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Difficulty: HardPascal's Principle and Hydraulics

A mechanic uses a hydraulic jack to lift a 2,000-pound engine. The output piston supporting the engine has a diameter that is exactly 4 times larger than the diameter of the input piston. Assuming an ideal system, if the mechanic pushes the input piston down a total distance of 16 inches, what is the required input force, and how far will the engine be lifted?

  1. 125 pounds of force; the engine is lifted 1 inchAnswer
  2. B
    500 pounds of force; the engine is lifted 4 inches
  3. C
    125 pounds of force; the engine is lifted 256 inches
  4. D
    32,000 pounds of force; the engine is lifted 1 inch

Answer

125 pounds of force; the engine is lifted 1 inch
In a hydraulic system, the mechanical advantage is determined by the ratio of the piston areas. Since the output piston's diameter is 4 times the input piston's diameter, its area is 424^2, or 16 times larger. This provides a mechanical advantage of 16. To lift a 2,000-pound engine, the required input force is 2,000÷16=1252,000 \div 16 = 125 pounds. Furthermore, because work is conserved (Fin×din=Fout×doutF_{in} \times d_{in} = F_{out} \times d_{out}), the input piston must move 16 times farther than the output piston. Therefore, pushing the input piston down 16 inches will lift the engine 16÷16=116 \div 16 = 1 inch.

Step-by-Step Solution

1
Determine the area ratio between the two pistons.
Area Ratio = 42=164^2 = 16.
In hydraulics, mechanical advantage depends on the cross-sectional area of the pistons. Because area scales with the square of the diameter (A=πr2A = \pi r^2), a piston with 4 times the diameter has 424^2, or 16 times, the area.
2
Calculate the required input force using Pascal's Principle.
Input Force = 2,000÷16=1252,000 \div 16 = 125 pounds.
The hydraulic system provides a mechanical advantage of 16. Therefore, the input force required is 1/161/16 th of the output load.
3
Calculate the distance the output piston (and engine) will move.
Output Distance = 16 inches÷16=1 inch16 \text{ inches} \div 16 = 1 \text{ inch}.
By the principle of conservation of work (Work In = Work Out), if force is multiplied by 16, the distance traveled must be divided by 16 to keep the total volume of fluid displaced equal.

Key Concept

Pascal's Principle and Mechanical Advantage

Alternative Method

You can verify your answers using the principle of conservation of energy (Work In = Work Out). The work done by the input piston is 125 lbs×16 inches=2,000 in-lbs125 \text{ lbs} \times 16 \text{ inches} = 2,000 \text{ in-lbs}. The work done on the engine is 2,000 lbs×1 inch=2,000 in-lbs2,000 \text{ lbs} \times 1 \text{ inch} = 2,000 \text{ in-lbs}. Since the work values match, the force and distance calculations are correct.
Estimated Time:1m 30s
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Topics

Pascal's Principle and Hydraulics

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