A force, F1, of 200 N is applied on a small piston of a hydraulic system with a diameter of 5.046 × 10^-2 m. The area of a large piston, F2, is 5.25 m², as shown in the diagram below. 5.1 State Pascal's law in words. 5.2 Calculate the force, F2, on the large piston. 5.3 The distance between the two pistons is decreased by using a shorter pipe. How will it affect the answer to QUESTION 5.2? Write only INCREASE, DECREASE or REMAIN THE SAME. 5.4 Write down TWO applications of hydraulic systems. 5.5 The diagram below shows a steel rod of length 1 m and a diameter of 0.02 m. A force, F, stretches the rod and produces a strain of 0.16 × 10^-2. Young's modulus of steel is 2 × 10¹¹ Pa. Calculate the 5.5.1 Stress in the rod 5.5.2 Force, F, applied on the rod.

NSC Technical Sciences Hydraulics: A force, F1, of 200 N is applied

A force, F1, of 200 N is applied on a small piston of a hydraulic system with a diameter of 5.046 × 10^-2 m - NSC Technical Sciences - Question 5 - 2023 - Paper 1

Question 5

A force, F1, of 200 N is applied on a small piston of a hydraulic system with a diameter of 5.046 × 10^-2 m. The area of a large piston, F2, is 5.25 m², as shown in ... show full transcript

Worked Solution & Example Answer:A force, F1, of 200 N is applied on a small piston of a hydraulic system with a diameter of 5.046 × 10^-2 m - NSC Technical Sciences - Question 5 - 2023 - Paper 1

Step 1

5.1 State Pascal's law in words.

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Answer

Pascal's law states that in a confined fluid at rest, an applied pressure change will be transmitted undiminished throughout the fluid. This means that any change in pressure will affect every part of the fluid equally.

Step 2

5.2 Calculate the force, F2, on the large piston.

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Answer

To calculate the force on the large piston, we can use the relationship given by Pascal's law:

$rac{F_1}{A_1} = rac{F_2}{A_2}$

Where:

$F_1 = 200 ext{ N}$
$A_1 = rac{ ext{π} imes (5.046 imes 10^{-2})^2}{4} ightarrow 1.9998 imes 10^{-3} ext{ m}^2$
$A_2 = 5.25 ext{ m}^2$

Now, substituting the values:

ightarrow F_2 = 525 ext{ N}$$

Step 3

5.3 The distance between the two pistons is decreased by using a shorter pipe. How will it affect the answer to QUESTION 5.2?

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Answer

The effect of decreasing the distance between the two pistons does not affect the force calculated in question 5.2. So the answer is: REMAIN THE SAME.

Step 4

5.4 Write down TWO applications of hydraulic systems.

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Answer

Hydraulic brakes in vehicles.
Hydraulic jacks for lifting heavy objects.

Step 5

5.5.1 Stress in the rod

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Answer

Stress ( $au$ ) is calculated by the formula:

$au = rac{F}{A}$

Where:

Young's modulus (E) relates stress and strain:

ightarrow 2 imes 10^{11} ext{ Pa} = rac{ au}{0.0016} $Thus, Stress =$ au = E imes ext{strain} = 2 imes 10^{11} ext{ Pa} imes 0.0016 = 3.2 imes 10^8 ext{ Pa}$$

Step 6

5.5.2 Force, F, applied on the rod.

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Answer

Using the area of the rod:

ightarrow 3.142 imes 10^{-4} ext{ m}^2$$ The force can then be calculated as: $$F = au imes A = (3.2 imes 10^8 ext{ Pa}) imes (3.142 imes 10^{-4} ext{ m}^2) ightarrow F = 100.544 ext{ N}$$

A force, F1, of 200 N is applied on a small piston of a hydraulic system with a diameter of 5.046 × 10^-2 m - NSC Technical Sciences - Question 5 - 2023 - Paper 1

Worked Solution & Example Answer:A force, F1, of 200 N is applied on a small piston of a hydraulic system with a diameter of 5.046 × 10^-2 m - NSC Technical Sciences - Question 5 - 2023 - Paper 1

5.1 State Pascal's law in words.

5.2 Calculate the force, F2, on the large piston.

5.3 The distance between the two pistons is decreased by using a shorter pipe. How will it affect the answer to QUESTION 5.2?

5.4 Write down TWO applications of hydraulic systems.

5.5.1 Stress in the rod

5.5.2 Force, F, applied on the rod.

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