Fig. 97 represents the case of motion up a plane when the coefficient of friction is one half, and the angle of elevation just great enough to make the weight border upon overhauling. The vertical arrow ka represents the weight W of 10 lbs., and we know also the line of action of the reaction R of the plane, which must make an angle with the left, or down-hill side, of the normal equal to 9, the angle of repose, whose tangent is the coefficient of friction. From k draw the indefinite line kl parallel to this line of action and complete the triangle of forces by drawing the power P in any desired direction, as af, ag, ah, aj; the figure shows four such triangles-afk, agk, ahk, ajk—whose sides represent the various values of P and R for the weight of 10 lbs. Suppose now the weight to move up the plane a distance ab = 10; if P is parallel to the plane it moves over the same distance; and for other directions of P the distances ac, ad, ae, moved over are found by dropping from 6 the perpendiculars bc, bd, be, upon the lines of action of P. The lengths of these lines being measured or calculated, the following results are obtained for the angle of elevation a = angle of repose, tan-1.5 sin-1 .447, and a weight W= 10 moving up over a distance 10. = Vertical distance passed over by W = 4.47. When Pis horizontal: = P= af = 13.33 and moves over ac = 8.95. When Pis parallel to the plane : = 1.12. Per cent. of applied energy wasted = 50. Velocity ratio Per cent. of applied energy wasted = 37.5. When P is 2 p above the plane: P = aj = 8.95, and moves over ae = 6. = Per cent. of applied energy wasted = 16.7. = 1.25. Velocity ratio=1.34. Mechanical efficiency = 1.12. Corresponding calculations have been made for a coefficient of friction = .375, with results to be given presently, and an experi mental verification thereof has been obtained by means of the model illustrated in Fig. 98. To obtain the desired coefficient of friction an iron weight sliding upon brick is used, and to secure a uniform pressure between the weight and the brick the point of application of P is at the intersection of the vertical through, the center of gravity of W with the reaction R, drawn through the center of the lower surface of W. The direction of P can be varied somewhat by experimenting upon different parts of the plane, or it can be changed by placing the pulley in any one of a number of grooves in the upright support. The following are the results for the angle of elevation a = angle of repose, tan-1 .375 sin-1 .35, and a weight W-2.17 lbs. moving up over a distance of one inch. Vertical distance passed over by W = .35. When P is horizontal: P = 1.90 moving over .94. Work done by P = 1.90 × .94 = 1.78 inch-lbs. Work wasted in friction = 1.78 - .76 = : 1.02. Per cent. of applied energy wasted = 57. Velocity ratio = 2.66. Mechanical efficiency Experiment gives: = 1.14. P1.95. Per cent. 58. Mechanical efficiency = 1.11. When P is parallel to the plane: P=1.53 moving over 1.00. Work done by P = 1.53 × 1.00 = 1.53 inch-lbs. Work wasted in friction = 1.53 — .76.77. Per cent. of applied energy wasted = 50. Velocity ratio 2.84. Mechanical efficiency = Experiment gives: = 1.42. P1.64. Per cent. = 53. Mechanical efficiency When P is op above the plane: P=1.43 moving over .94. Work done by P 1.43 × .94 = 1.34. = Work wasted in friction 1.34.76.58. = Per cent. of applied energy wasted = 48. Velocity ratio 1.07. Mechanical efficiency = 1.32. = 1.52. Experiment gives: P=1.54. = = 1.41. When Pis 2p above the plane: Work done by P = 1.53 × .75 = 1.15. 1.15.76 = .39. = 1.42. Per cent. of applied energy wasted = 34. Velocity ratio 2.14. Mechanical efficiency = Experiment gives: P = 1.64. Per cent. 38. Mechanical efficiency = 1.32. In the above tables the values of the reaction of the plane have not been included, inasmuch as they are, of course, proportional to the wasted work, of which the reaction is the direct cause. The reactions in the first case are, however, given in the figure so that they may be compared with the wasted work in the table. Of course there has been no attempt to obtain experimental results agreeing exactly with those of theory; such work would be time wasted so far as this discussion is concerned. Evidently a single instance of a non-overhauling mechanical power, wasting less than fifty per cent. in friction, is sufficient to disprove the proposed law; in making the models and experiments, it was therefore more important to allow everywhere a margin in favor of the law, and to obtain a conclusive disproof thereof in spite of the allowances, than to run any risk of leaning the other way for the sake of getting results nearer to those of calculation. ዋ It is to be noted that the velocity ratio is a maximum when P is parallel to the plane and is the same for any two directions of P at equal angles above and below the plane, while the mechanical efficiency is a maximum when P is at an angle o above the plane and is the same for any two directions at equal angles above and below this direction. The greatest efficiency does not, therefore, correspond with the greatest velocity ratio; in fact the maximum velocity ratio has no greater mechanical efficiency than the smallest ratio given, so that the latter has the decided advantage, because mechanical efficiency is the desirable thing in raising weights, while usually a large velocity ratio is intrinsically a dis advantage, submitted to only to get the efficiency. It is easy to obtain a general expression, both analytical and graphical, for the per cent. lost in friction. If W and R be resolved into components parallel and perpendicular to the plane the former alone appear in the equations of energy, there being no motion in the direction of the latter. In Fig. 99, ka W and lk or l'k = R, while al or al' P. kk' being normal to the plane, ma and km are the two components of W, and lp and pk those of R. (It would be cumbersome to continue the demonstration for both cases when, evidently, all that is proved when P makes an angle for P at an angle = = with the plane can easily be duplicated therewith.) Suppose now P to be divided into two parts at n, then the components of the first part are am and mn and of the second, pl and np; the component am is equal and opposite to, and therefore balances, the component ma of the weight, while pl balances lp of the reaction; consequently am performs the useful work of raising the weight and pl the useless work of overcoming the friction. The following proportion must therefore be true: Used energy wasted energy = am : pl = an: nl, and by composition Total applied energy wasted energy an+ nl : nl = al : nl. Consequently, Wasted energy = nl al |