TO FIND THE HORSE-POWER OF ANY WORKING AGENT. 13 ANSWER.—The rate of doing work, as measured in horse-power, is equivalent to 33,000 foot-pounds of work done per minute. Ist. I cubic foot of water weighs 62 lbs. ... 40 cubic feet of water weigh 40 × 621 = 2500 lbs. Note. Students will find it a great advantage, as well as a saving of time not to multiply figures together until the last stage of the answer has been reached, and then to cancel all common factors in numerator and denominator. For example, in the answer to the above question we might proceed thus— Ist. 40 cubic feet of water = 40 × 62 lbs. 2nd. Work done per minute = 40 × 62 × 330 ft.-lbs. The process consists in this-the factor, 330, can be cancelled from numerator and denominator, leaving 100 as the denominator. The factor, 10, can then be divided out of 40 in the numerator and from the 100 in the denominator, thus leaving 4 × 62 as the numerator and 10 as the denominator. The remainder of the work is evident. LECTURE II.-QUESTIONS. 1. Define the unit of work. What name is given to this unit? In drawing a load a horse exerts a constant pull of 120 lbs. ; how much work will be done in 15 minutes, supposing the horse to walk at the rate of 3 miles an hour? (S. and A. Exam. 1891.) Ans. 475,200 ft.-lbs. 2. How is the work done by a force measured? The resistance to traction on a level road is 150 lbs. per ton of weight moved; how many footpounds of work are expended in drawing 6 tons through a distance of 150 yards? Ans. 405,000 ft.-lbs. 3. Distinguish between force and work done by a force. How is each respectively measured? A traction engine draws a load of 20 tons along a level road, the tractive force on the load being 150 lbs. per ton. Find the work done upon the load in drawing it through a distance of 500 yards. (S. and A. Exam. 1888.) Ans. 4,500,000 ft.-lbs. 4. Find the number of units of mechanical work expended in raising 136 cubic feet of water to a height of 20 yards. The weight of a cubic foot of water is 62 lbs. Ans. 510,000 ft. -lbs. 5. A weight of 4 tons is raised from a depth of 222 yards in a period of 45 seconds; calculate the amount of work done. Ans. 5,967,360 ft.-lbs. 6. 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. Ans. 3360 ft.-lbs. 7. How is work done by a force measured? Give some examples. Set out a diagram of the work done in drawing a body weighing 10 lbs. up a smooth incline 4 feet high, marking dimensions. (S. and A. Exam. 1889.) 8. A train of 12 coal waggons weighing 133 tons is lifted by hydraulic power (two waggons being raised at a time) through 20 feet in 12 minutes. Estimate the work done in foot-tons. Taking the average of work done, how many foot-pounds are done per minute? (S. and A. Exam. 1890.) Ans. 2660 ft.-tons; 496,533.3 ft.-lbs per minute. 9. The plunger of a force-pump is 8 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, and plot out a diagram of work to scale. Ans. 7516.5 ft.-lbs. 10. A chain 30 feet long, and weighing 100 pounds per yard, lies coiled on the ground. Find by calculation and by a scale diagram of work how many units of work would be expended in just raising it by the top end from the ground. Ans. 15,000 ft.-lbs. et II. A chain, weighing 30 lbs. to the fathom, is employed to lift I torn to a height of 30 ft. by winding the chain on a barrel. Find by calculation and by a scale diagram of work, how many units of work will be exper ded (a) when the outer end of the chain is brought home to the barrel; (b) when 18 feet of it are still hanging free with the weight at the end of it. 12. Define the following mechanical terms :-Force, work, unit of wed ork, power, activity, and horse-power. A horse drawing a cart at the rate of 2 miles per hour exerts a traction of 156 lbs. ; find the number of united, of work done in one minute. Ans. 27,456 ft.-lbs. 13. In what way is the rate of doing work measured in horse-pows, ver? If 100 cubic feet of water be raised per minute through 330 feet, wn hat horse-power of engine will be required, supposing that there is no losse by friction or other resistances? Ans. 62.5 h.p. a 14. If a horse, walking at the rate of 23 miles per hour, draws 104 bs. out of a well by means of a cord going over a wheel, how many units work would he perform in one minute? Ans. 22,880 ft.-lbs. of LECTURE III. CONTENTS.-The Moment of a Force-Principle of Moments applied to the Lever-Experiments I. II. III.-Pressure on and Reaction from the Fulcrum-Equilibrant and Resultant of two Parallel Forces-Couples -Centre of Parallel Forces or Position of Equilibrant and ResultantCentre of Gravity-Examples of Centre of Gravity-The Lever when its weight is taken into Account-Examples I. II.-Position of the Fulcrum-Example III.-Questions. The Moment of a Force is equal to the force multiplied by the perpendicular distance from a point on its line of action. Then, For example, suppose a body to be resting on the point O, and a force, P, to be applied to the body in the direction PA. if the perpendicular distance from O on the line of action of the force be OA, the moment of the force P, tending to turn the body about the point O, is Px AO. If the force be reckoned in pounds, and the perpendicular distance in feet, the product will be in pounds-feet. The student must therefore avoid confusing the answer with ft.-lbs. of work. MOMENT OF A FORCE. Principle of Moments. If any number of forces act in one plane on a rigid body, and if these forces are in equilibrium; then the principle of moments asserts that the sum of the moments of those forces which tend to turn the body in one direction about a point, is equal to the sum of the moments of the forces which tend to turn the body in the opposite direction about the same point. Or, to state the principle more concisely, the opposing moments about the point are equal. If the moments of those forces which tend to turn the body to the right hand (i.e., in the direction of the motion of the hands of a clock) be called positive (+), and the moments of the remaining set of forces which tend to turn the body to the left hand (i.e., in the opposite direction to the movement of the hands of a clock) be called negative (−), then the algebraical sum of the moments of the forces which act in one plane, and which are in equilibrium about a point, is zero. Principle of Moments applied to the Lever.-A lever is simply a rigid rod, bar, or beam, capable of turning about a fixed B W F B A W B LEVERS IN EQUILIBRIUM. point called the fulcrum (F). Acting on the lever in one direction is a force or set of forces which we shall term the pull or pressure (P), and in the other direction there is the resistance or set of resistances to be overcome, which we shall term the weight (W). The pressure, P, and the weight, W, produce a reaction at the fulcrum, which is called the equilibrant (E). The parts of the lever between the fulcrum and the pressure and between the fulcrum and the weight are called the arms of the lever. The accompanying three figures show three ways in which F, P and W may be arranged with a straight In each case, the opposing moments about the fulcrum are equal, when the lever is in equilibrium. lever.* satisfies the conditions for equilibrium in the case of a lever. EXPERIMENT I. To prove the foregoing statements, take a rigid homogeneous bar, AB, of uniform section. Let the bar be of yellow pine, I inch deep, inch broad, and 32 inches long. Attach to the ends, A and B, light flexible cords with small hooks at their lower ends, and attach to the middle of the bar at F another light flexible cord, and pass this cord over a pulley having a minimum of friction at its bearings. Fix such a weight to the free end of this middle cord as will just counterpoise the bar and cords. Test the accuracy of this preliminary adjustment by *The levers represented by the above three figures are assumed to be without weight. A force, P, acts through a perfectly flexible, weightless cord at A, and another force, W, acts also through an exactly similar cord at B, with the fulcrum at F in each case. In the second and third case the cord attached at A passes over a frictionless pulley in order to give the necessary direction to the force P. These three relative positions of P, W and F used to be termed the first, second and third order of levers; but there is no necessity for any such distinction, since all the student has to remember is this, that when equilibrium exists the opposing moments Р BF W AF about the fulcrum are equal, i.e., (P x AF=W × BF), or, or = W AF' Р BF The ratio W to P is termed the theoretical advantage of the lever. PRINCIPLE OF MOMENTS APPLIED TO THE LEVER. 17 observing whether the bar hangs horizontal, and, if pulled down or up a little, whether the weight balances the bar and cords. Now affix equal weights, P, of, say, 4 oz., to the cords hanging from the ends A and B, and add an equilibrating weight, E, of 8 oz. to the end of the central cord. You will find that the bar will come to rest in a horizontal position, thus proving that If P and P are now removed from the ends A and B, and a single weight, R, of 8 oz. be hung from F (as represented by the vertical dotted line and arrow), the result as far as the balancing of the system is concerned will be unaffected. Consequently, i.e., Ꭱ = E = Р + Р Or, the resultant of two equal parallel forces acting in the same direction is equal to the sum of the two forces, and acts midway between them and parallel to them-i.e., at the same point as the equilibrant, and in the same line therewith, but in the opposite direction. B |