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.
| Number of Experiment. | R. Load in lbs. | Observed power in lbs. | P. Calculated power in lbs. | Difference of the observed and calculated 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.
THE SCREW BOLT AND NUT.
Fig. 45.