/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 The piston of a hydraulic automo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(\mathrm{kg}\) ? Now express this pressure in atmospheres.

Short Answer

Expert verified
The required gauge pressure is approximately 166463.91 pascals or 1.643 atmospheres.

Step by step solution

01

Calculate the area of the piston

The area of the piston, which is circular, is given by the formula \( A = \pi r^2 \). First, find the radius of the piston by dividing the diameter by 2: \( r = \frac{0.30}{2} = 0.15 \) meters. Now, calculate the area: \( A = \pi \times (0.15)^2 \approx 0.0707 \) square meters.
02

Calculate the force exerted by the car

The force exerted by the car is calculated using the formula \( F = mg \), where \( m = 1200 \) kg is the mass of the car and \( g = 9.81 \) m/s² is the acceleration due to gravity. Thus, the force is \( F = 1200 \times 9.81 = 11772 \) newtons.
03

Calculate the gauge pressure required

Pressure is defined as force divided by area. Use the formula \( P = \frac{F}{A} \) to find the gauge pressure needed to lift the car: \( P = \frac{11772}{0.0707} \approx 166463.91 \) pascals.
04

Convert pressure from pascals to atmospheres

1 atmosphere (atm) is equivalent to 101325 pascals. Use this conversion to express the pressure in atmospheres: \( P = \frac{166463.91}{101325} \approx 1.643 \) atm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Pressure calculation
Understanding pressure is key in hydraulic systems. Pressure is simply the force applied distributed over a certain area. It is calculated using the formula:\[ P = \frac{F}{A} \]where
  • \(P\) is the pressure
  • \(F\) is the force applied
  • \(A\) is the area over which the force acts.
In the case of a hydraulic lift, the pressure has to be calculated to know the force necessary to lift a given weight. The pressure needs to be enough to support the mass of the car. Always remember that the units matter greatly here. Typically, in Physics, pressure is measured in pascals (Pa). One pascal is defined as one newton per square meter. By using the given force and area of the piston, you can determine how much pressure is required to lift something heavy like a car. This concept is vital for so many everyday applications such as car brakes, airplanes, and even your own super adjustable office chair!
Force and area relationship
The relationship between force and area is foundational in understanding pressure. Whenever a force is exerted over an area, pressure is born. Sometimes a small object like the point of a needle can exert a lot of pressure due to its tiny area. In hydraulic lifts, the area of a piston and the force exerted by a car's weight are crucial to designing a system that can lift effectively. To find the force that a car exerts on a piston, multiply the car's mass by gravity (9.81 m/s²), as seen in this formula:\[ F = m \times g \]where
  • \(F\) is the force
  • \(m\) is the mass
  • \(g\) is the acceleration due to gravity.
The area, important for calculating pressure, is usually a known value based on system requirements. The effective pressure can then be adjusted to design hydraulic systems to perfectly balance the force of the car on the lift.
Unit conversion
Unit conversion is a crucial step that allows comparison and communication of scientific ideas. In everyday life, you might need to convert units for recipes, but in science, precision is even more important. Working with pressures, pascals might need to be converted into atmospheres for a practical understanding.Here's how simple conversions can be:
  • 1 atmosphere (atm) = 101325 pascals (Pa)
To convert the calculated pressure from pascals to atmospheres:\[ \text{Pressure in atm} = \frac{\text{Pressure in Pa}}{101325} \]Understanding conversion factors and applying them correctly ensures that values remain consistent and meaningful during various calculations. This small but vital skill can make all the difference in accurate problem-solving, especially in an exam or real-world scenario.
Pascal and atmosphere
Understanding different units of pressure is essential, especially between pascals and atmospheres, given their prevalence in scientific fields. Pascal is the standard unit of pressure in the International System of Units (SI), named after Blaise Pascal, a French mathematician, and scientist.
  • 1 Pascal (Pa) = 1 Newton per square meter
Meanwhile, the atmosphere unit (atm) is a standard measurement heavily utilized in meteorology and various practical scientific fields, representing the air pressure at sea level. Hence, understanding
  • 1 atm = 101325 Pa
helps bridge the gap between theoretical calculations and real-world applications conveniently. By understanding this relationship, you will be better armed to navigate between different contexts where different units may be used.
Physics problem solving
When it comes to solving physics problems, especially those involving hydraulic systems, break the problem into small, manageable parts. Start by identifying known quantities and relationships, like pressure, force, and area.Follow these general tips:
  • Identify what needs solving — finding pressure, force, or area, for example.
  • Apply the relevant formulas, in this case, using: \( F = m \times g \) and \( P = \frac{F}{A} \).
  • Convert units as necessary to ensure consistency. Units can often hint at which aspects need calculation.
  • Check if the solution makes sense logically and mathematically.
This systematic approach helps cultivate a clear path to the solution and builds a strong foundation for solving more complex problems in future studies. Understanding each step fosters better handling of the wrath of physics and develops essential problem-solving skills handy in everyday life situations and future careers.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What speed must a gold sphere of radius 3.00 \(\mathrm{mm}\) have in castor oil for the viscous drag force to be one-fourth of the weight of the sphere? The density of gold is \(19,300 \mathrm{kg} / \mathrm{m}^{3}\) and the viscosity of the oil is 0.986 \(\mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\)

A hot-air balloon has a volume of 2200 \(\mathrm{m}^{3} .\) The balloon fabric (the envelope) weighs 900 \(\mathrm{N} .\) The basket with gear and full propane tanks weighs 1700 \(\mathrm{N}\) . If the balloon can barely lift an additional 3200 \(\mathrm{N}\) of passengers, breakfast, and champagne when the outside air density is \(1.23 \mathrm{kg} / \mathrm{m}^{3},\) what is the average density of the heated gases in the envelope?

The cross-sectional area of the aorta is \(3 \mathrm{cm}^{2},\) and the average velocity of blood leaving the heart into the aorta is 30 \(\mathrm{cm} / \mathrm{s}\) . If the combined effective cross sectional area of the body's capillaries is \(600 \mathrm{cm}^{2},\) what is the average flow rate in a capillary? $$\begin{array}{l}{\text { A. } 1 \mathrm{cm} / \mathrm{s}} \\ {\mathrm{B} .2 \mathrm{cm} / \mathrm{s}} \\ {\mathrm{C} \cdot 0.01 \mathrm{cm} / \mathrm{s}} \\\ {\mathrm{D}, 0.15 \mathrm{cm} / \mathrm{s}} \\ {\mathrm{E} .0 .2 \mathrm{cm} / \mathrm{s}}\end{array}$$

Landing on Venus. One of the great difficulties in landing on Venus is dealing with the crushing pressure of the atmosphere, which is 92 times the earth's atmospheric pressure. (a) If you are designing a lander for Venus in the shape of a hemisphere 2.5 \(\mathrm{m}\) in diameter, how many newtons of inward force must it be prepared to withstand due to the Venusian atmosphere? (Don't forget about the bottom!) (b) How much force would the lander have to withstand on the earth?

A liquid is used to make a mercury-type barometer, as described in Section \(13.2 .\) The barometer is intended for spacefaring astronauts. At the surface of the earth, the column of liquid rises to a height of 2185 \(\mathrm{mm}\) , but on the surface of Planet \(\mathrm{X},\) where the acceleration due to gravity is one-fourth of its value on earth, the column rises to only 725 \(\mathrm{mm} .\) Find (a) the density of the liquid and (b) the atmospheric pressure at the surface of Planet \(\mathrm{X}\) .

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.