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_.
+-----------+----------+----------+----------+--------------+
| | R. |Power in | P. |Difference of |
|Number of |Total load|lbs. which|Calculated| the observed |
|Experiment.| on slide |just draws| value of |and calculated|
| | in lbs. | up slide.|the power.| 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.