A hydraulic press has an input cylinder of cross-sectional area 5.0 cm² and an output piston of area 200 cm². If a force of 40 N is exerted on the smaller piston, what maximum thrust force is generated at the larger piston?
Answer and explanation
Correct answer: B. 1,600 N
- Step 1: State Pascal's principle: F1 / A1 = F2 / A2. Step 2. Substitute values: 40 / 5.0 = F2 / 200 => 8.0 = F2 / 200. Step 3. Solve: F2 = 8.0 × 200 = 1,600 N. Answer: 1,600 N. Why not the others: 1,200 N uses area ratio 30; 800 N halves the output force; 2,000 N assumes input force 50 N.
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