5.1 Calculate the output force, F₂ of this hydraulic system - NSC Technical Sciences - Question 5 - 2023 - Paper 1

YES. The output force F₂ (25,051 N) is greater than the required input force of 20 kN (20,000 N). Therefore, the output force is sufficient to press the work piece flat.

Stress is defined as the internal restoring force per unit area of a material. It quantifies the intensity of internal forces within a material as it resists deformation.

To calculate the stress experienced by the bar, we use the formula:

$\sigma = \frac{F}{A}$

First, we calculate the cross-sectional area A of the round bar:

$A = \pi r^2 = \pi (0.015 \, \text{m})^2 \approx 7.0686 \times 10^{-4} \, \text{m}^2$

Then, substituting the force F = 4000 N:

$\sigma = \frac{4000 \, \text{N}}{7.0686 \times 10^{-4} \, \text{m}^2} \approx 5.66 \times 10^6 \, \text{Pa}$

Strain (ε) is defined as the change in length divided by the original length. It can be calculated using:

$\varepsilon = \frac{\Delta L}{L_0}$

Where:

• ( \Delta L = 200 , \text{mm} - 188 , \text{mm} = 12 , \text{mm} = 0.012 , \text{m} )
• ( L_0 = 200 , \text{mm} = 0.2 , \text{m} )

Thus,

$\varepsilon = \frac{0.012 \, \text{m}}{0.2 \, \text{m}} = 0.06$

Young's modulus (E) can be calculated using the formula:

$E = \frac{\sigma}{\varepsilon}$

Substituting the values for stress (5.66 × 10^6 Pa) and strain (0.06):

$E = \frac{5.66 \times 10^6 \, \text{Pa}}{0.06} \approx 9.43 \times 10^7 \, \text{Pa}$

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