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Mechanical Systems in Static EquilibriumThese notes cover the first eight lectures of PHY 61. Each chapter cor-

responds to one lecture and should be read before attending the lecture.

c Joshua Socolar, Duke University, 2005

Contents

1 Introduction 3

1.1 Two kinds of fundamental laws . . . . . . . . . . . . . . . . . 3

1.2 Getting started: Classical physics of static systems . . . . . . 4

2 The concept of force 6

2.1 Types of forces . . . . . . . . . . . . . . . . . . . . . . . . . . 72.1.1 General discussion of gravity and contact forces . . . . 72.1.2 General discussion of stress and rigid bodies . . . . . . 92.1.3 Looking ahead . . . . . . . . . . . . . . . . . . . . . . 10

2.2 Force is a vector quantity . . . . . . . . . . . . . . . . . . . . 102.2.1 Vector addition . . . . . . . . . . . . . . . . . . . . . . 112.2.2 Vector notation expressing force balance . . . . . . . . 122.2.3 Representing vectors with Cartesian coordinates . . . 122.2.4 Units of force . . . . . . . . . . . . . . . . . . . . . . . 14

2.2.5 Adding several vector forces . . . . . . . . . . . . . . . 142.3 The magnitude of a vector . . . . . . . . . . . . . . . . . . . . 15

3 Newtons third law and force balance equations 17

3.1 Newtons Laws . . . . . . . . . . . . . . . . . . . . . . . . . . 183.1.1 Implications of Laws 1 and 2 for static systems . . . . 183.1.2 Newtons Third Law . . . . . . . . . . . . . . . . . . . 193.1.3 Force is generated by an interaction . . . . . . . . . . 20

3.2 Describing physical systems with equations . . . . . . . . . . 203.2.1 Example: Analyzing a stationary pulley system . . . . 21

3.3 Force diagrams Internal and external forces . . . . . . . . . 23

3.4 Some useful math . . . . . . . . . . . . . . . . . . . . . . . . . 233.4.1 Unit vectors and components of vector equations . . . 233.4.2 The dot product . . . . . . . . . . . . . . . . . . . . . 243.4.3 Counting equations and unknowns . . . . . . . . . . . 25

3.5 Strategy for analyzing static systems . . . . . . . . . . . . . . 26

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4 The concept of torque 274.1 What is this thing called torque? . . . . . . . . . . . . . . . . 284.2 The direction of the torque vector . . . . . . . . . . . . . . . 294.3 The magnitude of the torque vector . . . . . . . . . . . . . . 304.4 Computing the torque using the cross product . . . . . . . . . 314.5 A nontrivial example of the computation of torques . . . . . . 32

5 Torque balance in static systems 36

5.1 In static systems, the choice of origin is arbitrary . . . . . . . 365.2 Example of torque balance analysis . . . . . . . . . . . . . . . 37

5.2.1 Start with a diagram . . . . . . . . . . . . . . . . . . . 385.2.2 Count the unknowns and equations . . . . . . . . . . . 38

5.2.3 Choose an origin and Cartesian axes . . . . . . . . . . 395.2.4 Write the force and torque balance equations . . . . . 395.2.5 Solve for the desired quantity . . . . . . . . . . . . . . 405.2.6 Check you answer using limiting cases . . . . . . . . . 40

6 Torques due to gravity and contact forces 42

6.1 Center of gravity and center of mass . . . . . . . . . . . . . . 436.2 Torque due to gravity . . . . . . . . . . . . . . . . . . . . . . 446.3 Example: A beam supported by columns . . . . . . . . . . . . 44

6.3.1 Treating the beam as the system of interest . . . . . . 456.3.2 What does a negative value ofF mean? . . . . . . . . 46

7 Frictional forces and associated torques 47

7.1 A simple approximate law for static friction . . . . . . . . . . 487.2 Example: A block on a ramp . . . . . . . . . . . . . . . . . . 497.3 Torque from a distributed force . . . . . . . . . . . . . . . . . 507.4 Criterion for tipping over . . . . . . . . . . . . . . . . . . . . 517.5 A nontrivial example: Pulling a chair . . . . . . . . . . . . . . 51

7.5.1 Draw a diagram and count unknowns and equations . 527.5.2 Identify the precise question of interest . . . . . . . . . 53

8 Static fluids: pressure and buoyancy 54

9 Concepts, Terms, and Techniques to Know 55

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1 Introduction

Physics is the study of the fundamental laws of nature and their conse-quences. Much of the beauty of the discipline derives from the fact thata few simple laws can be invoked to explain a tremendous range of physi-cal phenomena. The simplicity of the fundamental laws, however, does notguarantee that we can easily understand their consequences. Physical sys-tems can exhibit complicated behavior or consist of huge numbers of piecesthat collectively do things we would never have guessed were hidden in thesimple laws.

1.1 Two kinds of fundamental laws

When we speak of fundamental laws, we really have two different kindsof law in mind. First, we have the general framework that determines themeaning of the fundamental concepts of mass, force, distance, time, the na-ture of the building blocks of matter (subatomic particles), and the mannerin which particles behave when subjected to forces of unspecified origin. Inthe 20th century, the classical, or Newtonian framework was supersededby the relativistic, quantum mechanical framework. We now know that un-der extreme conditions matter and energy behave in ways that are highlycounterintuitive and can only be accurately described with the aid of rathersophisticated mathematical constructs. Quantum mechanics, special rela-tivity, and general relativity teach us that very small, very fast, and verymassive systems behave in ways that are not captured correctly by the sim-ple laws elucidated by Newton in the 17th century. Quantum mechanics isof particular importance for engineering, because it plays a crucial role indetermining the properties of familiar materials compressibility, electricalconductivity, etc.

The second type of fundamental law is a specific theory describing theforces that arise between particles. These are the laws of gravity, electromag-netism, the weak nuclear force, and the strong nuclear force. At present, it

is well established that electromagnetic and weak nuclear forces are differentaspects of a single electroweak force, just as electrical and magnetic forcesare joined in the theory of electromagnetism. There are also clear indicationsthat the strong nuclear force and electroweak force may be manifestations ofa unified theory. Gravity is proving more difficult to understand, but recentadvances in string theory suggest that all forces, including gravity, maybe related at the most fundamental level.

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The fundamental laws of physics express deep insights into the natureof space, time, matter, and physical observation. Unfortunately, it is quitedifficult to communicate these insights directly, as they require a certainmathematical and physical vocabulary that takes years to learn. And evenif we could pass right to the relativistic quantum mechanical descriptionof the world, we would find that the overwhelming majority of familiar,everyday phenomena from human to geological scales are most efficientlydescribed using the good old classical theories developed between 1600 and1900. The corrections predicted by the true theories are almost alwaystoo small to measure.

1.2 Getting started: Classical physics of static systems

So we need to make some choices about how to approach the tremendousvariety of phenomena that fall within the domain of physics. In this course,we will take a practical perspective. We will learn the classical descriptionsof the forces acting on objects and study the manner in which the objectsmove in response to those forces. Here are three reasons for taking thisapproach:

1. Historically, this is how physics as a discipline began, which meansthat many of the more modern theories are built on the conceptual

foundation that we will study.

2. For pedagogical purposes, it helps to study first those phenomena thatare familiar to all of us, without having to be distracted by the exacttheory of tiny effects that are quite difficult to observe and measure.At least that way we can have some intuitive clue about whether ourconclusions make sense, and we will get the benefit of seeing the rele-vance of the topics we study to our daily experience of the world.

3. With the notable exceptions of recent developments in nanotechnologyand materials engineering, the vast majority of engineering applica-tions are well served by the classical theories.

Now even the subject of classical physics turns out to require multipleconcepts that build on each other in important ways. We cannot simply pickwhatever physical phenomenon most interests us and start with that. Forreasons that may not become entirely clear until you have completed severalmore physics and engineering courses, it makes sense to begin with the

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analysis of simple mechanical systems balls, ramps, beams, tubs of water,springs, planets, etc. We leave such objects as electric and magnetic fields,light, heat, flowing fluids, and more complex materials for later semesters.

Almost all modern introductory physics textbooks, including CLRC, be-gin with the description of the motion of single particles, then move onto systems of particles and more complicated rigid bodies. Though thisapproach has its merits, we will take a different one in the interest of em-phasizing physics concepts of particular importance in engineering. We willfirst develop the mathematical language for dealing with static systems,systems in which all the forces on each object are balanced so that nothingmoves. After learning how to analyze the forces at play in static systems,

some of which may be rather complex, we will turn to the question of dy-namical processes that result when forces do not balance and follow thebooks presentation.

This handout presents the topics we will cover in the first seven lecturesof the course. The topics covered here can also be found in CLRC, but therethey are distributed over several different chapters.

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2 The concept of force

Key points

Section 2.1: You need to know what sorts of forces to look for when you en-counter a mechanical system whose behavior you want to understand.Here they are:

Fundamental forces

Gravity: The downward force (toward the center of the earth) we

know as weight.

Electrical and magnetic forces: For present purposes, these referonly to familiar electric and magnetic phenomena, such as theattractive force between a magnets and a piece of steel or betweena balloon containing a static electricity charge and a wall.

Effective forces arising from atomic scale electromagnetic interactions:

Normal force: The force perpendicular to a surface that preventsanother object from passing through or denting the surface.

Static or sliding friction: The force parallel to a surface that resiststhe sliding of another object along that surface.

Buoyant force: The force on a (partially) submerged object due topressure variation in a fluid.

Drag: The force that tends to slow an object down as it movesthrough a fluid (gas or liquid).

Thats all. Look for examples of these sorts of forces as you walkaround campus, and ask your friends or instructor about any situationin which the forces at play dont seem to be classifiable as one of the

above.

Section 2.2: Force is a vector quantity (unlike mass, for example, which isa scalar). An object will sit still if and only if the vector sum of theforces acting on it is zero. If an object is stationary when there are N

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distinct forces acting on it, we may writeN

n=1

Fn = 0. (1)

To work with an equation of this type, we often choose a particularCartesian coordinate system and write separate equations for the x,y, and z components of the force.

Explanations

2.1 Types of forces

The above list includes four types of forces classified as effective forces asopposed to fundamental. The distinction is important for understandingthe unity and simplicity of physical law. Effective forces represent the neteffect of the actions of vast numbers of interacting particles. A completetheory of them would reveal how the fundamental electromagnetic forcesoccurring between atoms gives rise to the readily observable forces familiarto us, such as the force a spring exerts when compressed or stretched. Yourphysics professor would love to talk over such theories with you, but the

discussion requires a certain physics and mathematics vocabulary that itwill take some time to learn. We begin by clarifying precisely what we meanwhen we speak of forces and learning how to identify where they are actingin simple mechanical systems.

2.1.1 General discussion of gravity and contact forces

Think for a minute about a parked car just sitting there on a flat road.What determines how strong its parts have to be just to keep it from col-lapsing under its own weight? Obviously, they need to be strong enough to

withstand the forces acting on them. But what are those forces? How canwe give a quantitative description of them and a theory for how to predictthem without measuring every single one?

Well come to the quantitative description soon, but first lets just seeif we can even name the quantities we are hoping to describe. Perhaps themost obvious force at play in this system is gravity. This is the force that

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pulls the car downward, toward the center of the earth, and is ultimatelyresponsible for the fact that most parts have to withstand any force at all.We will have more to say about gravity later in the course. For now, wejust note that each part of the car is pulled downward by gravity and thestrength with which it is pulled is called its weight.

Lets focus now on one particular part, the rim of one wheel for example.We know it is being pulled down by gravity. So what holds it up? Atthe most basic level, the answer is electric and magnetic forces that actbetween the molecules of tire and those of the rim. To attempt to analyze thewhole car in terms of its constituent molecules would be terribly inefficient,however, so it is useful to make a catalog of the several different ways in

which electromagnetic forces make themselves felt at a more macroscopiclevel.1

The first place to look for the effects of the electromagnetic force is inobjects that are touching each other. Since strong electrical forces betweenatoms prevent the two objects from passing through each other, there is acontact force between them. For example, there is a contact force betweenthe tire and the rim, and this force somehow prevents the gravitational forceon the rim from pulling it downward. There are also contact forces betweenthe rim and the axle it supports, between the axle and the shock absorbersit supports, between the shock absorber and the chassis it is attached to,etc. Ultimately all of these forces conspire to prevent each individual partfrom moving under its gravitational weight.

But there is a lot more going on in the car. There are actually two dif-ferent kinds of contact force at play: the normal force that prevents objectsfrom passing through each other; and a frictional force that resists the ten-dency of the objects to slide past one another. Frictional forces are crucialfor holding the car together, as they are what prevent bolts from unscrewingwhen pulled on. The threads of the bolt are pressed against the threads ofthe nut that holds it on. If it werent for the frictional force that counteractsthe tendency of the bolt to turn, it would simply slide out, twisting as itgoes, whenever it experienced a force pulling it out of the nut. The ability

to identify all of the normal and frictional forces at play in a given physicalsystem is one important skill that this course aims to develop.

1The system under consideration happens to be one in which there are no significantmacroscopic electrical or magnetic forces. Systems with strong macroscopic electromag-netic forces do exist, however, and will be the source of much wonder and excitement nextsemester.

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2.1.2 General discussion of stress and rigid bodies

Gravitational and contact forces are measures of how strongly one objectpulls or pushes on another. This is not quite all we need to consider, though.It will also be useful to be able to talk about the state of the interior of anobject that is subject to some forces. The term we use to describe theinternal state is stress.

The simplest example of stress is the tension in a string that is beingpulled from both ends. The tension at a given point on the string is a mea-sure of the strength of the forces that pull the mass at that point in oppositedirections. For an ideal string by which we mean one with negligibly small

mass the tension is the same at all points along the string. Another simplecase is the compression in a rod being pushed from both ends or a columnholding up a weight.

Fundamentally, we know that what holds the string together is electro-magnetic forces within and between molecules. When we pull on one end ofthe string, the molecules at that end stretch and move ever so slightly andthereby exert forces on the neighboring molecules above them. This messof intermolecular forces is conveniently described at the macroscopic levelby saying that the string is under tension. Similarly, when we say that acolumn holding up a roof beam is under compression, we are summarizingthe macroscopic effect of all the molecular interactions within the column.

All real materials deform from their unstressed state when held undertension or compression. In many cases, the deformation is small enoughthat useful explanations of the objects behavior can be found under thesimplifying assumption that there is no deformation at all. When we makethis approximation, we say that we are describing a rigid body. We will treatall of the objects we discuss this semester as rigid bodies, with three notableexceptions: springs, which undergo substantial changes in length when heldunder compression or tension; strings, which can bend around obstacles buthave fixed length; and fluids, which take the shape of their container andcan be moved out of the way by floating or submerged solid objects.

A third example of stress is the pressure in a gas that is being confinedby a container, say a balloon. The force the balloon skin exerts on the gasfrom all directions results in a pressure. The concept of pressure will allowus to analyze situations like a boat floating in water or a stone feeling lighterwhen it is submerged in a fluid. The force that a fluid exerts on objects dueto the variation of the pressure within the fluid is call a buoyant force.

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2.1.3 Looking ahead

How would you describe the way that a tire holds up a rim using the con-cepts of tension, compression, and pressure? This is a tricky question! Toanalyze the situation fully you have to have a detailed theory of the materialproperties of the tire, the rim, and the air inside the tire. For example, youhave to know how the air pressure varies from point to point. We wontworry about that for now, but you might enjoy arguing about what holdsthe rim up if you and a friend think you understand pressure already.

We will discover that for the purpose of answering many interestingquestions about complicated physical systems like our car, there are useful

theoretical shortcuts. We can say alot about the car without calculating thedetailed deformation of the tire, for example. It is worth keeping in mind,however, that in some circumstances it is precisely these sorts of details thatare the focus of an engineers attention. (Suppose you were asked to designthe precise structure of the part of the tire that attaches to the rim.)

Now reading prose descriptions of tires and rims and the forces betweenthem will only get us so far. In order to communicate with sufficient precisionand to be able to make useful calculations, we have to develop a vocabularyof mathematically precise terms and a system for representing the quantitiesof interest pictorially. Look back at the list of types of forces at the beginningof this section and think about how you could communicate the importantfeatures of a given force to a friend. Note that it is not good enough to sayonly how strong the force is; you also have to specify in what direction it acts.This may seem trivial when the force is gravity and you know your friendalready understands that gravity pulls things downward, but what aboutother forces? If your car is parked on a hill, for example, how would youdescribe the force of friction that keeps it from sliding down the hill? In thenext subsection, we begin building the necessary vocabulary for discussingsuch situations.

2.2 Force is a vector quantity

To fully describe a given force we have to specify both its strength and itsdirection. The appropriate mathematical construct for representing a forceis a vector. A vector is a single entity, often pictured as an arrow, thathas both a magnitude and a direction. When we use a vector to describea force, the magnitude tells you how strong the force is and the direction

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+ =

(b)

(a)

Figure 1: Adding vectors. (a) The sum of two vectors. The solid arrow onthe right represents the sum of the two dotted arrows. (b) A case where the

sum of two forces is zero, so that the object the forces are acting on remainsstationary.

tells you which way the forces pushes or pulls. For example, when a car sitson a level road, it experiences a gravitational force (due to its gravitationalinteraction with the earth) pulling it downward and a normal force (dueto the interactions between its tires and the road) pushing it upward. Theforces have the same magnitude (or strength), but they act in oppositedirections and hence correspond to different vectors.

2.2.1 Vector addition

We use boldface letters with little arrows over them to designate vectorquantities. The symbol Fg, for example, might specify the force due togravity on the car. The F reminds us that the vector in question representsa force; the subscript g here is a label that reminds us we are referring onlyto the force that gravity exerts on the rim, which may be one of many forcesacting on it.

The great thing about using vectors to represent forces is that there isan easy rule for determining the net effect of two forces acting on the sameobject. To add the two force vectors, we simply place the tail of one arrowat the head of the other, then construct the arrow that goes from the tailof the latter to the head of the former, as illustrated in Figure 1(a). In thecase of the tug of war shown in Figure 1(b), the two forces have the samestrength and exactly opposite directions, making the result of adding themtogether may be a vector of zero length.

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2.2.2 Vector notation expressing force balance

The crucial physics point is that the simple definition we have chosen doescorrectly describe the combined effect of two forces acting at the same timeon one object. Force is a vector quantity. Most important for us in ourstudy of statics is that any object will sit still if only if the forces acting onit cancel, by which we mean that the vector sum of all of the forces actingon it is exactly a vector of magnitude zero. If an object is stationary whenthere are N distinct forces acting on it, we may write

F1 + F2 + F3 + . . . + FN = 0; (2)

orN

n=1

Fn = 0. (3)

The rest of this section explains mathematical techniques for turning equa-tions like the ones above into algebraic equations involving ordinary variablesand numbers rather than vectors.

2.2.3 Representing vectors with Cartesian coordinates

The first construct we will need is a way to specify a vector using numbers

rather than just a picture of an arrow. We begin by choosing a Cartesiancoordinate systemconsisting of three mutually perpendicular coordinate axes the x-, y-, and z-axes. To fully describe a vector (or arrow), we write atriple of real numbers representing the x, y, and z coordinates of the headof the arrow assuming that its tail is at the origin, (0, 0, 0). For example, todescribe the forces acting on the rim of a car, we might choose the z-axis tobe vertical, the x-axis to align with the direction the car is facing, and they-axis to align with the direction of the axle, as shown in Figure 2.

Now we have to decide which direction along a given axis correspondsto increasing the coordinate. Does x increase as we move to the right or

to the left? For the most part, the choice is arbitrary. We just have tomake sure to stick to it throughout the entire analysis. There is one feature,however, that will make our life easier later on. We will always make surethe coordinate system is right-handed. To determine whether a Cartesiancoordinate system is right-handed, place the thumb of your right hand inthe direction of increasing z with your palm flat and fingers aligned withthe direction of increasing x. Now, without moving your palm, curl your

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z

y

x

Figure 2: A Cartesian coordinate system. With the thumb of your right

hand aligned with the z-axis and your fingers aligned with the x-axis, youcan curl your fingers to align with the y-axis.

fingers toward the y-axis. The direction they point should be defined as thedirection of increasing y. In our car example, if we define z as increasingupwards and x as increasing in the direction the car is facing (to the left),then y should be defined as increasing as you come away from the rim,outside the car, as shown in the figure.

Given a Cartesian coordinate system with directions labeled x, y, andz, the separate movements along the x, y, and z directions required to get

from the tail of some vectorF to its head are called the components of

F.Let these components be denoted Fx, Fy, and Fz. We can then specify the

vector F by giving its three components in the following notation:

F = (Fx, Fy, Fz). (4)

In the example of the gravitational force on the rim, we have

Fg = (0, 0,W) (5)

and we say that the z-component of Fg is W and the x- and y-components are both 0.

The utility of specifying vectors using Cartesian coordinates is that itmakes addition of vectors extremely easy. Given two vectors a = (ax, ay, az)

and b = (bx, by, bz), we have

a + b = (ax + bx, ay + by, az + bz). (6)

Its that simple.

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2.2.4 Units of force

Note that the numerical value of W in the above example will depend onthe units in which force is measured pounds, ounces, etc. For now, we willsimply make up a unit of force and give it the name Newton in honor ofour hero. We might say, for example, that the force required to hold up aliter of water at the surface of the earth is 9.8 Newtons, or 9.8N for short.(We will see later that the Newton is closely related to the kilogram, themeter, and the second.)

2.2.5 Adding several vector forces

In the case of the stationary ball hanging from a string, we can conclude thatthe force the string exerts on the ball is Fs = (0, 0, Wball), where Wball is theweight of the ball, because we know that the string force has to cancel thegravitational force Fg = (0, 0,Wball). The situation would not be staticunless Fs + Fg were equal to the zero vector. (This assumes, of course, thatthere are no other forces acting on the ball.)

We can, of course, combine more than two forces at once. As Figure 3shows, the order in which we arrange the arrows makes no difference. Vectoraddition follows the same rules as ordinary addition of numbers.

Figure 3: The order in which vectors are added does not matter. Each setof three arrows of the same style consists of the same three vectors addedin a particular order. All the different orders give the same result, indicatedby the dashed arrow with the large head.

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Figure 4 shows three strings attached to a ring. If the ring is knownto be stationary (i.e., the situation is static), the forces that the stringsexert on the ring must sum to zero. In order to answer questions aboutthe magnitudes and directions of the forces, we would like to express thisgeometric relation as a set of algebraic equations we can work with.

3

1F2

F

F

Figure 4: Left: A picture of a real system of three strings attached to a ringand held under tension. The hanging weights exert forces on the stringson the left and right. The peg at the top exerts a force on the verticalstring. The forces pulling on the strings must lie in the directions of thestrings themselves. (If the strings were stiff beams, forces perpendicular to

the beams could also exist.) Right: A free body diagram of the ring.

2.3 The magnitude of a vector

We close this section with one more definition. For the mathematics orphysics purist, there is a deeper meaning of the term vector quantity thatimplies the correct addition rule. In analyzing a given physical system, weare free to choose the Cartesian x and z axes directions. (The y directionis then fixed.) For a vector quantity, a different choice of axes changes thenumbers used to specify it in a particular way. For example, a force specified

by the vector (0, 0,W) when z-axis is chosen to be vertical, might berepresented by the vector (W, 0, 0) in a system where the x-axis is vertical.

In contrast to vector quantities, there are some physical quantities thatare independent of coordinate choices of axes. These are called scalars.Perhaps the simplest example is the mass of an object, which clearly doesnot depend on the choice of coordinate directions. Another example of a

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scalar is the strengthof a force, which youll recall is designated by the lengthof the vector representing that force. This quantity is often denoted withsimple italics; the strength of a force F is F. We also say that the scalarF is the magnitude of the vector F. Using the Pythagorean theorem twice,you should convince yourself that

F =

F2x + F2

y + F2

z . (7)

Suppose that F1 and F2 in Figure 4 are perpendicular to each other. Canyou prove that F3 must be greater than F1 and also greater than F2? If so,you are probably getting the point. If not, talk this over with your TA orinstructor.

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3 Newtons third law and force balance equations

Key points

Action-reaction pairs: Any force that is exerted on an object must arisebecause of the objects relation in space to some other object. Becausethe force exerted on object A and the force exerted on object B areboth consequences of the same relation between objects A and B,thetwo forces themselves are closely related: if at some instant in timeobject A exerts a force FAB on object B, then at that same instantobject B must be exerting a force FBA on object A, and it must be

true thatFBA = FAB. (8)

This is a mathematical expression of what is known as Newtons thirdlaw. You must understand exactly how it applies to all of the kindsof objects you care to contemplate.

Force balance equations: For static systems, we can construct systems ofequations relating various external forces acting in the system. Solu-tions of these equations then represent possible configurations of forces.The most common type of equation we encounter is one stating thata certain set of forces sum to zero:

F1 + F2 + F3 + . . . = 0. (9)

To analyze sets of equations of this type, it is useful to separate outindividual components of the vectors. A mathematical operation thataccomplishes this is the dot product.

F G = FxGx + FyGy + FzGz; (10)= F G cos(). (11)

The component of F in a given direction is given by F i, where i is aunit vector in the direction of interest.

Explanations

Isaac Newton was the first to formulate a coherent and useful theory offorces. One of his great achievements was to recognize that forces obey therules of vector addition (though he did not use that term). Another was

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the articulation of three laws that govern the motion of rigid objects andthe effects that forces produce on them. Since these laws proved to haveextraordinary predictive and explanatory power, we speak of them as lawsof nature rather than just Newtons theories. As mentioned above, we nowunderstand them to be consequences of even deeper principles concerningthe nature of space and time, but that is a topic for another time and place.

3.1 Newtons Laws

Here are Newtons laws as he presented them in his masterwork, The Math-ematical Principles of Natural Philosophy (commonly referred to as The

Principia), published in 1687. After a few pages concerning the definitionsof terms like quantity of matter (now called mass), quantity of motion(now called momentum), and force (now called force) he states what hecalls the Laws of Motion:2

Law 1: Every body perseveres in its state of rest, or uniform motion in aright line, unless it is compelled to change that state by forces im-pressed thereon.

Law 2: The alteration of motion is ever proportional to the motive forceimpressed; and is made in the direction of the right line in which that

force is impressed.

Law 3: To every action there is always opposed an equal reaction: or themutual actions of two bodies upon each other are always equal, anddirected to contrary parts.

We will put off for a few weeks the discussion of the aspects of these lawsthat deal with motion and continue to focus on static systems.

3.1.1 Implications of Laws 1 and 2 for static systems

There are two important implications of Laws 1 and 2 for static systems.First, a system can be thought of as static if all its constituent parts areeither completely stationary or moving together at a constant speed in astraight line. That is, a parked car is a static system, and so is a car parked

2These translations from the Latin are from The Principia, translated by AndrewMotte (Prometheus Books, 1995).

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on a moving train, as long as the train is moving at constant speed in astraight line. Second, if the forces acting on a static object were to becomeunbalanced, resulting in a net force that is not zero, some sort of motion ofthe car will result.

3.1.2 Newtons Third Law

For now, we focus on the third law, as it is the one that will really help usidentify all of the forces acting on any given object in static equilibrium.The meaning of the word action in Law 3 is is not entirely clear withoutthe definitions that precede the statements of the Laws in The Principia.

Roughly speaking, it means force. After stating the law, Newton offersthis clarification:

Paragraph directly following Law 3: Whatever draws or presses an-other is as much drawn or pressed by that other. If you press a stonewith your finger, the finger is also pressed by the stone. If a horsedraws a stone tied to a rope, the horse (if I may so say) will be equallydrawn back towards the stone, for the distended rope, by the sameendeavor to relax or unbend itself, will draw the horse as much towardthe stone, as it does the stone towards the horse, and will obstruct theprogress of one as much as it advances the progress of the other. . . .

It is absolutely critical that you understand Law 3, as it describes atruly fundamental feature of the world and is used over and over again inthe analysis of mechanical structures of all sorts. Here is another version:

Law 3 in modern language: When one body exerts a force on another,the second exerts a force on the first. The two forces are always equalin magnitude and opposite in direction.

Let us use the notation Fij to refer to the force exerted by body i on bodyj. 3 A more succinct version of the third law is this:

Law 3 in vector form: F12 = F213Note that the subscripts here have a very different meaning than the g we used before.

In the present case, we are using the subscripts to designate which object is exerting theforce and which is being subjected to that force. The subscript g was used to denotethe type of force being exerted in a situation where there was no ambiguity about whichobject was the subject of the force.

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3.1.3 Force is generated by an interaction

Newtons third law governs the two forces that arise from a single interac-tion. The contact interaction between a finger and a stone, for example,produces two forces: one on the finger and one on the stone. Likewise, thegravitational interaction between the earth and the moon produces both aforce on the moon and a force on the earth. It is because the forces arise fromone and the same physical interaction between the two objects in question

that they must have the same strength and opposite directions.

Imagine that you press on a stone with two fingers at the same time.What would Law 3 say about the forces between you and the stone? What

it really says is that the force each individual finger applies to the stone isequal in magnitude and opposite in direction to the force the stone appliesto that particular finger: Ff1,s = Fs,f1 and Ff2,s = Fs,f2. By summingthe two forces we can see that the third law also applies to the total forcebetween you and the stone: Ff1,s + Ff2,s = [Fs,f1 + Fs,f2] We will seelater, however, that the position where the force is applied to an object canbe very important, so unless you are sure that the context does not requirethis information, you should analyze action-reaction pairs in terms of theindividual interactions that produce them.

3.2 Describing physical systems with equations

In developing a physical description of a mechanical system, we first needto identify the constituent objects of interest and the forces that act onthem. Any portion of the full system can in principle be treated as a singleobject. In describing a car, for example, we might think of a wheel as asingle object even though we know that it is made of many parts tire, rim,hubcap, bolts, etc.

Our next goal is to be able to formulate a set of equations relating theforces in a given situation. We can then analyze the equations using familiartechniques of algebra to learn what we can about any unknown quantities

in the equations. To do this, we have to be able to consider a situationlike that shown on the left in Figure 5 and construct from it a picture likethat shown on the right. We can then clearly identify the relevant relationsbetween forces and then formulate equations relating them.

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FA

Fg

FN

FB

A

Figure 5: A simple pulley. One end of a string is attached to a block and theother is pulled downward. The mass of the string is very small comparedto the mass of the block. Left: A picture of a real system. Right: Adiagram identifying the quantities relevant for answering questions aboutthe strengths of the forces acting under various conditions. FA representsthe downward force on the rope (applied by the hand); FB represents theupward force the string exerts on the block; FN represents the normal forcethe ground exerts on the block; and Fg represents the force the earth exertson the block due to gravity.

3.2.1 Example: Analyzing a stationary pulley system

For example, we could ask What is the largest downward force that can beapplied at point A without lifting the weight? Having already drawn anappropriate diagram, we now list the physical facts that we know about thesystem:

Since the string is assumed to be massless, the tension along it is thesame everywhere, which means the force exerted by the hand must

have the same strength as the force exerted by string on the block.(Note the use of Newtons third law here. We assume, almost withoutthinking about it, that the force exerted by the block on the string isthe same as the force exerted by the string on the block.)

Since the block is assumed to be static, the vector sum of the forcesthat act on it must be zero.

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Because the floor is not sticky, the normal force it exerts on the blockmust be upward. It cannot hold the block down if the upward forcefrom the string is strong enough to lift it.

The hand is pulling the string straight down, not off at some angle. We may assume that the ground is level and the string segments are

vertical, so there are no horizontal forces that need to be analyzed.

These physical facts can now be translated into the language of mathe-matical equations:

|FA| = |FB|; (12)FB + FN + Fg = 0; (13)

FN z > 0; (14)FA = FAz, (15)

or

FAk = FBk; (16)FBk + FNk + Fgk = 0; (17)

FN > 0; (18)

where in the second set of equations we have explicitly exhibited the as-sumption that the x and y components of all forces are zero. Any solutionof these equations is a possible static configuration of the system.

The question we have posed for ourselves is now stated in mathematicalterms as What is the largest value ofFA for which all of the above equationscan be simultaneously satisfied? We have achieved our goal of trans-lating a physical question into a math problem. Finding the answermay or may not be easy, but we have made a great leap by formulating theproblem in mathematically precise terms. This will be a recurring theme inthe course. Almost all of the difficulty involved in solving physics problemswill arise in the translation of physical questions into math problems. Oncewe have the math problem in front of us, we can haul out whatever machin-ery is necessary to solve it, from pencil and paper to sophisticated softwarepackages like Matlab. If you can write down correct and complete sets ofequations relevant to the physical situations presented to you this semester,you will surely do well in this course. Solving them will turn out to be mucheasier than setting them up.

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3.3 Force diagrams Internal and external forces

Now lets think more carefully about the physics involved in passing from apicture of a real system to a useful idealized diagram. In cataloging the forcesthat act on any given object, we first must distinguish between internal

forces and external forces. The distinction here is not a fundamental one.It depends completely on what part of the world we take to be the objectof interest. External forces are the forces on the object that are generatedby an interaction with some part of the world that is not part of the object.Internal forces are the forces from interactions between one part of the objectand another part of the same object. In the example of the wheel, the force

the road exerts on the tire is an external force because the road is not partof the wheel. The force of the tire on the rim is an internal force becauseboth the tire and the rim are parts of the wheel we have taken as the objectof interest. (If our object of interest were the rim only, the force of the tireon the rim would be an external force.)

The reason for making a distinction between internal and external forcesis this: If we want to know the net force on an entire object, we only have toadd up the external forces. Newtons third law guarantees that the internalforces will cancel. That is, if we list all of the forces acting on a compoundsystem like the wheel, we have to include all the forces acting on the rimand all the forces acting on the tire, but the third law tells us that the sum

of the force of the rim on the tire and the force of the tire on the rim iszero. All internal forces will come in action-reaction pairs that cancel. Incontemplating the behavior of a particular object, or a particular part of amechanical system, it is essential to identify first all of the external forcesacting on that object. In the example of the wheel, the external forceswould be the contact force of the road on the tire, the gravitational forceon the entire wheel (which comes from its interaction with the earth), andthe contact force the axle exerts on the rim. (Technically, there is also abuoyant force due to the atmosphere around the wheel, but that is verysmall compared to the others.)

3.4 Some useful math

3.4.1 Unit vectors and components of vector equations

Finally, we consider some mathematical constructs that are helpful in work-ing with vector equations. In order to work efficiently with the external

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force vectors once we have identified them, it is useful to define a few specialvectors along with a notation for them. Any vector of length 1 is calleda unit vector, and we denote the three unit vectors (1, 0, 0), (0, 1, 0), and(0, 0, 1) by the symbols i, j, and k, respectively. We can therefore write

F = Fxi + Fyj + Fzk. (19)

Note that i, j, and k are defined with respect to a particular choice of axes.The direction ofk is not necessarily up. It simply designates that directionin space that has been chosen to be the z-axis.

Now suppose we have a vector equation such as

A + B = C. (20)

By expressing all of the vector components with respect to a given set ofCartesian coordinates, we can write this as three separate equations for thecomponents:

Ax + Bx = Cx;

Ay + By = Cy; (21)

Az + Bz = Cz.

That is, if the vector equation holds, there are three equations for the com-

ponents that must hold separately.

3.4.2 The dot product

It is useful to define a mathematical operation that picks out a particularcomponent of a vector. We define the dot product of two vectors such thatF i = Fx, F j = Fy, and F k = Fz. Thus, for example, the firstEquation (21) can be written:

A + B

i = C i. (22)

Notice that A i is a scalar quantity. It gives the size of the componentof A along the i direction. More generally, the dot product between anytwovectors can be defined as

S R SxRx + SyRy + SzRz. (23)

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You should make sure you see that the example above ofF

i is a specialcase of this formula. (Recall that i (1, 0, 0).)

Though it may not be immediately obvious, it is also true that

S R S R cos , (24)

where is the (lesser) angle between S and R. Another useful relation isthat the dot product of a vector with itself gives the square of the magnitudeof the vector:

S S = S2x + S2y + S2z = S2. (25)In this case = 0, so cos = 1.

The dot product obeys all of the ordinary rules of multiplication. Forexample,

(S + R) T = S T + R T. (26)To gain some confidence in the dot product and/or refresh your memory oftrigonometry, consider the triangle formed by vectors A, B, and C, whereA + B = C. By calculating (C A) (C A), show that you recover theformula B2 = C2 + A22ACcos , where is the angle between C and A.

C

BA

Figure 6: A triangle to use for a dot product exercise.

3.4.3 Counting equations and unknowns

You will often encounter situations in which it is not immediately clearwhether a particular force is known or not. You have partial information

about a static configuration, but you are not sure whether that informationis enough to allow you to determine all of the forces in the system uniquely.The mathematical situation is analogous to having a bunch of equations fora bunch of unknowns and wanting to know whether the equations have aunique solution. In the math case, we know that if the equations are linear,the solution exists and is unique if the number of equations is the same asthe number of unknowns. The same idea works for force balance.

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If we write all the force balance equations in component form, there arethree unknowns for every action-reaction pair of forces. And every vectorequation relating those forces gives us three ordinary algebraic equations. Soin principle, if we have n (vector) forces in a static system, we can solve forall the forces if and only if we have n linearly independent vector equationsrelating them.

In the case of Figure 4, for example, we are given the directions of allthree forces and told that the disk is static. How many of the magnitudes dowe have to know in order to determine all of the forces uniquely? Just one.If we choose coordinate axes such that the k is perpendicular to the plane ofthe disk, then all the z-components of the forces vanish. The force balance

equation F1 + F2 + F3 = 0 then gives two nontrivial equations: one forthe x-components and one for the y-components. Thus we can solve for twounknowns, which in this case would be the two unspecified force magnitudes.If you are trying to find the strength of a component of a certain force anddiscover that you have more unknowns than equations, you have to go backover the physics of the situation until you find a good reason for writing anadditional equation relating the unknown quantities.

3.5 Strategy for analyzing static systems

In analyzing force balance in a static system, always carry out the followingsteps (in order!):

1. Draw a diagram indicating all of the relevant force vectors.

2. Make a list of all of the physical facts you know about the system andwrite a mathematical equation expressing each one.

3. Count the unknowns and equations. If there are enough equations todetermine a solution, solve the resulting math problem to determinethe quantities of interest.

Step 1 is a technique that is known to be tremendously helpful. All physicistsand engineers agree that it is the best way to start. The truly crucial step,however, is step 2. That is where your knowledge of physics is used totransform the physical question into a purely mathematical one. If step 2is done correctly, then step 3 will follow fairly easily, even if the algebra issometimes a bit tedious.

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4 The concept of torque

Key points

The external forces acting on an object create torques, which may causethe object to rotate even if the forces themselves sum to zero. The torqueassociated with a given force must be defined with respect to a specifiedreference point (or origin). For a force F applied at a point r, the torquemay be determined in several equivalent ways:

by direct evaluation of the cross product = r F; by using the formula = rF sin for the magnitude and the right-hand

rule to get the direction of.

The latter method can be carried out most easily by forming the right tri-angle shown in Figure 7 and taking F times the moment arm s.

Fs

r

Figure 7: A useful diagram for calculating the torque due to a force ofstrength F applied at a distance r from the origin.

Explanations

We have seen that for any object in a static system, the sum of all theforces acting on the object must equal the zero vector. It is easy to envisionsituations, however, where the forces sum to zero but the system will notremain static. Consider a stationary disk to which two forces are applied,as shown in Figure 8. The vector sum of the forces is zero, but the disk willnot remain in static equilibrium. The problem is that balancing the forces isnot sufficient for stopping the disk from rotating, even though it is sufficient

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to prevent the disk from translating (moving to the left or right, or up ordown).

[b]

Figure 8: A disk that will move even though it is experiencing no net force.

The motion will consist of rotation of the disk, with no translation.

4.1 What is this thing called torque?

The concept of force and its mathematical representation as a vector helpedus to understand static systems in which there was a possibility of trans-lational motion. We now introduce a new concept, called torque, that isnecessary for analyzing static systems in which rotational motion may alsobe an issue. Combinations of translations and rotations cover all possibletypes of motion, so once we understand torque balance, we will have a com-

plete description of static systems. (In engineering circles, what physicistscall a torque is often called a moment.)

Roughly speaking, torque is a measure of how strongly an object is beingtwisted, in the same sense that force is a measure of how strongly it is beingpushed or pulled. When you try to unscrew the lid of a jar, you dont applyany net force to it. There is no point in pushing or pulling the lid. What youhave to do is make it rotate, not move over. You have to apply a torque toit. If the torque you apply is strong enough to overcome the torque suppliedby friction, then the lid will begin to rotate and eventually come off. If thetorque you apply is not sufficiently strong, a situation of static equilibriumwill arise in which the torque your fingers apply to the lid is balanced by thetorque the bottle applies through friction. (Note that you ultimately supplythe balancing torque yourself by holding the jar with your other hand tokeep it from turning.)

Every force that acts on an object produces torque. Unlike the force,however, which has a completely unambiguous strength, the strength of the

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torque depends on the choice of a reference point for its measurement. Wewill always speak of the torque about a particular point, with the point beingchosen at our convenience. As far as the physics is concerned, the choice ofthe reference point is arbitrary. But a judicious choice can often simplifythe mathematical analysis substantially. Thus we may speak of the torqueon the jar lid about its center, or the torque on a door about a point on oneof its hinges. It may not be obvious to you why the following definitionsmake sense, but they correspond to quantities that obey mathematical rulesin a way that you will eventually appreciate. For now, just take them asdefinitions to be learned.

4.2 The direction of the torque vector

It turns out that torque, like force, is a vector quantity. Well begin with thedefinition of the direction of the torque vector, then deal with the questionof its magnitude. Physically, the direction of the torque vector specifiesthe axis about which the torque promotes rotation. Lets first discuss thespecial case of Figure 8. If the two forces shown are really the only forcesacting on the disk, it will begin to rotate in the clockwise direction aboutan axis passing through the center of the disk while the center of the diskstays in the same place. (In general, figuring out which point sits still canbe complicated, but for now we assume the mass of disk is symmetrically

distributed about its center and trust our intuition.) We say that there isan unbalanced clockwise torque acting on the disk about the rotation axis.

But the term clockwise is not precise enough! Suppose you were look-ing at the disk from the back (from behind the page instead of in front of it).Then the torque would appear to be counterclockwise. To carefully spec-ify the torque, we have to indicate which direction to look from as well aswhether the tendency to rotate will be clockwise or counterclockwise. Physi-cists and engineers have agreed to do this using a right-hand rule, similar inspirit to the convention for labeling Cartesian coordinate axes.

Direction of rotation

vector

(Out of the page.)

Torque

The right-hand rule for torque: Hold your right in

front of your face with the thumb pointing at you and thefingers curled. Your fingers indicate a direction of rotationthat is counterclockwise. We define the direction of yourthumb as being the direction of the torque vector, and wedefine a positive torque to be one that promotes rotation inthe direction indicated by your fingers.

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In Figure 8, the direction of the torque vector is into the page, sincethat is the direction your right thumb would have to point in order foryour fingers to curl in the direction the disk would rotate. Make sure youunderstand this! It might help to go over it with a friend. Stand face to faceand hold a disk between you, then argue about whether the disk is rotatingclockwise or counterclockwise. Resolve the argument by insisting that thedirection of the torque vector is determined by the right-hand rule. If youboth hold your right hand with your fingers curled to show which way thedisk will rotate when given forces are applied, and you define clockwiseand counterclockwise to refer to what the rotation looks like when youposition yourself so your thumb is pointing at you (rather than away fromyou), then you will agree and it doesnt matter whose right hand is used.

To check to see whether you understand how to find the direction ofthe torque vector, consider the following situation. Instead of forces beingapplied to the disk as shown in Figure 8, imagine the two forces to be appliedat the top and bottom of the disk with the force at the top directed into thepage and the force at the bottom directed out of the page (as if someonewere pushing from behind the page). These forces would initiate rotationof the disk about a horizontal axis in the plane of the page. (Picture therotation of a coin being flipped.) Question: Which direction is the torquevector produced by these forces? Work it out for yourself before you lookat the answer in the footnote.4

4.3 The magnitude of the torque vector

We now return to the question of the strength of the torque; i.e., the mag-nitude of the torque vector. It turns out that this can be computed if theforces producing it are known and we know where those forces are beingapplied to the object. To define the magnitude of the torque, we must firstspecify a reference point as the origin of our coordinate system. To everypoint on any object, there then corresponds a position vector r specifyingits location in space with respect to the origin. For example, the points onthe surface of a sphere of radius r centered on the origin correspond to allof the position vectors that have the same magnitude r.

When a force F is applied to an object at position r, a torque is createdabout the origin. Now if the force acts directly along r, it will not promoteany rotation about the origin. That is, if you want to get the object to

4Answer: The torque vector points to the left.

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rotate about a certain point, it doesnt help to push directly toward thatpoint or pull directly away from it. What does help is to push or pull in adirection perpendicular to the vector r; i.e., perpendicular to the line passingthrough the origin and the point of contact where the force is applied. It isonly the component of F perpendicular to r that generates a torque aboutthe origin. This component has a magnitude Fsin , where is the (lesser)angle between r and F. is always less than (less than 180), so sin isalways positive.

The magnitude of the torque due to a force F applied at position r isdefined to be r F sin . The direction of is defined to be perpendicularto both r and F, pointing in the direction the right thumb goes when the

right-hand fingers are arranged to curl from r to F. This is consistent withthe discussion above of Figure 8, where the origin is taken to be the centerof the disk.

4.4 Computing the torque using the cross product

This definition of the torque vector associated with a given force applied ata given position corresponds to a mathematical operation called the vectorproductor cross product, usually denoted with the symbol . The relationbetween the vectors , r, and F is written as

= r F, (27)

which is read equals r cross F. In plain English: the cross product oftwo vectors yields a third vector that is perpendicular to the first two andhas a magnitude equal to the product of the magnitudes of the first twotimes the sine of the angle between them.

Does this definition produce a unique answer? You might worry thatthere could be more than one direction perpendicular to the first two vectors.There are always at least two, of course, which are exact opposites, but thisambiguity is resolved by the right-hand rule. The only way there can be

more than two is if the first two vectors have exactly the same or exactlyopposite directions. But in either of these cases we have sin = 0, so thecross product is the zero vector and it makes no difference what directionwe assign to it.

It turns out that the cross product can also be written in terms of theCartesian components of the vectors involved, though the general formulas

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are not entirely trivial to derive:

x = ryFz rzFy;y = rzFx rxFz; (28)z = rxFy ryFx.

Notice that the second equation can be obtained from the first by changingx to y, y to z, and z to x everywhere. The third can be obtained from thesecond by doing this again. So all you really have to remember is the firstone. Note also that the minus sign makes it very important to keep straightwhich vector is which on the right side of the equation. Because of the signdifference between r

yF

zr

zF

yand F

yr

zF

zr

y, we see that r

F =

F

r.To be consistent with our use of the right-hand rule to define coordinate axesand torques, we must define the torque as specified in Equation (27). In aright-handed coordinate system with unit vectors i, j, and k along the x-,y-, and z-axes, respectively, we have

i j = k;j k = i;k i = j.

(29)

The order of vectors on the left side of these equations matters!

Some mathematical properties of the cross product can be found in theAppendix E of CLRC. For our present purposes, the most important one isthe distributive property over addition:

a (b + c) = a b + ac. (30)

4.5 A nontrivial example of the computation of torques

To illustrate the mathematical definition of torque and see that it matchesyour physical intuition about twisting forces, consider again the disk inFigure 9(a). The figure shows three forces, all having a strength of one

Newton, and we will assume the disk has a radius of one meter.

Lets calculate the torque due to each of the individual forces. The firstthing we have to do is choose an origin about which the torques will bemeasured. The natural choice in this case is the center of the disk. It willalso help to choose the x, y, and z directions of our coordinate system.This choice is arbitrary (except that it has to be a right-handed Cartesian

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(c)1

2

(b)1

2x

y

3

3

(a) 12

3

Figure 9: A disk subject to 3 forces. The arrows with open heads indicate theforces, with the arrow tip indicating where on the disk the force is applied.In (c), the arrows with solid heads indicate the position vectors of the pointof application of the forces with respect to center of the disk.

system). Since all the forces and positions where they are applied lie in asingle plane, the math will be simpler if we choose two axes in that planeand the third perpendicular to the plane. Figure 9(b) shows such a choice.The circle with a dot in the middle indicates an arrow pointing out of theplane of the page (towards you).

Next, we exhibit the position vectors explicitly, as shown in Figure 9(c).We can get the directions of the torques fairly easily now. To find thedirection of 1, place the fingers of your right hand along r1 in such a way

that you can curl them toward the direction of F1. (Dont be fooled bythe drawing here! The direction of F1 is to the right, even though the linesegment representing the vector stretches to the left of the point r.) Withyour hand positioned this way, your thumb points into the page, which isthe direction of1.

Lets check this using the cross product definition of torque. We have

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r1 = (0, 1, 0)m andF1 = (1, 0, 0)N. Thus we get

1 = r1 F1; (31)= (r1yF1z r1zF1y,

r1zF1x r1xF1z, r1xF1y r1yF1x);= (0, 0, 1)Nm;

which indicates a torque of magnitude one Newton-meter in the k direc-tion.

What about 2? Before calculating the cross product, think about whatyou expect the answer to be. Will the force F2 generate a twisting action on

the disk? (If you think it will, which direction will the twist be clockwiseor counterclockwise?)

The calculation is straightforward. We have r2 = (cos , sin , 0)m andF2 = ( cos , sin , 0)N. Noting that all the z-components of both r2 andF2 are 0, we get

2 = r2 F2; (32)= (0, 0, r2xF2y r2yF2x);= (0, 0, cos ( sin ) sin ( cos ))Nm;= (0, 0, 0)Nm.

F2 generates no torque about the origin, as you may have expected.

Finally, we compute 3. Again, all the z-components of both r3 and F3are 0, so the x and y components of3 will vanish; i.e., 3 must be directedalong the z-axis. We have r2 = (1/2)(cos , sin , 0)m and F2 = (1, 0, 0)N.Note the point at which the force is applied is closer to the origin; r is halfa meter rather than one meter as in the other cases.

Again the z-components of the rs and and Fs vanish, so the onlynonzero component of the torque is its z-component. We get

3 = r3F3; (33)

= (0, 0, r3xF3y r3yF3x);= (0, 0, 0 1

2sin )1) N m.

F3 does generate a torque about the origin. The direction of3 is the positivek direction (out of the plane of the page), indicating that F3 promotes

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rotation in the counterclockwise direction as we look down the z-axis. Themagnitude 3 is equal to rF sin , as promised above.

The magnitude of 3 is reduced from 1 N m by two factors. First, it issmaller because r is smaller. In general, the closer to the origin a force isapplied, the smaller the torque about the origin will be. Second, it is smallerbecause the force is not perpendicular to the position vector. It is only thecomponent of F3 that is perpendicular to r3 that generates a torque. Thiscorresponds nicely to your experience in opening doors, for example. If youpush on the door near the hinges, rather than at the doorknob, you needa bigger force to get the door to rotate about its hinges. Also, if you wereto push the doorknob toward the hinges rather than at a right angle to the

plane of the door, you wouldnt generate any torque about the hinges, andmost people, having mastered the basics of how to apply larger torques withless force, would wonder what in the world you are doing.

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5 Torque balance in static systems

Key points

In static systems both the net force and the net torque must be zero. Tocompute the net torque, we may choose any convenient point as an originand take the vector sum of the torques about that point due to externalforces. If we know that an object is in static equilibrium, then we know thatthe following two vector equations must hold:

n

Fn = 0 (34)

n

n = 0, (35)

where the sums are taken over all of the external forces applied to the objectand the torques are all referred to the same origin.

Explanations

We have already seen what it means for forces to balance. The situationwith torques is entirely analogous. We just add up all the torques on eachobject as vectors. In order for the system to be static, the torques on anygiven object must sum to zero.

There does appear to be a problem, though. Since the torque producedby a given force depends on the choice of origin, how do we pick the correctorigin when checking for torque balance? The answer is both elegant andsurprising: it doesnt matter which point we choose as the origin. For astatic system, the torque is zero no matter which point is chosen as theorigin.. (The choice of origin is not arbitrary for a system that is not static.Well have much more to say about dynamic systems later.)

5.1 In static systems, the choice of origin is arbitrary

It is not hard to prove using simple algebra that the choice of origin has noeffect on the analysis of torque balance in a static system. Suppose we haveseveral forces F1, F2, . . . , Fn applied to an object at positions r1, r2, . . . , rn,where the position vectors refer to some particular choice of origin. Then

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we have for the net torque on the object about that origin:

=n

i=1

ri Fi

. (36)

Now suppose we decided to use a different point as the origin, and lets saythe new origin lies at position R with respect to the old origin. Then inthe new coordinates, the point of application of Fi will be at the positionr = r R. (See Figure 10.) So with respect to the new origin we have

=n

i=1 ri Fi

(37)

=n

i=1

(ri R) Fi

(38)

=n

i=1

ri Fi

ni=1

R Fi

(39)

= R

ni=1

Fi

, (40)

where we have used the distributive property of the cross product to get thethird line, then used it again to get the second term on the fourth line. Butif the object is static, then we know the sum of the forces on it must be zero.

Thus the second term in the last equation vanishes and we have shown that = as long as there is no net force. So for analyzing static systems, itmakes no difference which point we choose as a reference for measuring thetorques on an object, as long as we refer all the position vectors to the sameorigin. If we have several objects in our system, we can use a different originfor each of the different objects.

5.2 Example of torque balance analysis

To illustrate the way in which torque balance for enters the analysis of

static systems, we consider first a single rigid beam. If all of the points ofcontact of the forces are given, then solving for the forces is a simple linearproblem. We just write down the force balance and torque balance equationsin component form and we have a set of linear equations for the componentsof the forces.

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r

R

r

O

O

P

Figure 10: The relation between the position vectors referred to two differentorigins. Point P corresponds to the vector r in the coordinate system withits origin at O, and to r in the coordinate system with its origin at O. R

is the displacement of O

from O.

5.2.1 Start with a diagram

The situation can be a little trickier, however, if we are given different in-formation. Suppose that we are told that the beam is subject to two forcesas shown in Figure 11. (There is no gravity in the world of this beam.) Weare then asked to figure out how to apply a single additional force that willbalance the two shown so that the beam will remain static.

45

O

1 N F1

2 m 2 m

0.75 m

1 N

90O

F2

2.5 m

Figure 11: A beam that remains static because the forces and torques onit are balanced. One of the forces is not pictured. The problem is to figureout what it is and where it is applied.

5.2.2 Count the unknowns and equations

It may not be entirely clear that it is possible to balance the force and torquesimultaneously, so it behooves us to count the number of unknowns we have

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to determine and the number of equations they have to satisfy. We havetwo unknown vector quantities, the desired force F and the position r of itsapplication. We also have two vector equations to satisfy, one expressingforce balance and the other expressing torque balance. So it appears thatthere should be a solution. We have to be careful, though, because thetorque equation involves products of the unknowns in the form ofr F itis not a linear equation. Lets plow ahead and see what happens.

5.2.3 Choose an origin and Cartesian axes

First, we are free to choose whatever origin we like for computing the torques.

A good choice here might be the point where F2 is applied to the beam, sincethat will guarantee that 2 is zero, making the torque balance equationsimpler.

Next, we choose our axes. Since all of the action lies in the plane of thepage, we take the z-axis to point out of the page. To make the math assimple as possible, we can choose one of the axes to align with one of thegiven forces. A natural choice in this case might be that j (the y direction),aligns with F2, so i points to the right.

5.2.4 Write the force and torque balance equations

Now we express the forces and points of application in terms of their com-ponents: F1 = (F1/

2, F1/

2, 0)N; F2 = (0, F2, 0)N; r1 = (0.5, 1, 0);

and r2 = (0, 0, 0). Computing the torques using the cross product, we have1 = (0, 0, 0.5F1/

2)Nm and 2 = (0, 0, 0).

Force balance requires

F3 = F1 F2 (41)= (F1/

2, F2 F1/

2, 0) N. (42)

Torque balance requires

3 = 1 2 (43)= (0, 0, 0.5F1/

2) N m. (44)

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5.2.5 Solve for the desired quantity

We now have to find r3 such that 3 = r3 F3. Will the solution be unique?Clearly not, because we can always add to r3 any vector parallel to F3 with-out changing the torque generated by F3. To find this family of solutions,note first that r3z must be zero, which guarantees 3x = 3y = 0. The onlyremaining equation is the z-component of torque balance, which reads

3z = r3xF3y r3yF3x or (45)0.5F1

2= r3x

F2 F1

2

+ r3y

F12

. (46)

From this we derive the relation

r3y = 0.5

2F2F1

1

r3x, (47)

which is the equation of a line with intercept 0.5 and slope 1 2F2/F1.To achieve static equilibrium, the force F3 can be applied anywhere alongthis line.

5.2.6 Check you answer using limiting cases

Having obtained an answer to the question posed, it is now absolutely es-sential that we think about it to see whether it makes sense and what general

insights we can obtain from it. One type of analysis that is often useful is tolook at limiting cases. Consider for example the situation ifF2 is vanishinglysmall. In that case we would be trying to balance the single force F1. Todo so, we would expect to have to apply an equal and opposite force, and toavoid creating a net torque we would expect that force to be applied alongthe line defined by F1 as it is pictured in the figure. Does our mathematicsagree with this simple analysis? If we set F2 = 0 in Equation (42), we find

F3 = (F1/

2, F1/

2, 0) N, (48)

which is indeed a force equal and opposite to F1. If we set F2 = 0 inEquation (47), we find

r3y = 0.5 + r3x, (49)which is, as expected, the equation defining the line in the direction of F1and containing the point at which F1 is applied. You should analyze foryourself the case where F1 = 0. You will need to use Equation (46) rather

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than (47) to avoid division by zero. Before you do the math, though, lookat the picture and formulate an idea of what to expect. The general lessonhere is that in order for static equilibrium to obtain in a system subjectedto two forces, the forces must be equal and opposite and applied at pointsa long a single line that is parallel to the force vectors.

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6 Torques due to gravity and contact forces

Key points

To properly account for the torque exerted on an object due to its grav-itational interaction with the earth, we take the overall gravitational forceFg to be applied at the center of mass of the object. The center of masscan be computed as a sum over all points in the object, with the position ofeach point weighted by the density at that point:

rcm 1M

V

r dr. (50)

When the centers of mass of different pieces of the object are known, thecenter of mass of the composite object is given by a weighted sum over thedifferent pieces:

rcm =

Ni=1(Mircmi)N

i=1 Mi. (51)

The torque on the object due to gravity (with respect to whatever originwas chosen to define the position of center of mass) is

g = rcm Fg. (52)

This torque enters the equations for torque balance in exactly the same wayas any other torque.

Explanations

So far we have discussed torque only in situations involving contact forcesapplied at well defined positions. Gravitational forces do not fit this mold.The gravitational force on an object is distributed over the entire volumeoccupied by the object. Each little piece of the object, down to the sub-atomic scale, contributes its weight to the total Fg. To calculate the torqueassociated with the gravitational force requires summing over all the tinymass elements that make up the object. Letting V represent the volumeoccupied by the object, we can write

Fg =

V

w dr and (53)

g =

V

(r w) dr. (54)

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Here w is the weight per unit volume of the object at each point in spaceand the symbol dr means drx dry drz. w can vary from point to point eitherbecause the object is more dense in some places than others, because gravityis stronger in some places than others, or because the direction of gravity isdifferent in some places than others. (The latter two effects are importantonly if the object is really big. Taking the continent of Africa as an object,for example, we see that gravity, which points toward the center of the earth,has a significantly different direction in Tunisia than it does in Botswana.)

6.1 Center of gravity and center of mass

For convenience, we define a position rcg in space such that the torque dueto gravity, computed using Equation (54) is given by the simple expressiong = rcg Fg. This position is called the center of gravity of the object. Inall of the situations we will encounter henceforth in this course, the variationof the gravitational interaction over volumes the size of the object will benegligible. In that case, we can write w as g, where is the mass density,which may vary from point to point, and g is a constant vector indicatingthe strength and direction of gravity. (Well have more to say about theconcepts of mass and the vector g later.)

By definition, the integral of the density over the volume of the whole

object is just the mass of the object:

V dr = M. (Recall that an integralis just a convenient way of denoting a sum of contributions from all the tinypieces of the object.) We will also find it useful, especially when we discussthe dynamics of rotating objects, to define the center of mass of an objectas the position

rcm 1M

V

r dr. (55)

In a sense, the center of mass is the average position of all the mass inan object. In many cases, the center of mass is easy to guess. In general,for symmetric objects like uniform density disks, rectangles, cubes and such,the center of mass is at the center of the object. Keep in mind, though, thatthe center of mass does not actually have to lie within the object at all. Thecenter of mass of a donut is in the middle of the hole, for example.

The calculation of the center of mass can often be simplified by dividingthe object of interest into different parts whose centers of mass can be com-puted easily. Given two constituent objects, if we know their masses andthe positions of their centers of mass, the center of mass of the composite

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object is given byrcm =

M1rcm1 + M1rcm2M1 + M2

. (56)

More generally, for a composite object consisting of N parts, we have

rcm =

Ni=1(Mircmi)N

i=1 Mi. (57)

Equation (55) is just the continuum version of this sum.

6.2 Torque due to gravity

Now lets return to the computation of the torque on an object due to gravity.Because the cross product has the distributive property over addition andbecause is a scalar, we can rewrite Equations (53) and (54) as

Fg = g

V

dr = Mg (58)

and

g =

V

(r) dr

g (59)

= Mrcm

g (60)

= rcm Fg. (61)

This last line shows that in most applications of interest the torque due togravity will be properly taken into account if we assume that the weight ofthe object is concentrated at the center of mass.

6.3 Example: A beam supported by columns

Consider a beam resting on two supports as shown in Figure 12. We assumethe beam has a uniform density. Just for practice, lets begin our analysis

by identifying all of the forces in this system. Each object (beam or column)is subject to a constant gravitational force, which we can model as actingat the center of mass of the object. Each column exerts a vertical contactforce on the ground, and force balance applied to the entire beam-columnsystem (together with Newtons third law) tells us that the net force on theground must equal the weight of the whole system. Each column is alsosubject to a vertical contact force necessary for balancing the weight of the

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W

F

F

1

2

x1 x20x

Figure 12: A beam supported by two columns. The arrows near the beamshow the forces acting on the beam. The weight is marked as acting atthe center of mass of the beam (not the whole beam-column system) andrepresents only the weight of the beam (not the whole system). The arrowsbelow show the forces the columns exert on the ground.

beam, otherwise the beam would not be static. Is the contact force at the

top of a column equal in magnitude to the force that column exerts on theground? Think about this before looking at the footnote.5

6.3.1 Treating the beam as the system of interest

Focusing now on the beam alone, lets see if we can calculate the contactforces at the two columns. As is often useful, we choose our origin at thecenter of mass of the object of interest. This makes the torque due to gravityzero because it is applied at position r = (0, 0, 0) with respect to the centerof mass.

The two physical facts we know are: (1) the net external force on thebeam must be zero; and (2) the net external torque on the beam must bezero. We now have to express these in the form of mathematical equations.

5No. The contact force on the ground is larger because it must balance the weightof the column itself in addition to the compression force transmitted from the top of thecolumn.

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To be (obsessively) thorough, we would write full vector equations for bothquantities. But in this case, one sees that the forces are all in the z direction,since they come from gravity or contacts at surfaces perpendicular to thez-axis. Also, there is no need here to worry about the beam rotating aboutits long axis (the x-axis), so we only need the equation for the y-componentof the torque. For the two contact forces, we then have two equations:

F1 + F2 W = 0 and (62)x1F1 + x2F2 = 0. (63)

We have used x1 and x2 instead of r1 and r2 since we know the magnitudeof the torque depends only on the component ofr that is perpendicular to

F. Note that x1 is negative while x2 is positive..

These equations are easily solved for F1 and F2, yielding

F1 =

x2

x2 x1

W and F2 =

x1

x1 x2

W. (64)

As usual, in addition to checking these answers to make sure they satisfy theforce and torque balance requirements, you should check to see whether theymake sense in limiting cases. For example, suppose column 1 is positioneddirectly under the center of mass of the beam. In that case, x1 is zero andit appears that there is no force on column 2 at all. Does this make sense?

Yes, because if column 1 is right under the center of mass, then the beamwill be balanced on it and we could remove column 2 completely withouthaving the beam move.

6.3.2 What does a negative value of F mean?

What about if columns 1 and 2 are both on the same side of the centerof mass? In that case F2 becomes negative. Does this make sense? Yes,because in that situation column 2 will have to pull down on the beam tokeep it from tipping to the left, rather than push up on it. Notice that whenx2

x1 is small compared to both x1 and x2, both forces get very large (butin opposite directions). This corresponds to cases where the two columnsare near each other but far from the center of mass of the beam. You canget a direct physical confirmation of the fact that large forces are needed insuch a case by trying to hold a horizontal rod steady gripping it only nearone end. You should confirm this by experimenting with a yardstick, tennisracket, baseball bat, or something similar.

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7 Frictional forces and associated torques

Key points

Static friction is a force that prevents an object from sliding across asurface. The strength of this force is determined by the situation (i.e., theother forces at play) but can never exceed a certain fraction of the normalforce between the surface and the object:

Fstatic friction Fnormal. (65)

The coefficient of static friction, , is a material property of the interface

between the two materials in contact. does not not depend on the size orshape of the contact area.

In the most general situation, in order to compute the torque due tostatic friction (or due to normal forces), we need to know exactly how theforce is distributed over the contact area. In some cases, however, the rele-vant component of the frictional torque turns out to be independent of howthe force is distributed. When the distribution of normal forces matters, theambiguities are often resolved in limiting cases such as when an object isjust about to tip over or just about to slip.

Explanations

Static frictional forces occur when one object is prevented from slidingacross another. When a block rests on a ramp, for example, as shown inFigure 13, there is a frictional force directed along the ramp. The strengthof this force must be just what is needed to balance the sum of the downwardgravitational force exerted by the earth and the normal force exerted by theramp. Note that the normal force has a horizontal component, so it is notpossible for the normal force and gravity to cancel each other. The normalforce and frictional force must somehow combine to cancel the gravitationalforce if the block is to remain stationary.

The physics of friction is extremely subtle. Like the normal force, thestrength of the frictional force adjusts itself to the situation. The strengthof the normal force is as large as it needs to be in order to prevent the twocontacting objects from passing through each other. Similarly,