6.1 Define the following terms: 6.1.1 Stress 6.1.2 Strain 6.2 A steel bar experiences a stress of 250 MPa. The modulus of elasticity is 190 GPa. The bar has a diameter of 60 mm and is 220 mm long. Calculate the: 6.2.1 Strain on the bar. 6.2.2 Force exerted on the bar. 6.3 What is the effect of an increase in temperature on the viscosity of a fluid? 6.4 Define a perfectly plastic body. 6.5 Give TWO examples of perfectly plastic bodies. 6.6 State Pascal's law in words. 6.7 A hydraulic system is used to lift a 20 000 N vehicle in a workshop. If the vehicle sits on a piston of area 0.8 m², and a force is applied to another piston of area 0.05 m², what is the minimum force that must be applied to lift the vehicle? 6.8 Define the thrust of a liquid.

NSC Technical Sciences Viscosity: 6.1 Define the following terms:

6.1 Define the following terms: 6.1.1 Stress 6.1.2 Strain 6.2 A steel bar experiences a stress of 250 MPa - NSC Technical Sciences - Question 6 - 2020 - Paper 1

Question 6

6.1 Define the following terms: 6.1.1 Stress 6.1.2 Strain 6.2 A steel bar experiences a stress of 250 MPa. The modulus of elasticity is 190 GPa. The bar has a dia... show full transcript

Worked Solution & Example Answer:6.1 Define the following terms: 6.1.1 Stress 6.1.2 Strain 6.2 A steel bar experiences a stress of 250 MPa - NSC Technical Sciences - Question 6 - 2020 - Paper 1

Step 1

6.1.1 Stress

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Answer

Stress is defined as the force per unit area applied to a material. It is mathematically expressed as:

σ = \frac{F}{A}

where $σ$ is the stress, $F$ is the applied force, and $A$ is the cross-sectional area.

Step 2

6.1.2 Strain

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Strain is a measure of deformation representing the displacement between particles in a material body. It is often defined as the change in length divided by the original length, expressed as:

ε = \frac{ΔL}{L_0}

where $ε$ is strain, $ΔL$ is the change in length, and $L_0$ is the original length.

Step 3

6.2.1 Strain on the bar

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To calculate strain ( $ε$ ), use the formula:

ε = \frac{σ}{E}

Substituting the given values:

Stress ( $σ$ ) = 250 MPa = $250 \times 10^6$ Pa
Modulus of Elasticity ( $E$ ) = 190 GPa = $190 \times 10^9$ Pa

Thus, the strain is:

ε = \frac{250 \times 10^6}{190 \times 10^9} = 0.00131579 \approx 1.32 \times 10^{-3}

Step 4

6.2.2 Force exerted on the bar

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The force exerted on the bar can be calculated using the formula:

F = σ A

First, calculate the area ( $A$ ) of the bar using the diameter:

A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{60 \, mm}{1000}\right)^2 \approx 2.827 \times 10^{-3} \, m^2

Now, substitute the stress and area:

F = (250 \times 10^6 \, Pa)(2.827 \times 10^{-3} \, m^2) \approx 70750 \, N

Step 5

6.3 What is the effect of an increase in temperature on the viscosity of a fluid?

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Answer

As the temperature of a fluid increases, its viscosity typically decreases. This is due to the increased kinetic energy of the molecules, which allows them to overcome intermolecular forces more easily, resulting in lower resistance to flow.

Step 6

6.4 Define a perfectly plastic body

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Answer

A perfectly plastic body is a type of material that does not return to its original shape after the removal of a load. It deforms plastically when subjected to stress beyond its yield point and continues to deform without any increase in stress.

Step 7

6.5 Give TWO examples of perfectly plastic bodies

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Examples of perfectly plastic bodies include:

Clay
Lead

Step 8

6.6 State Pascal's law in words

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Pascal's law states that in a confined fluid at rest, a change in pressure applied at any point in the fluid is transmitted undiminished throughout the fluid in all directions.

Step 9

6.7 If the vehicle sits on a piston of area 0.8 m², and a force is applied to another piston of area 0.05 m², what is the minimum force that must be applied to lift the vehicle?

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Answer

Using Pascal's principle, we have:

\frac{F_1}{A_1} = \frac{F_2}{A_2}

Where:

$F_1$ = force on the large area (0.8 m²)
$A_1$ = area of the large piston = 0.8 m²
$F_2$ = force on the small area (0.05 m²)
$A_2$ = area of the small piston = 0.05 m²

Rearranging gives:

F_2 = F_1 \frac{A_2}{A_1}

Substituting the known values:

$F_1$ = 20000 N (weight of the vehicle)
$A_1$ = 0.8 m²
$A_2$ = 0.05 m²

Calculate $F_2$ :

F_2 = 20000 \times \frac{0.05}{0.8} = 1250 \, N

Step 10

6.8 Define the thrust of a liquid

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Thrust of a liquid is defined as the force exerted by the liquid on a surface due to its pressure. It is generally the product of the pressure of the liquid and the area of the surface in contact with it.

6.1 Define the following terms: 6.1.1 Stress 6.1.2 Strain 6.2 A steel bar experiences a stress of 250 MPa - NSC Technical Sciences - Question 6 - 2020 - Paper 1

Worked Solution & Example Answer:6.1 Define the following terms: 6.1.1 Stress 6.1.2 Strain 6.2 A steel bar experiences a stress of 250 MPa - NSC Technical Sciences - Question 6 - 2020 - Paper 1

6.1.1 Stress

6.1.2 Strain

6.2.1 Strain on the bar

6.2.2 Force exerted on the bar

6.3 What is the effect of an increase in temperature on the viscosity of a fluid?

6.4 Define a perfectly plastic body

6.5 Give TWO examples of perfectly plastic bodies

6.6 State Pascal's law in words

6.7 If the vehicle sits on a piston of area 0.8 m², and a force is applied to another piston of area 0.05 m², what is the minimum force that must be applied to lift the vehicle?

6.8 Define the thrust of a liquid

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