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Difficulty: MediumPascal's Principle and Hydraulics

A mechanic uses a hydraulic floor jack to raise the axle of a truck. The jack's lifting cylinder (output) has a cross-sectional area that is 1212 times larger than the pump cylinder's (input) area. If the mechanic pushes the pump cylinder down a total fluid-displacing distance of 36 inches36 \text{ inches} (accumulated over several strokes), how far upward will the truck's axle be lifted?

  1. 3 inches3 \text{ inches}Answer
  2. B
    432 inches432 \text{ inches}
  3. C
    36 inches36 \text{ inches}
  4. D
    0.33 inches0.33 \text{ inches}

Answer

The truck's axle will be lifted 3 inches3 \text{ inches}.
In any simple machine, including a hydraulic system, work input equals work output (assuming no friction). Because a hydraulic jack multiplies force, it must sacrifice distance to conserve energy. Since the output piston's area is 12 times larger, it exerts 12 times the force but only travels 1/12 of the distance. Therefore, 36 inches36 \text{ inches} of input distance results in 36/12=3 inches36 / 12 = 3 \text{ inches} of output distance.

Step-by-Step Solution

1
Identify the relationship between fluid volume, piston area, and travel distance.
V=A⋅dV = A \cdot d
In a closed hydraulic system, the volume of fluid displaced by the input piston must exactly equal the volume of fluid that moves the output piston.
2
Set up the conservation of volume equation for both cylinders.
Ain⋅din=Aout⋅doutA_{\text{in}} \cdot d_{\text{in}} = A_{\text{out}} \cdot d_{\text{out}}
Equating the input and output volumes allows us to solve for the missing variable.
3
Rearrange the formula to solve for the output distance (doutd_{\text{out}}).
dout=din⋅(Ain/Aout)d_{\text{out}} = d_{\text{in}} \cdot (A_{\text{in}} / A_{\text{out}})
Isolating the target variable makes calculation straightforward.
4
Substitute the known values into the equation.
dout=36⋅(1/12)=3 inchesd_{\text{out}} = 36 \cdot (1 / 12) = 3 \text{ inches}
The output area is 12 times the input area, making the ratio 1/121/12. Multiplying the input distance by this ratio gives the final output distance.

Key Concept

Conservation of fluid volume and mechanical advantage trade-offs in hydraulics.
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Topics

Pascal's Principle and Hydraulics

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