Subtopic: Steering and hydraulic control systems used in tractors
Question Statement:
A single-acting hydraulic lift ram cylinder of a tractor three-point hitch has a piston diameter of $D = 80\text{ mm}$. The hydraulic system relief valve pressure is set to $P = 17.5\text{ MPa}$. Taking $\pi \approx 3.14159$, the maximum hydraulic thrust force that can be exerted by the lift ram in $\text{kN}$ is ________ (round off to 2 decimal places).
Answer Key:
87.96
Step-by-Step Solution & Explanation
```
Hydraulic Ram Lift Force:
Piston Diameter D = 80 mm = 0.080 m
Pressure P = 17.5 MPa = 17.5 * 10^6 N/m^2
Force F = P * (pi * D^2 / 4)
```
**Step-by-step Calculation:**
1. Piston cross-sectional area ($A$):
$A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (0.080\text{ m})^2 = 0.0050265\text{ m}^2$
2. Maximum thrust force ($F$):
$F = P \times A = (17.5 \times 10^6\text{ N/m}^2) \times 0.0050265\text{ m}^2 = 87964.6\text{ N} = 87.96\text{ kN}$