Note carefully in all problems on the wheel and axle that more force is required the faster the weight is lifted. Moreover, if the axle is made smaller, the weight will be lifted more slowly and less force will be required.
1. Figure 24 shows a common winch or hoist which is a good illustration of the wheel and axle; the crank is the wheel and the
Fig. 23. - Wheel and Axle.
6-in. drum is the axle. If a boy turns the handle P uniformly with a force of 50 lbs., what weight can he lift at W?
2. Suppose in problem 1, 15% were lost in friction, what would be the answer to the problem?
3. If 26 ft. 8 in. of rope were wound up on the drum in Fig. 24, how many turns and parts of turns did the crank P make? (Take z = 22/7.))
4. In Fig. 24, what is the ratio between the weight lifted and the force applied?
5. A wheel and axle has the wheel 24 in. in diameter and the axle 12 in. in diameter. If 10 ft. of rope are wound up on the wheel how many feet will be unwound on the axle?
Note. - To do this problem it is necessary only to consider the circumferences of the wheel and the axle. One turn of the wheel will wind up 3.1416 X 24 in. of rope and at the same time unwind 3.1416 X 12 in. of rope from the axle. This is the same as saying that the lengths of cord wound and unwound are proportional to the circumferences of the wheel and axle. But we already know that the circumferences of circles are proportional to their diameters and so we can say that the lengths of rope wound and unwound are proportional to the diameters of the wheel and axle and in the above problem we will have,
24 : 12 = 10 : rope unwound from axle.
or 12 X 10 /24 = 5 ft. rope unwound from axle.
A simple rule for this would read: To find the length of rope unwound from the axle multiply the length of rope wound on the wheel by the diameter of the axle and divide this by the diameter of the wheel.
If we wanted to find the length of rope wound up on the wheel the rule would read: To find the length of rope wound on the wheel multiply the length of rope unwound from the axle by the diameter of the wheel and divide by the diameter of the axle. Or in the above problem,
Fig. 24. - Hoist.
24 : 12 = x : 5
24 X 5 / 12 = 10 ft.
In the derrick (Fig. 25), the hoisting mechanism is a form of double wheel and axle in which the axle of the first works upon the wheel of the second by means of gears. It is used for raising heavy weights.