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Chapter 14: Problem 25

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

Short Answer

Expert verified
166448 Pa or 1.642 atm.

Step by step solution

01

Calculate the Force Required to Lift the Car

The force required to lift the car is equal to the gravitational force acting on it, which is the weight of the car. This can be calculated using the formula \( F = m \cdot g \), where \( m = 1200 \, \text{kg} \) is the mass of the car and \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity. Thus, \( F = 1200 \times 9.81 = 11772 \, \text{N} \).
02

Calculate the Area of the Piston

The diameter of the piston is given as \( 0.30 \, \text{m} \). First, find the radius \( r \) by dividing the diameter by 2, so \( r = 0.30 / 2 = 0.15 \, \text{m} \). The area \( A \) of the piston can be calculated using the formula for the area of a circle: \( A = \pi r^2 \). Therefore, \( A = \pi \times (0.15)^2 \approx 0.0707 \, \text{m}^2 \).
03

Calculate Gauge Pressure Required

Gauge pressure \( P \) is calculated using the formula \( P = F / A \). Substituting the values from Step 1 and Step 2, we have \( P = 11772 / 0.0707 \approx 166448 \, \text{Pa} \).
04

Convert Pressure to Atmospheres

To convert pascals to atmospheres, use the conversion factor \( 1 \, \text{atm} = 101325 \, \text{Pa} \). Therefore, the pressure in atmospheres is \( 166448 / 101325 \approx 1.642 \, \text{atm} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gauge Pressure
In the context of a hydraulic lift, gauge pressure refers to the pressure within the system above atmospheric pressure. It's crucial because it allows us to understand the additional pressure required to lift a load, like a car. The gauge pressure is calculated by dividing the force needed to lift the object by the area over which this force is applied. Since atmospheric pressure is not considered in gauge pressure, it uniquely measures the effectiveness of the lift in comparison to typical atmospheric conditions.
Force Calculation
The force required to lift an object is central to understanding the mechanics of a hydraulic lift. It is computed through the formula:
  • \( F = m \cdot g \)
where \( F \) is the force, \( m \) is the mass of the object, and \( g \) is the gravitational acceleration (approximately \( 9.81 \, \text{m/s}^2 \)).
This force essentially represents the gravitational pull on the object, requiring the hydraulic system to exert an equal and opposite force to perform the lift efficiently.
For a hydraulic lift lifting a car with a mass of 1200 kg, the required force would be 11772 N. Understanding this helps in configuring the hydraulic system correctly to provide sufficient lift power.
Area of a Circle
The piston of a hydraulic lift is often circular, making the area of a circle formula integral to calculate the part on which force will apply. The area is found using:
  • \( A = \pi r^2 \)
where \( A \) is the area, and \( r \) is the radius of the circle.
To find the radius when the diameter is known (as given as 0.30 m), divide by 2 to get 0.15 m. Squaring the radius and multiplying by \( \pi \) gives an area of approximately 0.0707 m虏.
This area calculation is necessary to determine how spread out the force is over the piston's surface, affecting the pressure required to lift the load.
Pressure Conversion
Pressure conversion is a practical necessity when differing units or standards are used, such as pascals and atmospheres. To convert the gauge pressure of 166448 Pa to atmospheres, use the conversion factor:
  • 1 atm = 101325 Pa.
Division of the pressure in pascals by this conversion factor gives the pressure in atmospheres. In this case, it's approximately 1.642 atm.
This conversion is useful for communicating pressure in terms familiar to many contexts, like standard atmospheric pressure often used in weather forecasting and general physics contexts.

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Most popular questions from this chapter

A closed container is partially filled with water. Initially, the air above the water is at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{Pa}\right)\) 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 \(\mathrm{Pa}\) . (a) What is the gauge pressure at the bottom of the water? (b) By how much must the water level in the container be reduced, by drawing some water out through a valve at the bottom of the container, to return the gauge pressure at the bottom of the water to its original value of 2500 Pa? The pressure of the air above the water is maintained at 1500 Pa above atmospheric pressure.

You drill a small hole in the side of a vertical cylindrical water tank that is standing on the ground with its top open to the air. (a) If the water level has a height \(H,\) at what height above the base should you drill the hole for the water to reach its greatest distance from the base of the cylinder when it hits the ground? (b) What is the greatest distance the water will reach?

The Great Molasses Flood. On the afternoon of January \(15,1919,\) an unusually warm day in Boston, a \(27.4-\mathrm{m}\) -high 27.4 - diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 9 -m-deep stream, killing pedestrians and horses, and knocking down buildings. The molasses had a density of \(1600 \mathrm{kg} / \mathrm{m}^{3} .\) If the tank was full before the accident, what was the total outward force the molasses exerted on its sides? (Hint: Consider the outward force on a circular ring of the tank wall of width \(d y\) and at a depth \(y\) below the surface. Integrate to find the total outward force. Assume that before the tank ruptured, the pressure at the surface of the molasses was equal to the air pressure outside the tank.)

A cubical block of density \(\rho_{\mathrm{B}}\) and with sides of length \(L\) floats in a liquid of greater density \(\rho_{\mathrm{L}}\) (a) What fraction of the block's volume is above the surface of the liquid? (b) The liquid is denser than water (density \(\rho_{\mathrm{W}} )\) and does not with it If water is poured on the surface of the liquid, how deep must the water layer be so that the water surface just rises to the top of the block? Express your answer in terms of \(L, \rho_{\mathrm{B}}, \rho_{\mathrm{L}},\) and \(\rho_{\mathrm{w}}\) . (c) Find the depth of the water layer in part (b) if the liquid is mercury, the block is made of iron, and the side length is \(10.0 \mathrm{cm} .\)

An object of average density \(\rho\) floats at the surface of a fluid of density \(\rho_{\text { fluid }}\). (a) How must the two densities be related? (b) In view of the answer to part (a), how can steel ships float in water? (c) In terms of \(\rho\) and \(\rho_{\text { fluid }}\) what fraction of the object is submerged and what fraction is above the fluid? Check that your answers give the correct limiting behavior as \(\rho \rightarrow \rho_{\text { fluid }}\) and as \(\rho \rightarrow 0 .\) (d) While on board your yacht, your cousin Throckmorton cuts a rectangular piece (dimensions \(5.0 \times 4.0 \times 3.0 \mathrm{cm} )\) out of a life preserver and throws it into the ocean. The piece has a mass of 42 g. As it floats in the ocean, what percentage of its volume is above the surface?
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