Question

Difficulty: HardInclined Planes

Rank the following inclined plane configurations in order from the LEAST effort force required to the GREATEST effort force required. Assume ideal conditions with no friction.

  1. 1A 500 lb500\text{ lb} cargo loaded using a 15 ft15\text{ ft} ramp to a height of 3 ft3\text{ ft}.
  2. 2A 600 lb600\text{ lb} cargo loaded using a 12 ft12\text{ ft} ramp to a height of 3 ft3\text{ ft}.
  3. 3A 400 lb400\text{ lb} cargo loaded using an 8 ft8\text{ ft} ramp to a height of 4 ft4\text{ ft}.
  4. 4A 750 lb750\text{ lb} cargo loaded using a 15 ft15\text{ ft} ramp to a height of 5 ft5\text{ ft}.

Answer

The correct sequence from least to greatest effort force is: the 500 lb load on a 15 ft ramp (100 lbs), followed by the 600 lb load on a 12 ft ramp (150 lbs), then the 400 lb load on an 8 ft ramp (200 lbs), and finally the 750 lb load on a 15 ft ramp (250 lbs).
The effort force required for an inclined plane is calculated by dividing the weight of the load by the Ideal Mechanical Advantage (IMA = Length / Height). Calculating this for each scenario yields: 100 lbs, 150 lbs, 200 lbs, and 250 lbs respectively. Organizing these from least to greatest provides the correct sequence.

Step-by-Step Solution

1
Calculate the effort force for the 750 lb750\text{ lb} cargo.
Force = 750×(5/15)=750×(1/3)=250 lbs750 \times (5 / 15) = 750 \times (1 / 3) = 250\text{ lbs}.
To determine the actual force needed, divide the load by the Ideal Mechanical Advantage (IMA = Length / Height).
2
Calculate the effort force for the 400 lb400\text{ lb} cargo.
Force = 400×(4/8)=400×0.5=200 lbs400 \times (4 / 8) = 400 \times 0.5 = 200\text{ lbs}.
The ramp's IMA is 2, meaning it halves the required lifting force.
3
Calculate the effort force for the 600 lb600\text{ lb} cargo.
Force = 600×(3/12)=600×0.25=150 lbs600 \times (3 / 12) = 600 \times 0.25 = 150\text{ lbs}.
The ramp has an IMA of 4, which divides the 600 lb600\text{ lb} load by 4.
4
Calculate the effort force for the 500 lb500\text{ lb} cargo.
Force = 500×(3/15)=500×0.2=100 lbs500 \times (3 / 15) = 500 \times 0.2 = 100\text{ lbs}.
This configuration provides an IMA of 5, which significantly reduces the force.
5
Order the calculated effort forces from least to greatest.
The correct order is 100 lbs100\text{ lbs}, 150 lbs150\text{ lbs}, 200 lbs200\text{ lbs}, 250 lbs250\text{ lbs}.
The question specifically requests the sequence from least effort to greatest effort.

Key Concept

Ideal Mechanical Advantage and Effort Force in Inclined Planes
Estimated Time:2m 0s
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