Machinery's Handbook, 31st Edition
Simple Mechanisms
177
Wheels and Pulleys
: : FW r R F R W r # # # F R W r W r F R R F W r r W F R # # # = = = = = =
The radius of a drum on which is wound the lifting rope of a windlass is 2 in. What force will be exerted at the periphery of a gear of 24 in. diameter, mounted on the same shaft as the drum and transmitting power to it, if one ton (2000 lbs) is to be lifted? Here W = 2000; R = 12; r = 2. F 12 2000 2 333 lbs # = =
r
R
F
W
A
B
D
C
R 2
R 1
R
Let the pitch diameters of gears A , B , C and D be 30, 28, 12 and 10 in., respectively. Then R 2 = 15; R 1 = 14; r 1 = 6; and r = 5. Let R = 12, and r 2 = 4. The force F required to lift a weight W of 2000 lbs, friction being neglected, is: F 12 14 15 2000 5 6 4 95 pounds # # # # # = =
r 2
r 1
r
W
F
A , B , C and D are the pitch circles of gears. F R R R W r r r W r r r F R R R 1 2 1 2 1 2 1 2 # # # # # # # # # # = =
F W 2 1 =
W 1 2
2 α
: FW F W W F 2 = = =
: sec sec cos 2 # # α
The velocity with which weight W will be raised is one-half the velocity of the force applied at F .
W 1 2
α
F
F
W
W
n = number of strands or parts of rope ( n 1 , n 2 , etc.). F n W 1 # = The velocity with which W will be raised equals n 1 of the velocity of the force applied at F .
In the illustration is shown a combination of a double and triple block. The pulleys each turn freely on a pin as axis, and are drawn with different diameters, to show the parts of the rope more clearly. There are 5 parts of rope. Therefore, if 200 lbs is to be lifted, the force F required at the end of the rope is: F 5 1 200 40lbs # = =
n 1 n 3 n 5
F
n 2 n 4
W
Note: The above formulas are valid using metric SI units, with forces expressed in newtons, and lengths in meters or millimeters. (See note on page 174 concerning weight and mass.)
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