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Ch 12: Fluid Mechanics
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 12: Fluid MechanicsProblem 24
Chapter 12, Problem 24

Hydraulic Lift II.The piston of a hydraulic automobile lift is 0.30 m in diameter. What gauge pressure, in pascals, is required to lift a car with a mass of 1200 kg? Also express this pressure in atmospheres.

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First, calculate the area of the piston using the formula for the area of a circle: \( A = \pi r^2 \), where \( r \) is the radius of the piston. Given the diameter is 0.30 m, the radius \( r = 0.30 / 2 \) m.
Next, determine the force required to lift the car. This force is equal to the weight of the car, which can be calculated using \( F = mg \), where \( m = 1200 \) kg is the mass of the car and \( g = 9.81 \) m/s² is the acceleration due to gravity.
Calculate the gauge pressure required using the formula \( P = \frac{F}{A} \), where \( F \) is the force calculated in the previous step and \( A \) is the area of the piston.
Convert the pressure from pascals to atmospheres. Use the conversion factor: 1 atmosphere = 101325 pascals.
Finally, express the gauge pressure in both pascals and atmospheres, ensuring to clearly distinguish between the two units.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Pascal's Principle

Pascal's Principle states that a change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of its container. This principle is fundamental in hydraulic systems, where a small force applied at one point is transmitted to exert a larger force at another point, allowing for the lifting of heavy objects like cars.
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Pressure Calculation

Pressure is defined as the force applied per unit area. In the context of a hydraulic lift, the pressure required to lift an object can be calculated using the formula P = F/A, where P is the pressure, F is the force (equal to the weight of the car, which is mass times gravity), and A is the area of the piston. This calculation is crucial for determining the gauge pressure needed to lift the car.
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Unit Conversion

Unit conversion is essential for expressing physical quantities in different units. In this problem, after calculating the pressure in pascals, it is necessary to convert it to atmospheres for a comprehensive understanding. The conversion factor between pascals and atmospheres is 1 atm = 101,325 Pa, which allows for the expression of pressure in a more commonly used unit.
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Related Practice
Textbook Question

Exploring Venus. The surface pressure on Venus is 92 atm, and the acceleration due to gravity there is 0.894g. In a future exploratory mission, an upright cylindrical tank of benzene is sealed at the top but still pressurized at 92 atm just above the benzene. The tank has a diameter of 1.72 m, and the benzene column is 11.50 m tall. Ignore any effects due to the very high temperature on Venus. What force does the Venusian atmosphere exert on the outside surface of the bottom of the tank?

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Textbook Question

BIO. The lower end of a long plastic straw is immersed below the surface of the water in a plastic cup. An average person sucking on the upper end of the straw can pull water into the straw to a vertical height of 1.1 m above the surface of the water in the cup. (a) What is the lowest gauge pressure that the average person can achieve inside his lungs? (b) Explain why your answer in part (a) is negative.

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Textbook Question

A rock has mass 1.80 kg. When the rock is suspended from the lower end of a string and totally immersed in water, the tension in the string is 12.8 N. What is the smallest density of a liquid in which the rock will float?

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Textbook Question

A closed container is partially filled with water. Initially, the air above the water is at atmospheric pressure (1.01×105 Pa) and the gauge pressure at the bottom of the water is 2500 Pa. Then additional air is pumped in, increasing the pressure of the air above the water by 1500 Pa. What is the gauge pressure at the bottom of the water?

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Textbook Question

BIO. There is a maximum depth at which a diver can breathe through a snorkel tube (Fig. E12.17) because as the depth increases, so does the pressure difference, which tends to collapse the diver's lungs. Since the snorkel connects the air in the lungs to the atmosphere at the surface, the pressure inside the lungs is atmospheric pressure. What is the external– internal pressure difference when the diver's lungs are at a depth of 6.1 m (about 20 ft)? Assume that the diver is in fresh-water. (A scuba diver breathing from compressed air tanks can operate at greater depths than can a snorkeler, since the pressure of the air inside the scuba diver's lungs increases to match the external pressure of the water.)


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Textbook Question

A 950-kg cylindrical can buoy floats vertically in sea-water. The diameter of the buoy is 0.900 m. Calculate the additional distance the buoy will sink when an 80.0-kg man stands on top of it.

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