Law of the Machine
For an ideal machine
P=a W
where: | |||
symbol |
description |
type |
units |
P | effort | vector | N=newton=kg*m/s2 |
W | load | vector | N=newton=kg*m/s2 |
a | slope | scalar | none |
For a machine with friction
P=a W + b
where: | |||
symbol |
description |
type |
units |
P | effort | vector | N=newton=kg*m/s2 |
W | load | vector | N=newton=kg*m/s2 |
a | slope | scalar | none |
b | static friction | vector | N=newton=kg*m/s2 |
Limiting Mechanical Advantage and Efficiency
M.A. = W/P
but P=a W + b so,
M.A. = W/(a W + b)
M.A. = 1/(a + b/W)
As W is increased b/W becomes smaller. At high values of W, b/W becomes practically zero. The M.A. is said to reach a limiting value when W is infinitely large.
Limiting M.A. = 1/a
Limiting efficency = Limiting M.A./VR = 1/(a * VR)
This is the limiting efficiency which can be attained by a machine
Example
A simple lifting machine gave the following results on test:
Load Raised (N) | 30.2 | 93.4 | 155.8 | 218 | 311.8 | 438 |
Effort Applied (N) | 8.45 | 16 | 25.1 | 32.2 | 46.7 | 63.7 |
Plot these values and find the law of the machine.
Draw the efficiency on a load base. What is the limiting efficiency? VR = 15.75
- efficiency = 30.2/(8.45*15.75) = 0.246 at 30.2N (load)
- efficiency = 93.4/(16*15.75) = 0.37 at 93.4N (load)
- efficiency = 155.8/(25.1*15.75) = 0.393 at 155.8N (load)
- efficiency = 218/(32.2*15.75) = 0.431 at 218N (load)
- efficiency = 311.8/(46.7*15.75) = 0.424 at 311.8N (load)
- efficiency = 438/(63.7*15.75) = 0.437 at 438N (load)