a moving body, the line described by its center of gravity is always taken as the path of the body. In finding the distance that a body has moved, the distance that the center of gravity has moved is taken. The definition of the center of gravity of a body may be applied to a system of bodies if they are considered as being connected at their centers of gravity. 3 FIG. 9. W If w and W, Fig. 9, are two bodies of known weight, their center of gravity will be at C. This point C may be readily determined as follows: Rule 10.-The distance of the common center of gravity from the center of gravity of the large weight is equal to the weight of the smaller body multiplied by the distance between the centers of gravity of the two bodies, and this product divided by the sum of the weights of the two bodies. EXAMPLE.-In Fig. 9, w = 10 pounds, W: 30 pounds, and the distance between their centers of gravity is 36 inches; where is the center of gravity of both bodies situated? |