Wrought iron screw, square thread, diameter 2",
pitch 2 threads to the inch, arm 33"; nut
brass, bearing surfaces oiled; velocity ratio
414; useful effect, 28 per cent.; mechanical
efficiency 116; formula _P_ = 0·66 + 0·0075 _R_.
+-----------+--------+---------+----------+-----------------+
| | | | P. |Difference of the|
| Number of | R. |Observed |Calculated| observed and |
|Experiment.| Load | power | power | calculated |
| |in lbs. | in lbs. | in lbs. | powers. |
+-----------+--------+---------+----------+-----------------+
| 1 | 112 | 1·4 | 1·5 | +0·1 |
| 2 | 224 | 2·2 | 2·3 | +0·1 |
| 3 | 336 | 3·3 | 3·2 | -0·1 |
| 4 | 448 | 4·1 | 4·0 | -0·1 |
| 5 | 560 | 5·0 | 4·9 | -0·1 |
| 6 | 672 | 5·7 | 5·7 | 0·0 |
| 7 | 784 | 6·5 | 6·5 | 0·0 |
| 8 | 896 | 7·4 | 7·4 | 0·0 |
| 9 | 1008 | 8·1 | 8·2 | +0·1 |
| 10 | 1120 | 9·0 | 9·1 | +0·1 |
+-----------+--------+---------+----------+-----------------+
292. It may be seen from the column of differences how closely the
experiments are represented by the formula. The power which is
required to raise a given weight, say 600 lbs., may be calculated by
this formula; it is 0·66 + 0·0075 × 600 = 5·16. Hence the mechanical
efficiency of the screw-jack is 600 ÷ 5·16 = 116. Thus the screw
is very powerful, increasing the force applied to it more than a
hundredfold. In order to raise 600 lbs. one foot, a quantity of work
represented by 5·16 × 414 = 2136 units must be expended; of this only
600, or 28 per cent., is utilized, so that nearly three-quarters of the
energy applied is expended upon friction.
293. This screw does not let the load run down, since less than 50 per
cent. of energy is utilised; to lower the weight the lever has actually
to be pressed backwards.
294. The details of an experiment on this subject will be instructive,
and afford a confirmation of the principles laid down. In experiment
10 we find that 9·0 lbs. suffice to raise 1,120 lbs.; now by moving
the pulley to the other side of the lever, and placing the rope
perpendicularly to the lever, I find that to produce motion the other
way—that is, of course to lower the screw—a force of 3·4 lbs. must
be applied. Hence, even with the assistance of the load, a force of
3·4 lbs. is necessary to overcome friction. This will enable us to
determine the amount of friction in the same manner as we determined
the friction in the pulley-block (Art. 207). Let X be the force
usefully employed in raising, and Y the force of friction, which acts
equally in either direction against the production of motion; then to
raise the load the power applied must be sufficient to overcome both
X and Y, and therefore we have X + Y = 9·0. When the weight is to be
lowered the force X of course aids in the lowering, but X alone is not
sufficient to overcome the friction; it requires the addition of 3·4
lbs., and we have therefore X + 3·4 = Y, and hence X = 2·8, Y = 6·2.
That is, 2·8 is the amount of force which with a frictionless screw
would have been sufficient to raise half a ton. But in the frictionless
screw the power is found by dividing the load by the velocity ratio. In
this case 1120 ÷ 414 = 2·7, which is within 0·1 lb. of the value of X.
The agreement of these results is satisfactory.