/*! 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 19 A cylindrical disk of wood weigh... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 19

A cylindrical disk of wood weighing \(45.0 \mathrm{~N}\) and having a diameter of \(30.0 \mathrm{~cm}\) floats on a cylinder of oil of density \(0.850 \mathrm{~g} / \mathrm{cm}^{3}\) (Fig. E12.19). The cylinder of oil is \(75.0 \mathrm{~cm}\) deep and has a diameter the same as that of the wood. (a) What is the gauge pressure at the top of the oil column? (b) Suppose now that someone puts a weight of \(83.0 \mathrm{~N}\) on top of the wood, but no oil seeps around the edge of the wood. What is the change in pressure at (i) the bottom of the oil and (ii) halfway down in the oil?

Short Answer

Expert verified
The gauge pressure at the top of the oil column is 0 Pa. The change in pressure at the bottom of the oil and halfway down in the oil after a weight of 83.0 N is added on top of the wood is the same and is calculated using Pascal's law.

Step by step solution

01

Determine gauge pressure at the top of the oil column

The gauge pressure at the top of the oil column is equal to the atmospheric pressure because there's no overlying fluid at that point, therefore the gauge pressure would be 0 Pa.
02

Calculate weight of the oil cylinder

First, calculate the volume (V) of the oil cylinder using the formula for the volume of a cylinder \(V = \pi r^2 h\) where \(r\) is the radius and \(h\) is the height. Half diameter to get the radius, use 0.15m for the radius and 0.75m for the height. Multiply the volume found by the oil density and then by the acceleration due to gravity to find the weight of the oil column.
03

Calculate the new total force acting on the oil

Add the initial weight of the wood, the weight of the oil column and the added weight to find the total force acting on the oil.
04

Calculate the change in pressure at the bottom of the oil

According to Pascal's law, the additional pressure applied to the surface of the oil by the added weight is transmitted undiminished to all portions of the fluid. Therefore, the change in pressure at the bottom of the oil is given by \( \Delta P = F/A \) where \( F \) is the total force from step 3 and \( A \) is the cross-sectional area of the oil/wood cylinder calculated by the formula \( A = \pi r^2 \).
05

Calculate the change in pressure halfway down in the oil

Since the pressure is transmitted undiminished in the fluid, the change in pressure halfway down in the oil would still be the same as the change in pressure at the bottom of the fluid because it's the same fluid and same cross-section area. Therefore, no additional calculations are needed here, the change in pressure halfway down is equal to the change in pressure at the bottom of the oil column.

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 91影视!

Key Concepts

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

Gauge Pressure
Understanding gauge pressure is fundamental when dealing with fluid mechanics. Gauge pressure is the pressure relative to the atmospheric pressure. In simpler terms, it's the difference between the absolute pressure inside a system and the atmospheric pressure outside. If there's no overlying fluid atop a column, as with the top of the oil column in our exercise, the gauge pressure is zero because the pressure inside and outside are the same.

When additional weight is placed on the wood floating on the oil, the pressure increases; however, the increase measured is the gauge pressure change, not the total pressure. The gauge pressure is practical in everyday applications because it allows us to ignore atmospheric pressure when it's not relevant to the problem at hand, which simplifies calculations and understanding.
Pascal's Law
Pascal's law, named after the French mathematician and physicist Blaise Pascal, states that in a fluid at rest within a closed container, any change in pressure is transmitted undiminished in all directions throughout the fluid. In the context of our exercise, when an additional weight is placed on the floating disk of wood, Pascal's law asserts that this extra pressure will spread throughout the entire volume of the oil without loss.

This means that no matter where you measure within the oil column鈥攚hether at the bottom or halfway down鈥攖he pressure increase would be the same as long as the cross-sectional area remains unchanged. Understanding Pascal's law is crucial for various applications such as hydraulic lifts and the study of fluid dynamics in engineering systems.
Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid due to the gravitational force acting on it. It is often referred to as the pressure due to the weight of the fluid above a given point. In our exercise, we calculate the weight of the oil to find the force exerted by the oil column, which when divided by the cross-sectional area of the oil cylinder, gives us the hydrostatic pressure at the base of the column.

The deeper you go in a fluid, the higher the hydrostatic pressure - this is why the pressure at the bottom of the ocean is much greater than near the surface. However, when additional force is applied on the top of the fluid, as with the wood and weight in our exercise, this increases the hydrostatic pressure uniformly throughout the fluid due to Pascal's law, which is a key point in understanding how pressures change within a static fluid.

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

A cubical block of wood, \(10.0 \mathrm{~cm}\) on a side, floats at the interface between oil and water with its lower surface \(1.50 \mathrm{~cm}\) below the interface (Fig. E12.33). The density of the oil is \(790 \mathrm{~kg} / \mathrm{m}^{3}\). (a) What is the gauge pressure at the upper face of the block? (b) What is the gauge pressure at the lower face of the block? (c) What are the mass and density of the block?

On the afternoon of January \(15,1919,\) an unusually warm day in Boston, a 17.7 -m-high, 27.4-m-diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 5-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 liquid flowing from a vertical pipe has a definite shape as it flows from the pipe. To get the equation for this shape, assume that the liquid is in free fall once it leaves the pipe. Just as it leaves the pipe, the liquid has speed \(v_{0}\) and the radius of the stream of liquid is \(r_{0}\). (a) Find an equation for the speed of the liquid as a function of the distance \(y\) it has fallen. Combining this with the equation of continuity, find an expression for the radius of the stream as a function of \(y\). (b) If water flows out of a vertical pipe at a speed of \(1.20 \mathrm{~m} / \mathrm{s},\) how far below the outlet will the radius be one-half the original radius of the stream?

A barrel contains a \(0.120 \mathrm{~m}\) layer of oil floating on water that is \(0.250 \mathrm{~m}\) deep. The density of the oil is \(600 \mathrm{~kg} / \mathrm{m}^{3}\). (a) What is the gauge pressure at the oil-water interface? (b) What is the gauge pressure at the bottom of the barrel?

You are designing a diving bell to withstand the pressure of seawater at a depth of \(250 \mathrm{~m}\). (a) What is the gauge pressure at this depth? (You can ignore changes in the density of the water with depth.) (b) At this depth, what is the net force due to the water outside and the air inside the bell on a circular glass window \(30.0 \mathrm{~cm}\) in diameter if the pressure inside the diving bell equals the pressure at the surface of the water? (Ignore the small variation of pressure over the surface of the window.)
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Engineering Physics

Read Explanation

Energy Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Waves Physics

Read Explanation

Solid State Physics

Read Explanation
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.