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

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5.1 Calculate the output force, F₂ of this hydraulic system. 5.1.2 Will the output force be sufficient to press the work piece flat? Write only YES or NO. Explain t... show full transcript

Worked Solution & Example Answer:5.1 Calculate the output force, F₂ of this hydraulic system - NSC Technical Sciences - Question 5 - 2023 - Paper 1

Step 1

5.1 Calculate the output force, F₂ of this hydraulic system.

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Answer

To determine the output force, we can use Pascal's principle, which states that pressure applied to an enclosed fluid is transmitted undiminished throughout the fluid.

First, we calculate the pressure in piston A:

P=F1A1=1000N1.96×103m2=510204.08PaP = \frac{F_1}{A_1} = \frac{1000 \, \text{N}}{1.96 \times 10^{-3} \, \text{m}^2} = 510204.08 \, \text{Pa}

Then, we define the output force F₂ using the area of piston B:

F2=PA2=510204.08Pa4.91×102m2=25,051.02NF_2 = P \cdot A_2 = 510204.08 \, \text{Pa} \cdot 4.91 \times 10^{-2} \, \text{m}^2 = 25,051.02 \, \text{N}

Thus, the output force F₂ is approximately 25,051 N.

Step 2

5.1.2 Will the output force be sufficient to press the work piece flat? Write only YES or NO. Explain the answer.

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Answer

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.

Step 3

5.2.1 Define the term stress.

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Answer

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.

Step 4

5.2.2 Stress experienced by the bar.

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Answer

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

σ=FA\sigma = \frac{F}{A}

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

A=πr2=π(0.015m)27.0686×104m2A = \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:

σ=4000N7.0686×104m25.66×106Pa\sigma = \frac{4000 \, \text{N}}{7.0686 \times 10^{-4} \, \text{m}^2} \approx 5.66 \times 10^6 \, \text{Pa}

Step 5

5.2.3 Strain in the bar.

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Answer

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

ε=ΔLL0\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,

ε=0.012m0.2m=0.06\varepsilon = \frac{0.012 \, \text{m}}{0.2 \, \text{m}} = 0.06

Step 6

5.2.4 Young's modulus of elasticity.

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Answer

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

E=σεE = \frac{\sigma}{\varepsilon}

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

E=5.66×106Pa0.069.43×107PaE = \frac{5.66 \times 10^6 \, \text{Pa}}{0.06} \approx 9.43 \times 10^7 \, \text{Pa}

Step 7

5.3 Write down ONE disadvantage of using monograde oil in modern cars.

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Answer

One disadvantage of using monograde oil is that it only performs effectively within a narrow temperature range. This limits its ability to function optimally in varying weather conditions, making it unsuitable for modern cars which require more versatile oil options.

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