Question Bank • QB_FP_169 NAT • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: I.C. engine systems – fuel, cooling, lubrication, ignition, electrical, intake and exhaust

Question Statement:

A tractor hydraulic lift cylinder has a bore diameter of $80\text{ mm}$. The hydraulic system relief valve is set at $16\text{ MPa}$. Neglecting friction and seal drag, calculate the maximum lifting force exerted by the piston in $\text{kN}$ (round off to 2 decimal places).

Answer Key:

80.42

Step-by-Step Solution & Explanation

Piston cross-sectional area: $A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.080\text{ m})^2 = 5.0265 \times 10^{-3}\text{ m}^2$ Maximum thrust force: $F = P \times A = 16 \times 10^6\text{ Pa} \times 5.0265 \times 10^{-3}\text{ m}^2 = 80424.8\text{ N} \approx 80.42\text{ kN}$ Acceptable range: $80.0 - 81.0\text{ kN}$.
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