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Difficulty: MediumSimple Machines

An inclined plane is set at an angle of 30∘30^\circ to the horizontal. An effort of 320 N320\text{ N} applied parallel to the plane is used to push a load of 480 N480\text{ N} up the incline at a constant speed. What is the efficiency of the inclined plane expressed as a percentage?

Answer: 75% / 75 / 75 percent

Answer

The efficiency of the inclined plane is 75%75\%.
For an inclined plane inclined at 30∘30^\circ to the horizontal, the velocity ratio (VR) is given by VR=1sin⁡30∘=2\text{VR} = \frac{1}{\sin 30^\circ} = 2. The mechanical advantage (MA) is MA=LoadEffort=480 N320 N=1.5\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{480\text{ N}}{320\text{ N}} = 1.5. Dividing MA by VR and multiplying by 100%100\% yields an efficiency of (1.52)×100%=75%\left(\frac{1.5}{2}\right) \times 100\% = 75\%.

Step-by-Step Solution

1
Determine the Velocity Ratio (VR) of the inclined plane from its angle of inclination
VR = \frac{1}{\sin 30^\circ} = \frac{1}{0.5} = 2
For an inclined plane with inclination angle \theta, the velocity ratio is equal to \frac{1}{\sin \theta}.
2
Calculate the Mechanical Advantage (MA)
MA = \frac{\text{Load}}{\text{Effort}} = \frac{480\text{ N}}{320\text{ N}} = 1.5
Mechanical advantage is defined as the ratio of load to effort force.
3
Calculate the efficiency of the machine
\text{Efficiency} = \frac{\text{MA}}{\text{VR}} \times 100\% = \frac{1.5}{2} \times 100\% = 75\%
Efficiency is the ratio of mechanical advantage to velocity ratio expressed as a percentage.

Key Concept

Efficiency of an Inclined Plane
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Topics

Simple Machines

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