14-4 Pascal’s Principle

14-4 Pascal’s Principle#

Prompts

  • State Pascal’s principle. Why must the fluid be incompressible?

  • How does a hydraulic lift amplify force? If the output piston has 10× the area of the input piston, how does the output force compare to the input force?

  • What is the trade-off? If you get a larger output force, what happens to the displacement? Relate to conservation of work.

  • Give everyday examples of Pascal’s principle (toothpaste tube, hydraulic brakes, Heimlich maneuver).

  • In a hydraulic lift, the input piston moves down 5 cm. If \(A_o/A_i = 4\), how far does the output piston move up?

Lecture Notes#

Overview#

  • Pascal’s principle: Law of instant and equal pressure: push one, push all.

  • Application: Hydraulic lifts, brakes, jacks—small input force can produce large output force.

  • Trade-off: Force amplification comes at the cost of displacement; work (energy) is conserved.

Pascal’s principle#

Pascal’s principle

A change in pressure applied to an enclosed incompressible fluid is transmitted undiminished to every part of the fluid and to the container walls. The extra pressure \(\Delta p_{\text{ext}}\) appears equally at every point in the liquid:

(20)#\[ \Delta p = \Delta p_{\text{ext}}, \]

independent of depth.

Why incompressible? An incompressible fluid cannot change volume, so a pressure change must be felt everywhere at once—otherwise the fluid would flow or deform. A compressible fluid can compress locally, so the pressure change is absorbed near the piston instead of being transmitted uniformly. Liquids are nearly incompressible, so Pascal’s principle applies well to them.

Examples:

  • Car brakes: Small pedal piston → brake fluid → larger wheel pistons; multiplies foot force to stop the car.

  • Heimlich maneuver: Compress abdomen → pressure transmitted to lungs/throat → pops out obstruction.

  • Toothpaste tube: Squeeze bottom → pressure transmitted throughout → paste exits from top.

Hydraulic lever (hydraulic lift)#

A hydraulic lever has two pistons of different areas \(A_i\) (input) and \(A_o\) (output), connected by an incompressible liquid. Same pressure \(p\) throughout:

(21)#\[ \frac{F_i}{A_i} = p = \frac{F_o}{A_o} \]

Force amplification:

(22)#\[ \frac{F_o}{F_i} = \frac{A_o}{A_i} \]

  • Small input force \(F_i\) on small area → large output force \(F_o\) on large area

  • The ratio of forces equals the ratio of areas

Displacement trade-off: The volume \(\Delta V\) of liquid displaced by the input piston equals that received by the output piston:

(23)#\[ A_i d_i = \Delta V = A_o d_o \]

  • \(d_i\): input piston displacement (down)

  • \(d_o\): output piston displacement (up)

So

(24)#\[ \frac{d_o}{d_i} = \frac{A_i}{A_o} \]

The output piston moves less when \(A_o > A_i\). You push the small piston through a large distance to lift the load a small distance.

Work conserved (No free lunch)

Same \(p\) throughout means same energy per unit volume. Work input = work output:

\[W_i = F_i d_i = p \cdot \Delta V = F_o d_o = W_o\]

You save effort but not work: the hydraulic system trades force for distance—small input force over large displacement produces large output force over small displacement.

Poll: Hydraulic lift

A hydraulic lift supports a car of weight 10 kN. The output piston has radius 10× that of the input piston. What force must you apply to the input piston to hold the car in equilibrium?

  • (A) 10 N

  • (B) 100 N

  • (C) 1000 N

  • (D) 10 kN

  • (E) None of these

Summary#

  • Pascal’s principle: Pressure change in enclosed incompressible fluid transmitted undiminished everywhere

  • Hydraulic force: \(F_o/F_i = A_o/A_i\)

  • Hydraulic displacement: \(A_i d_i = A_o d_o\); \(d_o/d_i = A_i/A_o\)

  • Work: \(F_i d_i = F_o d_o\) (no free lunch)