SECTION XXXIII.-WORK. ART. 156.-Absolute Units of Work. Work is done when resistance is overcome; the quantity of work done is proportional to the resisting force and to the distance through which it is overLet the unit of work be denoted by W, then come. 1 W = F resistance by L displacement. By "displacement" is meant the displacement in the direction of the force. The British absolute unit is denominated the foot-poundal, and is defined by 1 foot-poundal = poundal resistance by foot displacement. The C.G.S. absolute unit is denominated the erg, and is defined by 1 erg dyne resistance by cm. displacement. = ART. 157.-Gravitation Units. Work is also measured in terms of gravitation units, by taking the corresponding gravitation unit of force instead of the absolute unit of force. The principal British unit is the foot-pound, defined by 1 foot-pound = lb. by weight by foot displacement. A metric unit is the kilogrammetre, defined similarly, 1 kilogrammetre kilogramme by weight by metre. The principal C.G.S. unit is the gramme-centimetre, defined similarly. ART. 158.-Work done by a Pressure. Suppose that the resistance is a pressure uniformly distributed over the surface of application. Then and 1 F resistance == (F per S) by S surface, 1 W = (F per S) by S surface by L displacement. If 1 S = L2, then, since the displacement is normal to the surface of application, 1 W = (F per L3) by L3 volume displacement. EXAMPLES. Ex. 1. Reduce 1 foot-pound to ergs, taking the acceleration due to gravity at 981 cm. per second per second. .:. 1 foot-pound = lb. by weight by foot, 1 lb. 453.6 gm., weight 981 cm. per sec. per sec., 1 foot 30.48 cm., 1 foot-pound = 981 × 453.6 × 30·48 gm. by cm. per sec. per sec. by cm. 981 × 453.6 × 30.48 ergs, = 1.356 × 107 ergs. Ex. 2. A train of 120 tons runs on a level road, and the resistances to be overcome are 8 lbs. per ton. How many units of work must be expended in making a run of 40 miles, when there is no useless expenditure of steam. 8 lb. by weight, resistance = ton of mass, 120 tons, ... ... 120 x 8 lb. by weight, resistance. 1 foot-pound = lb. by weight by foot advance, 120 x 8 foot-pound = foot advance, 40 x 5,280 feet; 120 x 8 x 40 x 5,280 foot-pounds, i.e., 2.03 × 108 Ex. 3. How many units of work must be expended in raising from the ground the materials for building a uniform column 66 feet 8 inches high and 21 feet square, a cubic foot of brickwork weighing one hundredweight. As the column is uniform there is the same amount of matter in the different courses, but the height through which the matter of a course is raised increases uniformly from the bottom to the top. Hence the correct result will be got by taking the average height. Ex. 4. What is the amount of work done per stroke by an engine when the average pressure of the steam during the stroke is 38 lb. weight per square inch, the length of the stroke being 3 feet, and the diameter of the piston 14 inches; 1 foot-pound = lb. wt. per sq. in. by sq. in. by ft. 38 lb. wt. per sq. in. by 72 sq. in. by 3 ft. per stroke, 38 x 7 x 3 foot-pounds per stroke, 1. Reduce a kilogrammetre to foot-pounds, and reciprocally. 2. The resistance to traction on a level road is 150 lbs. per ton moved; how many foot-pounds of work are expended in drawing 6 tons through a distance of 150 yards? 3. A hole is punched through a plate of wrought-iron one half inch in thickness, and the pressure actuating the punch is estimated at 36 tons. Assuming that the resistance to the punch is uniform, find the number of foot-pounds of work done. 4. It is found, neglecting friction, that a force acting horizontally will move 10 lbs. up 5 feet of an incline rising 1 in 4. Find the work done, and find also the force parallel to the plane which will just support the weight of 10 lbs. 5. Calculate the amount of work done, independently of that lost through friction, in drawing a car of two tons weight, laden with 30 passengers averaging 11 stones each in weight, up a slope, the ends of which differ in level by 50 feet. 6. The inclination of a mountain path to the horizon is 30°; how much work is done against gravity by a man of 12 stones weight in walking a mile along the path? 7. How much work is done by means of a crane in raising from the ground the material required to build a stone wall 100 feet long, 36 feet high, and 2 feet thick, the density of the stone being 153 pounds per cubic foot. 8. A fly-wheel weighing 7 tons turns on a horizontal axle 1 foot in diameter. If the coefficient of friction between the axle and its bearing is 0.075, what number of foot-pounds of work must be done against friction while the wheel makes 10 turns? 9. The resistance of friction along an inclined plane is taken at 150 lbs. for each ton of weight moved. Find the work done in drawing 2 tons up 100 fect of an incline which rises 1 foot in height for 25 in length. 10. Weights of 10 lbs. and 8 lbs. are connected by a very fine thread which rests on a rough fixed pulley, so that they hang suspended; the heavier weight is found to be just not heavy enough to fall and draw the lighter weight up; if now we suppose the weights to move uniformly, so that one goes up and the other down through 12 feet, how many foot-pounds of work are done against friction during the motion? 11. The plunger of a force-pump is 83 inches in diameter, the length of the stroke is 2 feet 6 inches, and the pressure of the water is 50 lbs. per square inch; find the number of units of work done in one stroke. 12. Determine the unit of mass in order that the absolute unit of work may be equal to the foot-pound, the second and foot being the units of time and length, and 32-2 feet per sec. per sec. being the implied intensity of gravity. SECTION XXXIV.-KINETIC ENERGY. ART. 159.-Kinetic Energy. By kinetic energy is meant the equivalent of the work spent in increasing the speed of a body. The work spent is proportional to the mass of the body, to its initial speed, and to the increment of speed. It is evident that the increment of speed must be taken small, in order that the initial speed may not vary sensibly. Hence 1 W = M by (L per T) initial by (L per T) increment. To find the entire amount of kinetic energy in a body, we have to consider that the total increment is the final speed, and that the average initial speed is half the final speed; hence As particular cases W = M by (L per T final)2. Observe that the dimensions of each of the equivalents of W are the same. ART. 160.-Kinetic Energy of Rotation; Moment of Inertia. We have W = M by (L per T)2. Suppose that a rigid body revolves round an axis, and that the whole of the mass of the body is at nearly the same distance from the axis; then since it follows that 1 L per T = L radius by (radian per T), W M by (L radius)" by (radian per T), = or W per (radian per T)2 = M by (L radius). The idea M by (L radius)2 depends on the body only; it is called the moment of inertia of the body. From the above the unit is seen to be equivalent to half unit of work per square of unit of angular velocity. ART. 161.-Radius of Gyration. When the several portions of the body are at different distances from the axis, an equivalent radius can be found such that, were the whole mass situated at its end, the value of M by L2 would be the same as the sum of the actual values of M by L2 for the different portions. This equivalent radius is called the radius of gyration. Compare Arts. 31 |