Week 8 Homework Name 1 A Hydraul

Week 8 Hw Name 1 A Hydraul

Analyze and solve the following problems related to fluid mechanics, hydrostatics, and hydraulics. Provide clear, step-by-step explanations and calculations for each problem to demonstrate understanding of principles such as Pascal's law, pressure calculation in fluids, buoyancy, and force interactions between water and objects submerged or supported hydraulically.

Paper For Above instruction

Problem 1: Hydraulic Lift Force Calculation

A hydraulic lift is used to lift heavy machinery by exerting pressure through a hydraulic system. The system consists of two pistons with different areas: a large piston with an area of 5 m², which is used to support the heavy machine, and a smaller piston with an area of 1 m² where force is applied to operate the lift. If a force of 1000 N is applied on the smaller piston, what force can be exerted on the larger piston to lift the machine?

Using Pascal's Law, the pressure exerted on both pistons is equal:

Pressure = Force / Area

Force applied on small piston: F₁ = 1000 N

Area of small piston: A₁ = 1 m²

Pressure in system: P = F₁ / A₁ = 1000 N / 1 m² = 1000 Pa

Force exerted on large piston (F₂):

F₂ = P × A₂, where A₂ = 5 m²

F₂ = 1000 Pa × 5 m² = 5000 N

Therefore, the hydraulic lift applies a force of 5000 Newtons on the heavy machine.

Problem 2: Pressure at the Bottom of a Water Tower

A vertical pipe of height 22 meters is filled with water, open to the atmosphere at the top. What is the pressure at the bottom of the pipe?

Pressure at the bottom of the water column is due to the weight of the water above and is given by:

Pressure = atmospheric pressure + ρgh

Assuming atmospheric pressure (P₀) = 101,325 Pa, density of water ρ = 1000 kg/m³, acceleration due to gravity g = 9.81 m/s², and height h = 22 m:

Pressure due to water column: P_water = ρgh = 1000 kg/m³ × 9.81 m/s² × 22 m = 215,820 Pa

Total pressure at the bottom: P_total = P₀ + P_water = 101,325 Pa + 215,820 Pa = 317,145 Pa

Hence, the pressure at the bottom of the pipe is approximately 317,145 Pascals.

Problem 3: Force Experienced at the Depth of the Mariana Trench

The Mariana Trench reaches a depth of about 11,000 meters. Given the density of seawater as 1025 kg/m³, what is the approximate force per unit area (pressure) experienced at this depth?

Pressure at depth: P = ρgh

ρ = 1025 kg/m³, g = 9.81 m/s², h = 11,000 m:

P = 1025 kg/m³ × 9.81 m/s² × 11,000 m ≈ 111,024,750 Pa (or about 111 MPa)

This immense pressure would exert a force of approximately 111 million Pascals on every square meter at that depth.

Note: The actual force experienced by an object depends on its contact area, but the pressure calculation illustrates the intense pressure environment.

Problem 4: Depth to Double Atmospheric Pressure

Atmospheric pressure at sea level is approximately 1 atm (~101,325 Pa). How deep must one go underwater to double this pressure?

Rearranging the hydrostatic pressure formula:

h = (P_total - P₀) / (ρg)

Desired total pressure: P_total = 2 × 101,325 Pa = 202,650 Pa

Pressure increase due to water: ΔP = P_total - P₀ = 101,325 Pa

Using water density ρ = 1000 kg/m³:

h = 101,325 Pa / (1000 kg/m³ × 9.81 m/s²) ≈ 10.33 meters

Thus, one needs to go approximately 10.33 meters underwater for the pressure to double atmospheric pressure.

Problem 5: Force to Submerge a Tennis Ball

A tennis ball has a density of 0.084 g/cm³ and a diameter of 3.8 cm. What is the force required to fully submerge the ball in water?

First, convert density to SI units:

0.084 g/cm³ = 0.084 g / cm³

1 g = 1×10⁻³ kg, 1 cm³ = 1×10⁻⁶ m³, so:

Density ρ_ball = 0.084 g/cm³ = 0.084 × 10⁻³ kg / 10⁻⁶ m³ = 84 kg/m³

Calculate the volume of the sphere:

V = (4/3)πr³

Diameter d = 3.8 cm = 0.038 m, so radius r = 0.019 m

V = (4/3)π × (0.019)³ ≈ 2.87 × 10⁻5 m³

Mass of the ball:

m = ρ_ball × V ≈ 84 kg/m³ × 2.87 × 10⁻5 m³ ≈ 0.00241 kg

Weight of the ball (force due to gravity):

F_gravity = m × g ≈ 0.00241 kg × 9.81 m/s² ≈ 0.0236 N

Buoyant force (Archimedes' principle):

F_buoyancy = ρ_water × V × g = 1000 kg/m³ × 2.87 × 10⁻5 m³ × 9.81 m/s² ≈ 0.281 N

Force required to submerge the ball is the sum of the force to counteract gravity and to overcome buoyancy:

F_total = F_buoyancy + F_gravity ≈ 0.281 N + 0.0236 N ≈ 0.305 N

This is the approximate force needed to fully submerge the tennis ball in water.

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