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

To find the force on the large piston (F2), we can use Pascal's principle, which states that the pressure applied at one point in a hydraulic system is transmitted throughout the system.

We start by calculating the area of the small piston:

A_1 = rac{ ext{π} d^2}{4} = rac{ ext{π} (5.046 imes 10^{-2})^2}{4} \approx 1.9988 imes 10^{-3} m^2

Next, we calculate the pressure (P) using the small piston:

P = rac{F_1}{A_1} = rac{200 N}{1.9988 imes 10^{-3} m^2} \approx 100480 N/m^2

Now, we can find the force on the large piston (F2) using its area (5.25 m²):

$F_2 = P imes A_2 = 100480 N/m^2 imes 5.25 m^2 \approx 525 N$

REMAIN THE SAME

• Hydraulic brakes
• Hydraulic lifts

To calculate the stress (σ) in the rod, we use the formula:

ext{σ} = rac{F}{A}

First, we calculate the area of the rod:

A = rac{ ext{π} d^2}{4} = rac{ ext{π} (0.02 m)^2}{4} \approx 3.142 imes 10^{-4} m^2

Now substitute into the stress formula:

ext{σ} = rac{0.16 \times 10^{-2}}{3.142 \times 10^{-4}} \approx 3.2 \times 10^{8} Pa

Using the relationship of stress:

$F = ext{σ} imes A$

Substituting the known values:

$F = (3.2 \times 10^{8} Pa) \times (3.142 \times 10^{-4} m^2) \approx 100.54 N$