SECTION IV. WORK, ENERGY, POWER. CHAPTER XIV. WORK 163. An external force is said to be doing work on a mass when its point of application is passing over distance in the direction of the force. Examples. The weight of a mass falling freely does work on the mass. The pressure of the foot of a bicyclist does work on the treadle of his bicycle. The pull of a horse does work on the carriage it is drawing. When a horse is attached to a railway truck and pulls at it, then as long as the truck is at rest, we say the horse is leaning against the truck, and does no work; but when the truck moves and the horse continues his pull, he does work as long as his pull is continued. 164. DEF. The work done in an interval on a mass by an external force, varies directly as the force; and also directly as the resolved part in the direction of the force, of the distance passed over in that interval by the point of application of the force. Thus, the work done by a force applied to a mass in motion, is doubled, trebled, etc., when the force is doubled, trebled, etc., for the same distance; and the work done by a given force is doubled, trebled, etc., when the resolved part of the distance in the direction of the force is doubled, trebled, etc. It is convenient to refer to the resolved part in the direction of the force, of the distance passed over by the point of application of a force, as the distance through which the force works. I I NOTE. The work done by an external force on a moving mass is independent of the length of the interval occupied; it depends only on the force and the distance. 165. DEF. We shall choose as our unit work the work done by a poundal while its point of application moves over a foot in the direction of the poundal. We shall call this unit work a foot-poundal. Hence the measure of the work done by a uniform force, is the product of the number of poundals by the number of feet through which it works. Example. The work done by the weight of 1 lb. while the i lb. passes over i foot vertically downwards, is g foot-poundals. For, the weight of 1 lb.=g poundals, 166. When the distance through which the force works, is passed over by the point of application in the direction opposite to that of the force, this distance is negative; and in this case the work done by the force is said to be negative; or we say that work is done against the force. Work which is done in lifting mass under the action of gravity, is done against the weight of the mass. 167. The work done against gravity in raising 1 lb. through a vertical distance of 1 foot is g foot-poundals. 8 This amount of work is called a foot-pound, which is an abbreviation for a foot-poundweight. Hence a foot-pound is equal to g foot-poundals. Example. Find the work done by the weight of i cwt. while the cwt. falls a vertical distance of 10 feet. The force is 32 X 112 poundals. The distance passed over by the mass in the direction of the force, that is, the distance through which the weight works, is 10 ft. The work done is therefore 32 x 112 x 10 foot-poundals. 168. When two equal and opposité forces (as, for example, the action and reaction of a stress) act upon the same point the work done by one of the forces is equal and opposite to the work done by the other. Also, if two equal and opposite forces act upon different points which have no motion relative to each other, the work done by one of the forces is equal and opposite to that done by the other force. The total work done by two such forces is zero. Example i. The tension of an inextensible string does no work. A man in a railway carriage who presses with his feet against the opposite seat does no work, notwithstanding that the mass, to which he applies two equal and opposite forces, is in motion. Example ii. When the masses in an Atwood's machine are equal and are in motion the weight of one does as much work on its mass as the second mass does against its own weight. Example iii. I cwt. slides down an inclined plane of height 10 ft. and inclination a; find the work done by the weight of the mass when the mass goes from the top to the bottom. The distance passed over by the mass is the slant side of the plane ; the direction of the force is vertically downward; the resolved part of the distance in the direction of the force is the vertical distance passed over; this is, the height of the plane, viz. 10 feet. Hence, the work done is 10 x 112 x 32 foot-poundals ; that is, 1120 foot-pounds. Example iv. A horse draws a tram-car of 2 tons along a horizontal road with the uniform velocity of 71 miles an hour; supposing the resistance of the friction etc. is go of the weight of the car, find how many foot-pounds the horse is doing per second. The resistance of friction etc. is a force equal to the weight of sb of 2 tons; that is, the weight of 56 lbs. Therefore the horse must be applying a force of 56 xg poundals ; [for, since the velocity of the car is uniform, no additional force is required by the inertia of the car.] The point of application of this force is moving at the rate of 74 miles per hour ; that is, of 11 velos. Therefore the horse is doing 11 x 56 xg foot-poundals per second ; that is, 616 foot-pounds per second. 169. PROP: The number of foot-pounds in the work done by the weight of a mass while the mass is moved from one position to another is the product of the number of lbs. weight by the number of feet in the vertical distance of the second position below the first. For we may consider the path of the mass to be along a number of inclined planes; the work done by the weight in descending any one of these planes is the same as the work done in descending the vertical height [Example iii. above). The work done in ascending any inclined plane is that due to its vertical height and is negative; hence The work done by the weight of a mass in going by any path whatever from one position to another is the same as that done by the weight of the mass in descending from the first position to the horizontal plane in which the second position lies. Thus, the work done in raising 1 ton through a vertical distance 50 feet is the same, viz. 2240 x 50 x g foot-poundals, by whatever path the ton arrives at that height. It may be raised by screws; or by a system of pulleys; or by a horse pulling it up an incline in a cart; or it may be raised by a man carrying it piece-meal up a ladder. Find the number of foot-poundals and of foot-pounds done by the weights of the following. 1. 56 lbs. in descending 40 feet. 5. A train of 300 tons in descending an incline of 1 in 50, one mile in length. 6. A waggon and horses of 3 tons in descending a hill, whose inclination is I in 30, for a distance of 200 yards. I 7. Compare the work done in 1 hour against gravity (i) by a man who lifts 28 Ibs. a height of 3 ft. every minute, (ii) by a man of 10 stone who walks up an incline of 1 in 6 at the rate of 4 miles an hour. 8. Find the work done against friction in moving a ton 100 feet along a rough horizontal plane the horizontal resistance of the friction being t of the weight. 9. Find the number of foot-pounds per second required to draw a cart of i ton along a horizonal plane uniformly at the rate of 5 miles an hour, the resistance due to friction etc. being to of the weight. 10. Find the greatest uniform velocity with which an engine capable of producing 140,000 foot-pounds of work per second, can draw a train of 100 tons along a horizontal plane, the resistance due to friction etc. being so of the weight. 11. An engine is drawing a train of 120 tons up a smooth inclined plane of 1 in 60 at the rate of 24 miles an hour ; how much work is it doing per second (friction to be neglected)? 12. Supposing the resistance due to friction etc. in Question 11 to be equivalent to the weight of 1120 lbs. and to act opposite to the direction of motion, find at what rate the engine could draw the same train up an incline of 1 in 6o, doing the same amount of work per second as before. 13. Supposing the resistance due to friction in Question 12 to be złoth of the pressure of the train on the inclined plane, find at what rate the engine could draw the train up the inclined plane with the same expenditure of work per second. 14. Supposing that a man of 12 stone in walking raises his whole weight a vertical distance of 1 inch at every step, and that the length of each step is 21 feet, find how much work the man does in this way in walking i mile. 15. Compare the work done by the man in Question 14 with that done by another man of 12 stone going a mile with uniform speed on a bicycle weighing : cwt., the resistance due to friction etc. of the bicycle being ito of the weight. I |