pressure - plymouth university · pdf filethe force per unit area is called the pressure, p:...

Basic Physics

Pressure

R Horan & M Lavelle

The aim of this package is to provide a short selfassessment programme for students who want tounderstand pressure and some of its elementaryapplications.

Copyright c© 2005 [email protected] , [email protected] Revision Date: November 29, 2005 Version 1.0

http://www.plymouth.ac.uk/maths

mailto:[email protected] kern +.1667em elax ,protect kern +.1667em elax [email protected]

Table of Contents

1. Introduction (Density)2. Pressure3. Hydraulics4. Archimedes’ Principle5. Final Quiz

Solutions to ExercisesSolutions to Quizzes

The full range of these packages and some instructions,should they be required, can be obtained from our webpage Mathematics Support Materials.

http://www.plymouth.ac.uk/mathaid

Section 1: Introduction (Density) 3

1. Introduction (Density)Density (symbol ρ) is defined by

ρ =mass

volume=

m

V.

Example 1 A cubic metre of gold has a mass of m ≈ 2 × 104 kg.Hence the density of gold is given by mass over volume:

ρgold =m

V=

2× 104 kg1 m3

= 2× 104 kg m−3 .

Example 2 The density of water is 1 g cm−3, i.e., a cubic centimetreof water has a mass of one gram. This can be expressed in S.I. units(see the packages on Units and on Changing Units) as follows.One gram is 10−3 kg and one centimetre is 10−2 m. Therefore 1 cm3 =(10−2)3 = 10−6 m3. This implies (see the package on Powers):

ρwater =1 g

1 cm3=

10−3 kg10−6 m3

= 103 kg m−3 .

Section 1: Introduction (Density) 4

Exercise 1.(a) A rectangular metal block with mass 18 kg has dimensions

0.1 m× 0.15 m× 0.2 m. Calculate the volume of the block andthus its density.

(b) The density of lead is just over 1× 104 kg m−3. Find the mass ofa cylindrical rod of length ` = 0.5 m and radius r = 0.02 cm.(The volume of a cylinder is V = πr2`.)

(c) What is the volume of 4 kg of water in litres? (1 L = 10−3 m3.)

(d) The density of mercury is ρmerc = 13, 600 kg m−3, what is thevolume of 4 kg of mercury?

Quiz The specific gravity of a substance is the dimensionless ratio:

specific gravity =ρ

ρwater

where ρ is the density of the substance. Such ratios are just numbersand have no units. Select the specific gravity of mercury.

(a) 136 (b) 0.074 (c) 1.36× 103 (d) 13.6

Section 2: Pressure 5

2. PressurePeople walking on ice are advised to use snowshoes so as to spreadtheir weight (a force) over a larger area of ice. The force per unit areais called the pressure, P :

pressure =forcearea

.

Snowshoes decrease the pressure exerted on the ice. The S.I. units ofpressure are newtons per square metre (N m−2) which are also calledpascals (symbol Pa, see the packages on Forces and Units).

Example 3 If your mass is 70 kg and the total area of the soles ofyour feet is 0.2m2, what pressure would you exert on the ground?The force on the ground is your weight: F = mg = 70 × 9.8 =686 N (we take g, the acceleration due to gravity, to be approximately9.8ms−2). Hence to two significant figures the pressure is

P =F

A=

70× 9.80.2

= 3, 400 Pa .


Quiz Find the pressure from a force of 100 N on an area of 0.25m2?

(a) 400 Pa (b) 25 Pa (c) 4, 000 Pa (d) 2.5 Pa

Exercise 2.(a) If a pressure gauge measures an increase in 3× 104 Pa on an area

of 0.07 m2 what is the increase in the force applied to the area?(b) Find the pressure produced by a kilogram of lead on a horizontal

surface if the area it rests on is 0.02 m2?(c) A scuba diver measures an increase in pressure of around 105 Pa

upon descending by 10 m, what is the change in force per squarecentimetre on the diver’s body?

To understand the pressures in liquids it is important to know thatliquids are not easily compressible. This means that the densityof liquids does not change when forces are applied to them. (E.g., thedensity of sea water does not vary much with depth.)


To find the pressure at the bottom of a column of liquid ofdensity ρ, height H and area A, consider the diagram below:

Volume of liquid: V = HADensity of liquid, ρ, so its mass: m = V ρ = HAρ

A

H

Thus the pressure exerted by the liquid is

P =weightarea

=mg

A

=HAρg

A= Hρg

where g = 9.8 ms−2 is the acceleration due to gravity. Note that thearea of the column has cancelled – the pressure only depends on theheight of the column.


The column of atmosphere above us also weighs downwards and pro-duces atmospheric pressure. Atmospheric pressure on Earth is101.3 kPa. Most pressure gauges measure the pressure above or be-low atmospheric pressure. This difference is called gauge pressure.The total pressure is called absolute pressure.

Example 4 If the density of sea water is ρ = 1, 030 kg m−3, what isthe pressure at 10 m below sea level?

From P = Hρg, the pressure at 10 m is given by

P = 10× 1030× 9.8 = 100, 940 Pa .

Thus the gauge pressure (i.e., from the sea water) is P = 101 kPawhich is very close to the pressure exerted by the atmosphere.The absolute pressure 10 m under the sea is the sum of the gaugepressure and the atmospheric pressure:

Pabs = 101 + 103 = 204 kPa .


Exercise 3.(a) What is the absolute pressure 30 m below the surface of the sea?(b) At what depth below the surface of the sea is absolute pressure

three times atmospheric pressure?

(c) The density of mercury is ρmerc = 13, 600 kg m−3, what is thegauge pressure under ten metres of mercury?

(d) At what depth of mercury would the absolute pressure be twiceatmospheric pressure?

The origin of pressure in fluids: gases, such as our atmosphere,or liquids, such as the oceans, are made up of huge numbers of minuteatoms and molecules moving around very quickly. As well as collidingwith each other they hit the edges of any solid object (such as yourbody). These collisions exert a force upon the object and are thesource of air pressure or water pressure.

Section 3: Hydraulics 10

3. HydraulicsIn a hydraulic lift, when a force, F1,is exerted on a piston of area A1, thisproduces a force, F2, on a piston ofarea A2, given by:

F1

A1=

F2

A2.

F1

A1

F2

A2

Proof When a force, F1, is exerted on the left hand piston the extrapressure, ∆P , on the piston on the left is:

∆P =F1

A1

this is transmitted through the liquid to the other piston

∆P =F2

A2

which proves the relation.


Example 5 If a force F1 = 100 N isapplied on the left in the hydrauliclift shown, where A1 = 0.02 m2 andA2 = 0.08 m2, what force F2 will beproduced?

F1

A1

F2

A2

Since the pressure exerted by the force F1, on the left, is transmittedby the liquid through to the other piston, we have

F2

A2=

F1

A1

∴ F2 =F1A2

A1

=100× 0.08

0.02= 100× 4 = 400 N.

Thus the force on the right is four times larger.


Quiz If in a hydraulic lift, one of the areas is three times larger thanthe other and a force of 15 N is applied to the smaller area, what forcewill be measured at the larger area?

(a) 15 N (b) 5 N (c) 18 N (d) 45 N

Exercise 4. These questions refer tothe lift in the diagram.(a) If F1 = 350 N, A1 = 0.7 m2 and

A2 = 1.2 m2 what is F2?

(b) If F1 = 210 N, A1 = 0.3 m2 andF2 = 500 N what is A2?

(c) If F1 = 1, 000 N, A1 = 1m2 andA2 = 10, 000 cm2 what is F2?

(d) What must F1 be to exert a forceF2 = 10, 000 N if A1 = 0.75 m2

and A2 = 3m2?

F1

A1

F2

A2

Section 4: Archimedes’ Principle 13

4. Archimedes’ PrincipleIf an object is in a liquid its weight pulls it down, but the pressure ofthe liquid around the object exerts an upwards force on it.Archimedes’ principle states that:

The upthrust or buoyancy force on a submerged object isequal to the weight of the liquid displaced by the object.

There are three alternatives:1) Weight of displaced liquid is equalto weight of object: the object is inequilibrium and will not move.2) Weight of displaced liquid is greaterthan the weight of object: the objectwill rise.3) Weight of displaced liquid is lessthan the weight of object: the objectwill sink.

weight

buoyancy

Section 4: Archimedes’ Principle 14

Example 6 The mass of the Titanic was roughly 40, 000 tonnes, howmuch water did it displace?Since a tonne is 1, 000 kg, the mass of the Titanic, in kilograms, wasapproximately m = 40, 000× 1, 000 = 4× 107 kg. To float, its weighthad to be equal to the weight of displaced water, i.e., it needed todisplace its own mass of water. The density of water is ρwater =1, 000 kg m−3, thus the volume of water it displaced was

V =m

ρ=

4× 107 kg1, 000 kg m−3

= 4× 104 m3 .

Exercise 5.(a) The mass (or displacement) of the fully loaded Titanic was

53, 147 tonnes. How much extra water did it displace?(b) Find the upthrust on an object of volume 0.02 m3 submerged in

water. What would it be if it were submerged in mercury?(c) How heavy would this object need to be to sink in mercury?

(ρmerc = 13, 600 kg m−3.)

Section 5: Final Quiz 15

5. Final Quiz

Begin Quiz

1. What is the volume occupied by a kilogram of gold? (Recall thatρgold = 2× 104 kg m−3.)

(a) 5×10−3 m3 (b) 2×104 m3 (c) 5×10−5 m3 (d) 5×10−4 m3

2. One atmosphere of pressure is defined as 101.325 kPa. From theanswers below, select the closest approximation to the mass of airabove a square metre of the earth.(a) 104 kg (b) 105 kg (c) 102 kg (d) 1, 000 kg

3. In a hydraulic lift a force of 5, 000 N is applied to a piston witharea of 0.0125 m2. If the area of the other piston is 0.25 m2, selectthe force on the other piston.(a) 250 N (b) 1, 000 N (c) 105 N (d) 25, 000 N

End Quiz

Solutions to Exercises 16

Solutions to ExercisesExercise 1(a)

The volume V of a rectangular block with mass m = 18 kg, whosedimensions are 0.1 m× 0.15 m× 0.2 m is

V = 0.1 m× 0.15 m× 0.2 m = 0.003 m3 = 3× 10−3m3 .

Therefore its density ρ is

ρ =m

V

=18 kg

3× 10−3 m3

=183× 103 kg m−3 = 6× 103 kg m−3 .

Click on the green square to return �


Exercise 1(b)

A cylindrical rod of length ` = 0.5 m and radius

r = 0.02 cm = 2× 10−4m

has volume

Vcyl = πr2` = 3.14× 4× 10−8m× 0.5 m = 6.28× 10−8m3 .

It is made from lead (density is approximately ρlead = 1×104 kg m−3).Thus the mass of this rod equals to

m = Vcyl × ρlead

= 6.28× 10−8m3 × 1× 104 kg m−3

= 6.28× 10−4kg .



Exercise 1(c)

From Example 2 we know that the density of water is

ρwater = 1g cm−3 =10−3kg10−6 m3

= 103 kg m−3

The volume of 4 kg of water is thus

V =m

ρwater

=4 kg

103 kg m−3

= 4× 10−3 m3.

Since 1 L = 10−3 m3, this volume in litres is

V = 4× 10−3 m3 = 4× 10−3 × 103L = 4 L .

Click on the green square to return�


Exercise 1(d)

The volume of m = 4 kg of mercury whose density is

ρmercury = 13, 600 kg m−3 = 1.36× 104 kg m−3

can be calculated as follows

V =m

ρmercury

=4 kg

1.36× 104 kg m−3

= 2.9× 10−4 m3.



Exercise 2(a)

When a pressure gauge measures an increase ∆P = 3× 104 Pa on anarea A = 0.07 m2, it means that the increase in the applied force ∆Fis given by

∆P =∆F

A∴ ∆F = ∆P × A .

Inserting the given data, we find that the increase in force is

∆F = 3× 104 Pa× 0.07 m2

= 0.21× 104 Pa×m2

= 2.1× 103 N .

Recall that 1 Pa m2 = 1NClick on the green square to return �


Exercise 2(b)

A mass m = 1 kg of lead which rests on a horizontal surface of areaA = 0.02 m2 exerts a pressure P due to its weight:

P =weightarea

=m g

A

=1 kg × 9.8 ms−2

0.02 m2= 490 Pa .



Exercise 2(c)

When descending by 10 m the scuba diver measures an increase inpressure ∆P = 105 Pa. The increase in the force ∆F on an areaA = 1 cm2 of the diver’s body is given by ∆F = ∆P × A:

∆F = ∆P × A = 105 Pa× 1 cm2 = 105 Pa× 10−4m2 = 10 N .

Let us show that this force is just the extra weight W of the additionalh = 10m water column above the diver.Its volume is V = hA = 10m×10−4 m2 = 10−3 m3. From the densityof water, the mass of the additional column of water is thus:

m = V ρwater = 10−3 m−3 × 103 kg m−3 = 1kg .

Hence the weight of the water is mg where g = 9.8 ms−2. The weightof 1 kg is 9.8 N. (The difference between this and the 10 N result showsthat the diver did not measure the pressure change perfectly!)Click on the green square to return

�


Exercise 3(a)

To find the gauge pressure 30 m under the surface of the sea, we use

P = Hρg

with H = 30 m, ρ = 1, 030 kg m−3 and g = 9.8 m s−2. This gives

P = H = 30m× 1, 030 kg m−3 × 9.8 m s−2 = 303, 000 Pa

to three significant figures.

The absolute pressure 30 m under the sea is the sum of this gaugepressure and the atmospheric pressure:

Pabs = 303, 000 + 103, 000 = 406 kPa .

We used here the fact that liquids are hard to compress, so the densityof sea water does not increase much at greater depths. (For gases thedensity does change with altitude as gases are easy to compress.)Click on the green square to return �


Exercise 3(b)

If the absolute pressure Pabs = Pgauge + Patms below sea level is threetimes atmospheric pressure

Pabs = 3× Patms

thenPgauge = 2× Patms .

The depth H where this occurs can be found expressing the gaugepressure:

Hρg = 2× Patms , ∴ H =2Patms

ρg

Using the data we have

H =2× 101× 103 Pa

1.03× 103 kg m−3 × 9.8 ms−2≈ 20 m .



Exercise 3(c)

The gauge pressure under H = 10m of mercury is

Pmerc = Hρmercg

= 10 m× 1.36× 104 kg m−3 × 9.8 ms−2

= 1.33× 106 Pa .

This is more than ten times larger than atmospheric pressure!Click on the green square to return

�


Exercise 3(d)

When the absolute pressure of mercury Pabs = Pgauge +Patms is twiceatmospheric pressure

Pabs = 2× Patms

the gauge pressure isPgauge = Patms .

Therefore the corresponding depth H of mercury can be found as:

gHρmerc = Patms , ∴ H =Patms

gρmerc

Recalling the density of mercury ρmerc = 1.36×104 kg m−3, we obtain

H =101× 103 Pa

1.36× 104 kg m−3 × 9.8 ms−2= 0.76 m .



Exercise 4(a)

If in a hydraulic lift, a force F1 = 350N is exerted on a piston of areaA1 = 0.7 m2 and the area of the second piston is A2 = 1.2 m2 thenthe resulting force F2 can be found from

F2

A2=

F1

A1

∴ F2 =F1 ×A2

A1

This yields

F2 =350 N× 1.2 m2

0.7 m2=

350× 1.20.7

N = 600 N .



Exercise 4(b)

If in a hydraulic lift, the force applied to the piston of area A1 = 0.3 m2

is F1 = 210N and the resulting force is F2 = 500N then the area A2

of the second piston can be found as followsF2

A2=

F1

A1

∴ A2 =F2 ×A1

F1

=500 N× 0.3 m2

210 N= 0.71 m2 .



Exercise 4(c)

We are told that in a hydraulic lift, the force applied to the first pistonof area A1 = 1m2 is F1 = 1, 000 N and the area of the second piston isA2 = 10, 000 cm2. Since 1 cm2 = 10−4 m2 , we have A2 = 1m2. Thusthe force F2 produced on the second piston is given by

F2

A2=

F1

A1

∴ F2 =F1 ×A2

A1=

1, 000 N× 1 m2

1 m2= 103 N .



Exercise 4(d)

To find the force F1 which must be applied to a first piston of areaA1 = 0.75 m2 in order to produce a force F2 = 10, 000 N on a secondpiston of area A2 = 3 m2, we use the equation

F2

A2=

F1

A1

∴ F1 =F2 ×A1

A2

=10, 000 N× 0.75 m2

3 m2= 2, 500 N .



Exercise 5(a)

Since the mass of a boat is directly related to how much water itdisplaces, ship designers call the mass of a boat its displacement.

The mass of the fully loaded Titanic in kilograms was

mT = 53, 147× 1, 000kg ≈ 5.3× 107 kg .

The density of water is ρwater = 1, 000 kg m−3, thus the total volumeof water it displaced was

VT =m

ρ=

5.3× 107 kg1, 000 kg m−3

= 5.3× 104 m3 .

Comparing the total volume VT with the volume V displaced by thenon-loaded Titanic, we see that the extra water displaced was

∆V = VT − V = 5.3× 104 m3 − 4× 104 m3 = 1.3× 104 m3 .



Exercise 5(b)

The buoyancy force on a submerged object of volume V = 0.02 m3 isequal to the weight of the liquid displaced by the object, therefore

Fbuoyancy = mwater × g = V × ρwater × g

= 0.02 m3 × 1, 000 kg m−3 × 9.8ms−2

= 196 kg ms−2 = 200 N .

If the same object is submerged in mercury then the buoyancy forceis

Fbuoyancy = mmercury × g = V × ρmercury × g

= 0.02 m3 × 13, 600 kg m−3 × 9.8ms−2

= 2700N .

Both answers are given to two significant figures.Click on the green square to return �


Exercise 5(c)

An object of volume V = 0.02 m3 submerged in mercury sinks whenits weight is more than the weight of displaced mercury:

mobject × g > mmercury × g .

The mass of displaced mercury is

mmercury = V × ρmercury

= 0.02 m3 × 13, 600 kg m−3

= 272 kg .

Therefore, in order to sink, the mass of an object must be at least

mobject > 272 kg .


Solutions to Quizzes 34

Solutions to QuizzesSolution to Quiz:From the Example 1 and Exercise 1d we know that the density ofwater and mercury are

ρwater = 1 g cm−3 =10−3kg10−6 m3

= 103kg m−3 ,

ρmercury = 13, 600 kg m−3 = 1.36 kg m−3 × 104 .

Therefore the specific gravity of mercury is the dimensionless ratio

ρmercury

ρwater=

1.36× 104 kg m−3

103 kg m−3= 13.6 .

End Quiz


Solution to Quiz:

The pressure caused by a force of

F = 100 N

on an area A = 0.25 m2 is

P =F

A=

100 N0.25 m2

= 400 Nm−2 = 400 Pa .

End Quiz


Solution to Quiz:If in a hydraulic lift, one of the areas is three times larger than theother, say

A1 = 3×A2

and the force applied to the smaller area A2 is F2 = 15 N , then theforce F1 measured at the larger area can be found from

F2

A2=

F1

A1

∴ F1 =F2 ×A1

A2=

F2 × 3×A2

A2

= 3× F2 = 3× 15 N = 45 N.

End Quiz

pressure - plymouth university · pdf filethe force per unit area is called the pressure, p:...

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