# data[].attributes.body
Hydraulics is a key part of engineering that deals with the mechanical properties of fluids.

It revolves around the transmission of force through the use of fluids that are constricted/squeezed. Hydraulics are very important as it plays major roles within many different mechanical industries such as construction, automotive and aerospace.

# data[].attributes.headline
Real-World example 1 - Hydraulic Braking System in Vehicles

# data[].attributes.subject
Real-World example 1 - Hydraulic Braking System in Vehicles

# data[].attributes.body
Hydraulic braking systems in cars use the principles of hydraulics to transmit force from the brake pedal to the brake calipers, resulting in the application of braking force to the wheels.

Suppose a driver applies a force of 200 N to the brake pedal and that the master cylinder has a piston with a diameter corresponding to an area of 5 x 10−4 m².

Using the hydraulic equation P = F / A​, we can calculate the pressure (P) in the brake fluid:

P = 200 N​ / 5×10−4 m2 =400,000 Pa

# data[].attributes.body
The equation that the concept of hydraulics is derived from Pascal's law, which states that a change in pressure applied to a confined fluid is transmitted to all portions of the fluid and to the walls of its container. It can be expressed as followed:

P2 A ₁​ = P 2​A ₂

Where P is pressure and A is area. This proves that they're both interconnected with each other in hydraulic systems.

Buoyancy forces also play a role in hydraulics, which arise due to the displacement of fluid by submerged objects. The buoyancy force equation, also known as Archimedes principle goes as follows:

Fb​ = ρ x V x g

where ρ is density, V is volume, and g is the acceleration due to gravity. This equation correlates to hydraulics, as the density of the fluid affects the overall weight of the fluid column within the hydraulic system.

# data[].attributes.body
Hydraulics relies on the fundamental principle that fluids are virtually incompressible. This characteristic allows for the efficient transfer of force from one point to another within a closed system. The physics of hydraulics involves relationships between pressure, area, and force in fluid-filled systems.

# data[].attributes.headline
Real-world example 2 - Hydraulic Excavator

# data[].attributes.subject
Real-world example 2 - Hydraulic Excavator

# data[].attributes.body
Hydraulic systems are also used in excavators for digging and lifting operations. The hydraulic cylinders control the movement of the excavator arm, bucket, and other components.

For example, assume that the excavator arm exerts a lifting force of 5000 N and that the hydraulic cylinder has a piston with a diameter corresponding to an area of 2x 10−3 m².

Using the hydraulic equation P= F / A​, we can calculate the pressure (P) in the hydraulic cylinder:

P = 5000 N / 2×10−3 m² = 2,500,000 Pa

# data[].attributes.body
1. In a hydraulic system, a piston with a diameter of 5 cm is subjected to a force of 80 N. Calculate the pressure exerted on the piston. Use the hydraulic equation P = F/A​ and ensure that all units are in meters.

Convert Diameter to Radius:

r = 0.025 m

Calculate the Area (A):

A = π(0.025m)²

A ≈ 0.0019635 m²

Calculate Pressure (P):

P= 80 N​ / 0.0019635 m²

P ≈ 40779.96 Pa

2. In a hydraulic system, there are two pistons. Piston 1 has a diameter of 8 cm, and piston 2 has a diameter of 4 cm. If a force of 200 N is applied to piston 1, what force will be exerted by piston 2? Assume the hydraulic fluid is incompressible and neglect any friction or losses.

To calculate the force exerted by piston 2 (F 2​), we can use the hydraulic equation:

P₁A,= P₂A₂

Where:

P, is the pressure applied to piston 1,

A, is the area of piston 1,

P 2​ is the pressure on piston 2,

A 2​ is the area of piston 2.

The area of a piston is given by the formula for the area of a circle:

A = πr ₂

Where r is the radius of the piston.

1. Calculating the Area of Piston 1 (A 1​):

A₁ =π(0.04 m)²

A₁ = 0.005027 m²

2. Calculating the Pressure on Piston 1 (P 1​):

P1 =​ Force/A₁

P₁ = 200 / 0.005027

P1 = 39758 Pa

3. Using Pascal's Law to Find Pressure on Piston 2 (P ₂):

P2 = P1​

P₂ = 39758 Pa

4. Calculating the Area of Piston 2 (A ₂​):

A₂=π(0.02m)²

A₂ = 0.001256 m₂

5. Calculating the Force on Piston 2 (F ₂​):

F2 = P₂ x A ₂​

F₂ = 39758 Pa x 0.001256 m²

F₂ = 50 N