A hydraulic lift has piston areas 4.0 cm² and 240 cm². The input force is 180 N and the output platform weighs 1200 N. (a) Calculate transmitted pressure. (b) Calculate output force and maximum added load in equilibrium. (c) Explain the output displacement. (d) An off-centre load rests on two supports. State the complete equilibrium conditions and explain how support forces are found.

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Pressure 4.5 × 10⁵ Pa; output force 1.08 × 10⁵ N; added load 1.068 × 10⁵ N; volume displacement is conserved and both force and moment sums vanish. Convert input area: 4.0 cm² = 4.0 × 10⁻⁴ m². Pressure p = F/A = 180/(4.0 × 10⁻⁴) = 4.5 × 10⁵ Pa. Output area is 0.0240 m², so force = pA = 4.5 × 10⁵ × 0.0240 = 1.08 × 10⁵ N. Added load = 108000 − 1200 = 106800 N. For an incompressible liquid, displaced volumes approximately match: Aᵢdᵢ = Aₒdₒ. The sixty-times-larger output area therefore moves one-sixtieth of the input distance, preserving the work trade-off. Static equilibrium requires zero vector resultant force and zero resultant moment about any point. Either condition alone is insufficient. The two upward support forces sum to total weight. Taking moments about one support removes its unknown force and determines the other; an off-centre load generally gives unequal support forces.

Hydraulic lift and rotational balance

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