Answer:
Page 21
IN NATURAL PH2LOSOPf7l
between them from a pole io feet in length. Whe e should it
be suspended so that one will lift only 50 lbs.?
One lifts 50 lbs.; the other 200 lbs. The proportionate length of the arms
of the lever should be the same as the proportionate weights-i. e., 1 to 4.
10 + 5 = 2, the unit of measure. lIence one arm is 2 feet long and the other
S feet long. PRooF.-(See Prob. 7, p. 10.) 50 x 8 = 200 x 2. This is the
substance also of the equation P x Pd = XV x Wd.
5. In a lever of the first class, 6 feet long, where should the
F be fplaced so that a P of I lb. will balance a Woof 23 lbs.?
6 feet = 72 inches. 72 + 24 = 3, the unit of distance. The W must be
placed 3 in. and the P 69 in. from the F. PRooF. 23 x 3 = 1 x 69 (Prob. 4).
6. What P would be required to lift a barrel of fiork with a
windlass whose axle is one foot in diameter and handle 3 ft.
gong.
P: W: rad. of axle:: rad. of wheel.
x: 200 lbs:: 1/2 ft.: 3 ft.: = 331/s lbs.
7. What sized axle, with a wheel 6 feet in diameter, would
be required to balance a W of I ton by a P of Ioo lbs F
P: W:: rad. of axle: rad. of wheel.
100 lbs.: 2000 lbs.:: x: 3 ft.
x = /20o ft. = the rad.; hence the diameter = 3/10 ft.
8. WHhat number of movable pzulleys would be required to;ift a W of 200 lbs. with a P of 25 lbs.?
W= P x twice the no. of mov. pulleys; hence p-= twice the no. of mov. pul's.
200 + 25 = 8. 8 + 2 = 4 = the no. required.
9. How many lbs. could be lifted with a system of 4 zmovable
pulleys, and one fixed pulley to change the direction of the
force, by a P of oo lbs.?
W= P x twice the no. of mov. pulleys.
100 lbs. x (4 x 2) = 800 lbs. = the W.
io. What weight could be lifted with a single horse-15owe,
(33,000 lbs.) acting on the tackle-block? (Fig. 62.)
This block has 3 movable pulleys, and using the equation of the pulleys given
in the last two problems, we have, making no allowance for friction,
33,000 lbs. x (3 x 2) = 198,000 lbs.
Explanation:
