Fig. 5.1 shows a hydraulic car jack. The small piston has an area of 5.0 × 10⁻⁴ m² and the large piston has an area of 2.0 × 10⁻² m². A force of 50 N is app...

(a) Principle of the hydraulic jack [1]

Pressure applied to an enclosed liquid is transmitted equally in all directions (Pascal's principle). [1] This means the pressure at the small piston is the same as the pressure at the large piston, which is how a small input force can produce a large output force.

(b) Pressure on the small piston [2]

\(p = \dfrac{F}{A} = \dfrac{50}{5.0 \times 10^{-4}}\) [1]

\(p = \mathbf{100\,000 \text{ Pa} \;(1.0 \times 10^5 \text{ Pa})}\) [1]

(c) Force on the large piston [2]

The pressure throughout the oil is the same, so:

\(F = p \times A = 100\,000 \times 2.0 \times 10^{-2}\) [1]

\(F = \mathbf{2000 \text{ N}}\) [1]

The hydraulic jack provides a force multiplier of \(\dfrac{2000}{50} = 40\). The ratio of the output force to input force equals the ratio of the piston areas: \(\dfrac{A_2}{A_1} = \dfrac{2.0 \times 10^{-2}}{5.0 \times 10^{-4}} = 40\).

(d) Distance the large piston moves [2]

The liquid is incompressible, so the volume displaced by the small piston equals the volume received by the large piston:

\(A_1 d_1 = A_2 d_2\)

\(d_2 = \dfrac{A_1 \times d_1}{A_2} = \dfrac{5.0 \times 10^{-4} \times 0.16}{2.0 \times 10^{-2}}\) [1]

\(d_2 = \dfrac{8.0 \times 10^{-5}}{2.0 \times 10^{-2}} = \mathbf{0.004 \text{ m}}\) (4 mm) [1]

The large piston moves a much smaller distance than the small piston. Energy is conserved: a large force over a small distance equals a small force over a large distance (ignoring friction).