What effort will a machine of efficiency 90% apply to a load of 180N if its efforts arm is twice as long as its load arm?
Given:
- Load \( W = 180 \, \text{N} \)
- Efficiency \( \eta = 90\% = 0.9 \)
- Let the effort \( E \) be applied.
The efficiency of a machine is given by the formula:
\(\eta = \frac{\text{Mechanical Advantage (MA)}}{\text{Velocity Ratio (VR)}}\)
Mechanical Advantage (MA) is defined as the ratio of load to effort:
\(\text{MA} = \frac{W}{E}\)
Determine the Velocity Ratio (VR)
Given that the effort arm is twice the load arm, we can express the Velocity Ratio (VR) as:
\(\text{VR} = \frac{\text{Length of effort arm}}{\text{Length of load arm}} = \frac{2L}{L} = 2\)
Substitute into the efficiency formula
Substituting the expressions for MA and VR into the efficiency formula gives:
\(0.9 = \frac{\frac{W}{E}}{2}\)
Rearrange to find the effort \( E \)
Multiplying both sides by \( 2E \):
\(1.8E = W\)
Substituting \( W = 180 \, \text{N} \):
\(
1.8E = 180\)
Solve for \( E \)
Now, dividing by 1.8:
\(E = \frac{180}{1.8} = 100 \, \text{N}\)
Therefore, the effort applied by the machine is: \(\boxed{100 \, \text{N}}\).
There is an explanation video available below.
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