To find the minimum diameter of the piston, we need to relate the force exerted by the hydraulic cylinder to the pressure of the hydraulic fluid and the area of the piston.

Step 1: Understand the relationship between force, pressure, and area

The force exerted by the hydraulic cylinder is given by: $F = P \cdot A$ Where:

• $F = 100{,}000 \, \text{N}$ (force exerted by the cylinder)
• $P = 14{,}000 \, \text{kPa}$ (pressure of the hydraulic fluid, convert this to Pa by multiplying by $10^3$ since $1 \, \text{kPa} = 1{,}000 \, \text{Pa}$)
• $A$ is the area of the piston.

Step 2: Convert pressure to pascals

$P = 14{,}000 \, \text{kPa} = 14{,}000 \times 10^3 \, \text{Pa} = 14{,}000{,}000 \, \text{Pa}$

Step 3: Solve for the area of the piston

$A = \frac{F}{P} = \frac{100{,}000 \, \text{N}}{14{,}000{,}000 \, \text{Pa}} \approx 0.00714 \, \text{m}^2$

Step 4: Relate the area to the piston's diameter

The area of a circle is given by: $A = \frac{\pi d^2}{4}$ Where $d$ is the diameter of the piston. Solving for $d$: $d^2 = \frac{4A}{\pi}$ $d = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4 \times 0.00714 \, \text{m}^2}{\pi}} \approx 0.0953 \, \text{m}$

Step 5: Convert the diameter to millimeters

$d \approx 95.3 \, \text{mm}$

So, the minimum diameter of the piston is approximately 95 mm (considering the answer tolerance of ±1 mm).

Would you like further details or have any questions?

Here are 5 related questions to consider:

1. How does the piston diameter change if the required force is increased?
2. What is the impact of changing the hydraulic fluid pressure on the size of the piston?
3. How does the efficiency of a hydraulic system relate to the size and pressure applied to the piston?
4. What materials would be suitable for constructing a piston that needs to withstand high pressures?
5. How would the design of the hydraulic system change if the piston diameter needed to be minimized for space constraints?

Tip: When designing hydraulic systems, always ensure the piston diameter is sufficient to handle the required force without overloading the system, considering both safety and efficiency.