THE INCLINED PLANE WITH FRICTION.
265. The friction of the truck upon the glass plate is always very small, and is shown to have but little variation at those inclinations of the plane which we used. But when the friction is large, we shall not be justified in neglecting its changes at different elevations, and we must adopt more rigorous methods. For this inquiry we shall use the pine plank and slide already described in [Art. 117]. We do not in this case attempt to diminish friction by the aid of wheels, and consequently it will be of considerable amount.
266. In another respect the experiments of [Table XIII]. are also in contrast with those now to be described. In the former the load was constant, while the elevation was changed. In the latter the elevation remains constant while a succession of different loads are tried. We shall find in this inquiry also that when the proper allowance has been made for friction, the theoretical law connecting the power and the load is fully verified.
267. The apparatus used is shown in [Fig. 33]; the plane, is, however, secured at one inclination, and the pulley c shown in [Fig. 32] is adjusted to the apparatus, so that the rope from the pulley to the slide is parallel to the incline. The elevation of the plane in the position adopted is 17°·2, so that its length, base, and height are in the proportions of the numbers 1, 0·955, and 0·296. Weights ranging from 7 lbs. to 56 lbs. are placed upon the slide, and the power is found which, when the slide is started by the screw, will draw it steadily up the plane. The requisite power consists of two parts, that which is necessary to overcome gravity acting down the plane, and that which is necessary to overcome friction.
Fig. 42.
268. The forces are shown in [Fig. 42]. r g, the force of gravity, is resolved into r l and r m; r l is evidently the component acting down the plane, and r m the pressure against the plane; the triangle g l r is similar to a b c, hence if r be the load, the force r l acting down the plane must be 0·296 r, and the pressure upon the plane 0·955 r.
269. We shall first make a calculation with the ordinary law that the friction is proportional to the pressure. The pressure upon the plane a b, to which the friction is proportional, is not the weight of the load. The pressure is that component (r m) of the load which is perpendicular to the plane a b. When the weights do not extend beyond 56 lbs., the best value for the coefficient of friction is 0·288 ([Art. 141]): hence the amount of friction upon the plane is
0·288 × 0·955 R = 0·275 R.
This force must be overcome in addition to 0·296 R (the component of gravity acting down the plane): hence the expression for the power is
0·275 R + 0·296 R = 0·571 R.
270. The values of the observed powers compared with the powers calculated from the expression 0·571 R are shown in Table XIV.
Smooth plane of pine 72" × 11"; angle of inclination 17°·2; slide of pine, grain crosswise; slide started; formula P = 0·571 R.
| Number of Experiment. | R. Total load on slide in lbs. | Power in lbs. which just draws up slide. | P. Calculated value of the power. | Differences of the observed and calculated powers. |
|---|---|---|---|---|
| 1 | 7 | 4·6 | 4·0 | -0·6 |
| 2 | 14 | 8·3 | 8·0 | -0·3 |
| 3 | 21 | 12·3 | 12·0 | -0·3 |
| 4 | 28 | 16·5 | 16·0 | -0·5 |
| 5 | 35 | 20·0 | 20·0 | 0·0 |
| 6 | 42 | 24·2 | 24·0 | -0·2 |
| 7 | 49 | 28·0 | 28·0 | 0·0 |
| 8 | 56 | 31·8 | 32·0 | +0·2 |
271. Thus for example, in experiment 6, a load of 42 lbs. was raised by a force of 24·2 lbs., while the calculated value is 24·0 lbs.; the difference, 0·2 lbs., is shown in the last column.
272. The calculated values are found to agree tolerably well with the observed values, but the presence of the large differences in No. 1 and No. 4 leads us to inquire whether by employing the more accurate law of friction ([Art. 141]) a better result may not be obtained.
In [Table VI]. we have shown that the friction for weights not exceeding 56 lbs. is expressed by the formula F = 0·9 + 0·266 × pressure, but the pressure is in this case = 0·955 R, and hence the friction is
0·9 + 0·254 R.
To this must be added 0·296 R, the component of the force of gravity which must be overcome, and hence the total force necessary is
0·9 + 0·55 R.
The powers calculated from this expression are compared with those actually observed in Table XV.
Smooth plane of pine 72"× 11"; angle of inclination 17°·2; slide of pine, grain crosswise; slide started; formula P = 0·9 + 0·55 R.
| Number of Experiment. | R. Total load on slide in lbs. | Power in lbs. which just draws up slide. | P. Calculated value of the power. | Differences of the observed and calculated powers. |
|---|---|---|---|---|
| 1 | 7 | 4·6 | 4·7 | +0·1 |
| 2 | 14 | 8·3 | 8·6 | +0·3 |
| 3 | 21 | 12·3 | 12·5 | +0·2 |
| 4 | 28 | 16·5 | 16·3 | -0·2 |
| 5 | 35 | 20·0 | 20·1 | +0·1 |
| 6 | 42 | 24·2 | 24·0 | -0·2 |
| 7 | 49 | 28·0 | 27·8 | -0·2 |
| 8 | 56 | 31·8 | 31·7 | -0·1 |
For example: in experiment 5, a load of 35 lbs. is found to be raised by a power of 20·0 lbs., while the calculated power is 0·9 + 0·55 × 35 = 20·1 lbs.
273. The calculated values of the powers are shown by this table to agree extremely well with the observed values, the greatest difference being only O·3 lb. Hence there can be no doubt that the principles on which the formula has been calculated are correct. This table may therefore be regarded as verifying both the law of friction, and the rule laid down for the relation between the power and the load in the inclined plane.
274. The inclined plane is properly styled a mechanical power. For let the weight be 30 lbs., we calculate by the formula that 17·4 lbs. would be sufficient to raise it, so that, notwithstanding the loss by friction, we have here a smaller force overcoming a larger one, which is the essential feature of a mechanical power. The mechanical efficiency is 30 ÷ 17·4 = 1·72.
275. The velocity ratio in the inclined plane is the ratio of the distance through which the power moves to the height through which the weight is raised, that is 1 ÷ 0·296 = 3·38. To raise 30 lbs. one foot, a force of 17·4 lbs. must therefore be exerted through 3·38 feet. The number of units of work expended is thus 17·4 × 3·38 = 58·8. Of this 30 units, equivalent to 51 per cent., are utilized. The remaining 28·8 units, or 49 per cent., are absorbed by friction.
276. We have pointed out in [Art. 222] that a machine in which less than half the energy is lost by friction will permit the load to run down when free: this is the case in the present instance; hence the weight will run down the plane unless specially restrained. That it should do so agrees with [Art. 147], for it was there shown that at about 13°·4, and still more at any greater inclination, the slide would descend when started.