Sheaves cast iron 2"·5 diameter; plaited rope
0"·25 diameter; velocity ratio 6; mechanical
advantage 4; useful effect 67 per cent.; formula
_P_ = 2·36 + 0·238 _R_.
+-----------+---------+----------+----------+------------------+
| | | | P. | Discrepancies |
| Number of | R. | Observed |Calculated| between observed |
|Experiment.| Load | power | power | and calculated |
| | in lbs. | in lbs. | in lbs. | powers. |
+-----------+---------+----------+----------+------------------+
| 1 | 57 | 15·5 | 15·9 | + 0·4 |
| 2 | 114 | 29·5 | 29·5 | 0·0 |
| 3 | 171 | 43·5 | 43·1 | - 0·4 |
| 4 | 228 | 56·0 | 56·6 | + 0·6 |
| 5 | 281 | 70·0 | 69·2 | - 0·8 |
| 6 | 338 | 83·0 | 82·8 | - 0·2 |
| 7 | 395 | 97·0 | 96·4 | - 0·6 |
| 8 | 452 | 109·0 | 109·9 | + 0·9 |
+-----------+---------+----------+----------+------------------+
204. This table contains five columns; the weights raised (shown in the
second column) range up to somewhat over 4 cwt. The observed values of
the power are given in the third column; each of these is generally
about one-quarter of the corresponding value of the load. There is,
however, a more accurate rule for finding the power; it is as follows.
205. To find the power necessary to raise a given load, multiply the
loads in lbs. by 0·238, and add 2·36 lbs. to the product. We may
express the rule by the formula _P_ = 2·36 + 0·238 _R_.
206. To find the power which would raise 228 lbs.; the product of
228 and 0·238 is 54·26; adding 2·36, we find 56·6 lbs. for the power
required; the actual observed power is 56 lbs., so that the rule is
accurate to within about half a pound. In the fourth column will be
found the values of _P_ calculated by means of this rule. In the fifth
column, the discrepancies between the observed and the calculated
values of the powers are given, and it will be seen that the difference
in no case reaches 1 lb. Of course it will be understood that this
formula is only reliable for loads which lie between those employed
in the first and last of the experiments. We can calculate the power
for any load between 57 lbs. and 452 lbs., but for loads much larger
than 452 or less than 57 it would probably be better to use the simple
fourth of the load rather than the power computed by the formula.
207. I will next perform an experiment with the three-sheave
pulley-block, which will give an insight into the exact amount of
friction without calculation by the help of the velocity ratio. We
first counterpoise the weight of the lower block by attaching weights
to the power. It is found that about 1·6 lbs. is sufficient for this
purpose. I attach a 56 lb. weight as a load, and find that 13·1 lbs.
is sufficient power for motion. This amount is partly composed of the
force necessary to raise the load if there were no friction, and the
rest is due to the friction. I next gradually remove the power weights:
when I have taken off a pound, you see the power and the load balance
each other; but when I have reduced the power so low as 5·5 lbs. (not
including the counterpoise for the lower block), the load is just able
to overhaul the power, and run down. We have therefore proved that a
power of 13·1 lbs. or greater raises 56 lbs., that any power between
13·1 lbs. and 5·5 lbs. balances 56 lbs., and that any power less than
5·5 lbs. is raised by 56 lbs.
When the power is raised, the force of friction, together with the
power, must be overcome by the load. Let us call _X_ the real power
that would be necessary to balance 56 lbs. in a perfectly frictionless
machine, and _Y_ the force of friction. We shall be able to determine
_X_ and _Y_ by the experiments just performed. When the load is
raised a power equal to _X_ + _Y_ must be applied, and therefore
_X_ + _Y_ = 13·1. On the other hand, when the power is raised, the
force _X_ is just sufficient to overcome both the friction _Y_ and the
weight 5·5; therefore _X_ = _Y_ + 5·5.
Solving this pair of equations, we find that _X_ = 9·3 and _Y_ = 3·8.
Hence we infer that the power in the frictionless machine would
be 9·3; but this is exactly what would have been deduced from the
velocity ratio, for 56 ÷ 6 = 9·3 lbs. In this result we find a perfect
accordance between theory and experiment.