Question

Efficiency of a simple lifting machine is:

• A. $\frac{\text{Velocity ratio}}{\text{Mechanical advantage}}$
• B. $\frac{\text{Mechanical advantage}}{\text{Velocity ratio}}$
• C. $\text{Mechanical advantage} \times \text{Velocity ratio}$
• D. $\sqrt{\text{Mechanical advantage} \times \text{Velocity ratio}}$

Hint:

Remember: Efficiency = MA over VR. Since $MA$ represents the actual "gain" we get and $VR$ represents the theoretical "potential" based on geometry, efficiency tells us how much of that potential is actually realized.

Answer 1

The Correct Option is B

The performance of any lifting machine (like a pulley system, lever, or screw jack) is evaluated through its efficiency, which relates the work output to the work input. 1. Basic Definition of Efficiency: Efficiency ($\eta$) is defined as the ratio of useful work done by the machine to the total work put into the machine. $$\eta = \frac{\text{Output Work}}{\text{Input Work}}$$

2. Deriving the Relationship: Let $W$ be the load lifted, $P$ be the effort applied, $d_w$ be the distance the load moves, and $d_p$ be the distance the effort moves.

• Mechanical Advantage (MA): This is the ratio of load to effort: $MA = \frac{W}{P}$.

• Velocity Ratio (VR): This is the ratio of distance moved by effort to distance moved by load: $VR = \frac{d_p}{d_w}$.
Output Work = $W \times d_w$
Input Work = $P \times d_p$ $$\eta = \frac{W \times d_w}{P \times d_p} = \left(\frac{W}{P}\right) \times \left(\frac{d_w}{d_p}\right)$$ $$\eta = MA \times \frac{1}{VR}$$ $$\eta = \frac{MA}{VR}$$

3. Conclusion: The efficiency of a machine is always the ratio of its Mechanical Advantage to its Velocity Ratio. In an "Ideal Machine," friction is zero, so $MA = VR$ and efficiency is $100\%$. In real-world machines, $MA$ is always less than $VR$ due to friction, meaning efficiency is always less than $1$ (or $100\%$).