Question Bank • QB_SUB_FP_HYD_002 NAT • 2 Marks

Section 3: Farm Power

Tractors and Power Tillers

Subtopic: Steering and hydraulic control systems used in tractors

Question Statement:

A tractor hydraulic cylinder has a bore diameter of $70\text{ mm}$. If the relief valve limits system pressure to $18\text{ MPa}$, the maximum thrust force developed by the cylinder (in kN, rounded to 1 decimal place) is:

Answer Key:

69.3

Step-by-Step Solution & Explanation

Piston area: $A = \frac{\pi}{4} (0.070)^2 = 3.848 \times 10^{-3}\text{ m}^2$ Force: $F = P \times A = (18 \times 10^6) \times (3.848 \times 10^{-3}) = 69272\text{ N} = 69.3\text{ kN}$
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