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.