Newton’s laws - Piazza

Chapter 3

Newton’s laws

Up to now, we’ve only analyzed the motion of objects without regard towhy they move. We’ve simply said that an object had some function a(t)describing its acceleration or x(t) describing its position or that it hadsome initial speed v and launch angle θ. But how did it get there? Andhow does mass factor into everything?

All ventures into the realm of dynamics, which examines the causesbehind motion, begin with Newton’s three laws of motion. Though ar-guably not as elegant as Euclid’s five postulates, they provide all the toolswe need to build mechanics from the ground up.

3.1 Newton’s laws of motion

There are many ways to state Newton’s laws. Perhaps you’ve seen orheard them enough times that they’re ingrained into your mind in oneform or another, but it’s important to understand what they really meanbefore we start to show how they’re used.

First, we’ll give a couple definitions. A force is an influence that,when acting by itself, causes an object to undergo a nonzero acceleration.Roughly speaking, we might think of a push or a pull. Examples of forcesinclude the gravitational force that pulls you down, the normal force theground exerts on you to keep you from falling through the earth, and the

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3. NEWTON’S LAWS

intermolecular forces that bind all the atoms in your body together. Forceis a vector quantity since it has both a magnitude and direction.

The net force, then, is the (vector) sum of all the forces acting on anobject. In this book, unless otherwise indicated, we’ll try to stick with theconvention of using ~F (possibly with a subscript, if we need to differenti-ate between the net force acting on each of several objects) to denote thenet force acting on an object.1

Note that Newton’s laws only directly apply to point masses or par-ticles—that is, objects with negligible volume. We’ll see why Newton’slaws easily generalize to many objects with girth, however, in Section 4.6.Until then, take it for granted that Newton’s laws apply to systems (orunits of objects) that we’ll call free bodies: solid boxes and crates, balls,and many other rigid objects.

Newton’s first law

We give Newton’s first law as follows:

An object’s velocity is constant if the net force acting on it iszero.

Note that the object’s velocity may be zero, as in the case of an object atrest. In symbols, we can write

d~vdt

= 0 if ~F = 0, (3.1)

where ~v is the velocity of the object and ~F is the net force acting on theobject. Note that we could just have easily written ~a = 0 in place ofd~v/dt = 0.

The first law might seem redundant—after all, if you read further,you’ll find that it’s just a special case of the second law for which ~F = 0.Nonetheless, the first law is still significant for a couple of reasons. First,historically, Newton stated the first law separately from the second law toemphasize the inaccuracy of the centuries-old view that objects naturallycome to rest, as Aristotle and others had believed.

1Many textbooks use the sigma sign (∑~F) or the subscript “net” (~Fnet) to denote thenet force.

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3.1. Newton’s laws of motion

Secondly, the first law is used to define what we call an inertial frameof reference, a frame in which the first law holds true. (As we briefly men-tioned in 2.4, a frame of reference is more or less our choice of our coor-dinate axes.) It turns out that frames moving at constant velocities areinertial, which is consistent with the “constant speed” definition we gaveat the end of the previous chapter.

We say that Newton’s laws apply only in inertial frames. In other words,we can’t apply Newton’s laws if we observe an object accelerating with-out a nonzero net force acting on the object, as is the case in the interiorof a car undergoing a turn or the inside of a rotating cylindrical room. Toapply Newton’s laws to these so-called accelerated frames of reference, weneed to account for the fictitious forces2 that seem to be moving objectsin our frame. Needless to say, it gets a little complicated.3 But take abreath—all frames we use in this book will be inertial.

Newton’s second law

The most common statement of Newton’s second law is

The acceleration times the mass of an object is equal to themagnitude of the net force acting on the object.

Mathematically,~F = m~a, (3.2)

where~a is the acceleration of the object, m is the mass of the object, and~F is the net force acting on the object. In Section 4.4, we’ll give an alter-native statement of the second law.

Newton’s second law lays down a couple important things. First,it essentially defines the scalar quantity we call (inertial) mass.4 Someproperty of an object, which we term mass, determines the effect of a

2This is where the centrifugal, azimuthal, and Coriolis forces arise.3Changing to a non-inertial frame where appropriate can often simplify calculations

immensely. However, we won’t explore rotational motion terribly in depth in this book,which makes such situations much rarer.

4It might seem silly to note that mass is always positive, but strictly speaking, there’sno reason to assume that negative mass does not exist—just none has been discovered(yet).

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3. NEWTON’S LAWS

force on that object via a direct proportion. A large force applied to asmall mass results in a larger acceleration than that same force applied toa large mass. (Mass is sometimes also called inertia because mass can bethought of as the size of an object’s “resistance” to change its motion.)

The other idea that the second law introduces is that the magnitudeof the net force is related to the magnitude of an object’s acceleration bya simple direct proportion. Force is proportional to neither the object’svelocity nor the square of its acceleration, for instance. Furthermore, thenet force points in the same direction as the object’s acceleration. In otherwords, the net force vector is always parallel to the acceleration vector.

By multiplying the units on the right side of Equation 3.2, we cansee that the unit of force is kg ·m/s2, which is more commonly called anewton and symbolized with the letter N.

Newton’s third law

The most pithy of the three, Newton’s third law5 claims that

If one object exerts a force on a second object, the second ob-ject also simultaneously exerts a force, equal in magnitudeand opposite in direction, on the first object.

Where ~F1,2 is the force exerted on the first object by the second and ~F2,1 isthe force exerted on the second object by the first, we have

~F1,2 = −~F2,1, (3.3)

or, equivalently,~F1,2 + ~F2,1 = 0.

Sometimes these equal-magnitude pairs of forces governed by Newton’sthird law are referred to as action-reaction pairs: each “action” force ex-erted on another object must also result in an equal and opposite “re-action” force. Be careful, however, as some students are misled by this

5It is worth noting that while the third law holds for all situations in this book, New-ton’s third law is not true in general. Though it applies to all contact forces and evengravitational and electric forces, the third law does not work for magnetic forces sincea magnetic field can, roughly speaking, store energy and momentum. (Momentum,however, is still conserved.)

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3.2. Some types of forces

terminology. Action-reaction pairs exist simultaneously; the “reaction”force does not precede the “action” force.

As we’ll see in Section 4.4, the importance of Newton’s third law liesin the fact that we can use it to derive the law of conservation of momen-tum.

3.2 Some types of forces

In kind of a qualitative manner, let’s examine a few simple types of forcesthat appear in elementary mechanics. Our list below is by no meanscompletely descriptive (in terms of the actual fundamental nature of theforces) or exhaustive, but it more or less touches on everything we willencounter in this book. It’s important to note that what we say about thefollowing forces represent only relatively rough models, as opposed tothe more precise models we usually give.

Gravity

All objects on the earth are subject to a constant downward accelerationof magnitude g, approximately 9.81 m/s2. The cause of that acceleration,of course, is the earth, which exerts a force on each object equal to theproduct of the object’s mass and the acceleration due to gravity. We saythat the weight of an object is the magnitude of the gravitational forceexerted by the earth on the object. The weight of a 10-kilogram black,then, is about 98 newtons. Note that gravity always points downwardregardless of the angle at which an object is situated. For instance, ablock sitting on an inclined plane feels gravity pulling straight down, ofwhich one component is perpendicular to the plane and one parallel tothe plane. (We’ll talk about this in more detail in just a bit.)

Gravitational forces are subject to Newton’s third law—while the earthexerts a force on you, you also exert a force on the earth. Furthermore,the force the earth exerts on you is equal in magnitude and antiparallel tothe force you exert on the earth. We will address the (more) fundamentalnature of gravity in Section 6.2.

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3. NEWTON’S LAWS

We often refer to weight in pounds. Strictly speaking, then, the poundis a unit of force.6 Why, then do we equate it with the kilogram, which isa unit of mass? As we’ve just seen, an object’s weight is directly propor-tional to its mass. Therefore, for convenience, we can simply say that onekilogram is “equal” to its weight in pounds. Since the relationship be-tween weight and mass is linear, we can obtain a nice conversion factor,approximately 2.2 pounds per kilogram.

Normal force

A block sitting on the floor is pulled down by gravity, but it does notexperience a net force. Clearly, there must be some force, equal in mag-nitude to the object’s weight, pushing upward on the object. In this case,the floor provides a contact force on the box that stops it from fallingthrough the floor. If we start pushing down on the block and exertingadditional downward force, the floor instantaneously responds by pro-viding sufficient addition upward force to counter the downward force.7

The normal force is the physical contact force provided by a surfaceto prevent an object touching it from going through the surface. The sizeof the force provided is exactly enough to keep the object from fallingthrough, so it can never cause an object to repel from the surface. The nor-mal force always acts perpendicular to the surface; therefore, if a pointmass moves across a curved surface, the normal force constant changesdirection depending on the curvature of the surface. Generally, we firstfind the component of the force that acts in the direction of the surface.

The normal force also obeys Newton’s third law. In general, the nor-mal force exerted on one object by a surface is opposite to the normalforce exerted on the surface by the object. Note that the surface need notbe an immobile one—two moving boxes can exert equal-and-oppositenormal forces on each other if they remain in contact.

6The slug is the unit of mass in the customary system. Nonetheless, it is rarelyused, even by engineers who still use customary units, which occasionally creates someconfusion between units of force and mass.

7Strictly speaking, we’re making several idealizations here, one of which is the as-sumption that the floor and block are perfectly rigid.

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Friction

Push a heavy box with all your might and you might find that it doesn’tbudge. Intuitively, this seems obvious. Nonetheless, it appears that New-ton’s second law contradicts our intuition. Though the box has a largemass, shouldn’t it accelerate at least a little bit? It turns out that when-ever we slide two surfaces past each other, a resistive force called frictionis generated that opposes the object’s motion. It turns out that we canclosely model the friction force between two objects based on just twovariables: the types of surfaces involved—for instance, wood on woodor metal on concrete—and the the normal force acting on each of the twoobjects. For each type of surface, there is a specific friction coefficient µ

(pronounced “mew”) that linearly relates the magnitude of the normalforce n to the magnitude of the friction force f :

f ≤ µn. (3.4)

This is a nice model8 since it doesn’t depend on the physical dimen-sions of the object described. This might seem counter-intuitive sinceobjects with large surface area might appear to have a greater frictionforce, but in fact, we’ve accounted for those factors in our equation. Sup-pose we take a cube of putty and, keeping the same mass, squish it intoa flatter but wider shape. Though the squished putty has more surfacearea, the normal force per unit area, and thus the friction force per unitarea, decreases exactly enough to balance out the increase in surface area.Therefore, the friction force acting on the whole putty remains the same.

Note that we need a less than or equal sign because µn is the maxi-mum friction force generated. Otherwise, the friction force merely pro-vides the minimum amount of force to oppose the force acting against it;if you’re only pushing a tiny bit on a box, the friction force only needsto push back a little to stop the box from moving. It’s a common mis-take to label “ f = µn” without knowing that the maximum friction forceoccurs, which often leads to some bizarre situations—friction certainlywon’t make a box accelerate from rest!

8Note that our description of friction here is merely a rough model, unlike the fun-damental force laws that govern gravitation and electromagnetism. There are moreexact molecular descriptions of friction, but they won’t be useful for us here.

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3. NEWTON’S LAWS

There are two general types of friction9 that are relevant to our studyof physics. Static friction acts when an object is not moving. What wesimply called “friction” above is actually static friction. As long as theheavy box on the floor remains still, the force opposing your push is staticfriction.

However, if you apply enough force to dislodge the block and get itmoving, kinetic friction takes over. Kinetic friction acts between slidingsurfaces. For kinetic friction, it is the case that equality is always satisfiedin Equation 3.4. Since the block is moving, the friction force just remainsconstant.

µ is different for kinetic and static friction, even for the same surfaces.We denote the static friction coefficient as µs and the kinetic friction co-efficient as µk, but often only one of the friction types is relevant in aproblem, in which case we just use the unsubscripted letter. It is alwaysthe case that µk ≤ µs for a given material (why?).

Theoretically speaking, there is no upper bound on the size of a fric-tion coefficient, but friction coefficients typically lie between 0 and 1:static friction coefficients µs range from a few tenths for Teflon on Teflonall the way to a little over 1 for copper on cast iron, for instance.

Example 3.1 Starting from zero force, you gradually increase the force Fexerted on a block lying on a level surface. Given the coefficients of staticand kinetic friction µs and µk, respectively, determine the size of the resistiveforce f exerted on the block by friction as a function of F. Let the mass of theblock be m.

Where n = mg is the magnitude of the normal force exerted on the blockby the surface, initially your applied force is less than µsn. In this case, thestatic friction force is strong enough just to counter your force F. Just as theapplied force increases past the maximum static friction force µsn, the blockbudges. Finally the block is set in motion and the kinetic friction force µkntakes over, which causes a drop in the friction force (recall µk ≤ µs). Thuswe have

f (F) =

{F if F ≤ µsmg

µkmg if F > µsmg.

9Of course, there are other types, such as rolling friction, as well as more compli-cated models for friction, but we won’t worry about them here.

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Note that if you later reduced the applied force to less than the maximumstatic friction force µsn but greater than the kinetic friction force µkn, theblock would still keep moving and the friction force would be the kineticfriction force µkn. Thus, once you have overcome the initial static frictionto get the block moving, you can reduce the applied force and still keep theblock moving. �

Example 3.2 A driver in a car of mass m of initial velocity v0 sees an obstacleahead and immediately slams her brakes, locking the tires. Assume thatwe know the coefficient of kinetic friction µ between the tires and road Wemight want to know how far the car travels before it comes to a stop—inother words, the stopping or braking distance.

Gravity exerts a force downward of size mg on the car, so the upwardnormal force is of size mg as well. Taking the initial direction of the carto be positive, the force provided by kinetic friction must be −µmg; it isnegative because the force points opposite the direction of the car’s motion.Therefore, Newton’s second law tells us that the acceleration of the car mustbe −µg. Thus we can use Equation 2.10 to find the stopping distance d:

0 = v20 − 2µgd

d =v2

02µg

.

We can draw some conclusions from this simple formula. Stopping dis-tance is proportional to the square of the initial speed v0, so drivers shouldquadratically increase their spacing from the car in front as they increasetheir driving speed. This is only exacerbated by the fact that human driversrequire a non-negligible reaction time to hit the brakes in response to anunexpected event.

Furthermore, µ is obviously of importance. Rainy roads carry a lower µ

than dry roads do, and drivers need to anticipate a larger stopping distancein the case of heavy rain. Should icy conditions happen to decrease µ sig-nificantly, transportation officials might choose to spread salt or sand on theroad in an effort to increase µ. �

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Tension force

As we’ll soon see, in mechanics we’ll make a number of assumptionsto help us simplify our model. One of these—as silly as it may sound,right up there with “frictionless ice”—is massless string. The key reasonfor needing massless string is the fact that it allows us directly to applyNewton’s third law to the tension force experienced by two objects, oneon each end of a string. Why must the string be massless? If there wereany sort of mass on the string, we would have to use F = ma for eachunit mass or point mass on the string. We can quickly see that the tensionforce on each side of the string is not necessarily the same.

In general, a few other constraints also apply when we use masslessstring—no stretching is allowed. The length of the string must remainconstant. Usually, this fact allows us to relate the position (and velocityand acceleration) of one object to another. Finally, the direction of theforce provided on an object by a tension force is the same as the directionof the string where it attaches to the object.

Springs

Imagine a system with a spring connected to a wall on one end and amass on the other (Figure 3.1). Hooke’s law says that a mass stretcheda displacement x from the equilibrium position, the position of the masswhen the spring is relaxed, experiences a restoring force F given by

F(x) = −kx, (3.5)

where k > 0 is a constant called the spring constant, which is a propertyof the specific spring used. Note that the spring constant has units N/m.

Note that the magnitude of the force is always proportional to themagnitude of the displacement and that the force always acts in the op-posite direction as the displacement, as indicated by the negative sign. Ifthe spring is stretched, it creates some force to pull it back to its originalposition. Likewise, if the spring is compressed, a pushing force is exertedto restore itself to equilibrium.

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x

m

mF

k

Figure 3.1 Let the rightward direction be positive. At the top is a spring, oneend attached to a wall and the other attached to a mass m, at its equi-librium position in its relaxed state. There is no net force acting onthe mass. At the bottom is that same spring compressed with somedisplacement x (in this case negative). There is a net force F acting onthe mass as the opposite direction as the displacement (in this casepositive).

Example 3.3 Suppose we have a spring of spring constant k. If we cut thespring in half, what is the new spring constant j of each of the two half-springs?

Put the two half-springs back together to form a single spring of theoriginal length. If we pull the single spring by some displacement d, thenthe restoring force has magnitude F = kd by definition. However, each half-spring is only stretched by a displacement d/2, since the total stretch mustbe d. Therefore, the spring constant of the each of the half-springs is givenby

F = j(

d2

),

soj = 2k.

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3. NEWTON’S LAWS

Intuitively, it might seem that the spring constant of the half-springs andthe original should be the same, but in fact the spring constant is not anintrinsic property of the material that comprises the spring. In general, fac-tors such as the number of coils in the spring are significant, so it should notsurprise us that the relaxed length of the spring affects the spring constant.�

3.3 Using Newton’s laws

In the end, we usually want to be able to apply Newton’s second lawto find the acceleration of an object, and to do that, we need to find thenet force. Finding the net force acting on an object begins with care-fully labeling the various forces acting on an object, which may includea hodgepodge of different types of forces pointing in different ways. Thefree-body diagram is a visual tool that helps us to keep track of all the forcesacting on an object and their directions.

We usually begin by drawing a small box to represent an object. Theexact size and shape don’t matter, as we draw a point in the middle ofthe box and draw all the forces as vectors with tail on the point. Usually,we note the mass of the object as a variable or number in the box andlabel each of the vectors with its corresponding variable. Sometimes it’suseful for us to draw out the vertical and horizontal component vectors(especially for inclined plane problems, as we’ll soon see) using dottedlines, but make sure that you remember that they’re just components andnot additional vectors.

A very simple example in which we can apply this analysis is a sys-tem containing two blocks, one of mass m1 and the other of mass m2,connected by a taut, massless string (Figure 3.2). We tie another string tomass m1 and pull it right with a force F. We might want to determine theacceleration of the two blocks and the tension force in the string connect-ing the two blocks.

We take as our first free-body the block of mass m1. We draw a smallbox10 to represent the block. We then draw an arrow, size relative to ap-

10In principle, it doesn’t matter what shape you draw—it could be a squiggly line, amustache, or nothing at all (a small dot suffices).

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3.3. Using Newton’s laws

m1

Fm2

Figure 3.2 Two blocks are connected by a taut string with tension T. A force Fpulls the leftward block toward the right.

m1

F

N1

W1

Tm2

N2

W2

T

Figure 3.3 Free-body diagrams for m1 and m2. We end up ignoring the weightand normal force, but we draw them on the free-body diagram justin case.

proximate magnitude (or the putative magnitude, anyway), in the direc-tion of each force. We then label the four forces acting on the block: grav-ity (−mg), normal force (mg), pulling force (F), and tension force (T) fromthe string (Figure 3.3). For convenience’s sake, we often omit forces thatcancel out and provide little insight in solving the problem. In this case,we may simply want to ignore gravity and the normal force completely,which cancel each other out, since we’re only concerned with what hap-pens in the horizontal direction. Thus, we’re left with two forces: pullingforce and tension.

We can thus write the equation

F− T = m1a1.

We don’t have enough information to solve yet, so let’s move onto our

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3. NEWTON’S LAWS

second piece of information:

T = m2a2,

which gives the force acting on m2. It still seems as if we’re stuck: wehave three unknowns—T, a1, and a2—and just two equations. What dowe do?

It turns out that it must be the case that a1 = a2. Otherwise, the blockswould separate and the string would break! (In general, massless stringscannot stretch, or else they would have mass. This fact turns out to becritical in many situations.) Knowing this, we let a = a1 = a2, fromwhich we get

F−m2a = m1a,

soa =

Fm1 + m2

. (3.6)

We can also solve for the tension:

T = m2a =Fm2

m1 + m2.

If we had simply wanted to determine a, however, and not T as well,would it have been acceptable to consider both masses as one free body?The answer is yes. In fact, doing so is consistent with Equation 3.6. Often,there are multiple “systems” we might choose as a free body. Dependingon what we want to find, some are more convenient than others, as wejust saw here: if had chosen the system of both blocks to be our freebody, we could have immediately written down F = (m1 + m2)a andignored the tension entirely. Be careful, though, since we can’t always dothis. We’ll talk about when we can and cannot apply Newton’s laws toa single system as a whole in Section 4.6. For now though, just assumethat we have chosen valid systems as free bodies.

Example 3.4 A block of mass m1 sits on a tabletop as shown in Figure 3.4.It is connected, via a string that runs across a pulley, to a vertically hangingblock of mass m2. What is the magnitude of the acceleration of the twoblocks if the coefficient of kinetic friction between the block and tabletop isµ?

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m1

m2

Figure 3.4 Diagram for Example 3.4.

This is very similar to the worked example in the main text, except thatthe pulling force F is replaced by the downward force of gravity −m2g anda friction force has been added to the block on the tabletop.

First, we inventory all the forces acting on the tabletop block. The forceexerted by gravity −m1g is countered by the normal force n1 = m1g, sothere is no motion in the vertical direction. In the horizontal direction, thereis a tension force, of magnitude T, directed to the right, as well as a kineticfriction force, of magnitude µn1 = µm1g. Thus Newton’s second law tellsus that

m1a = T − µm1g

T = m1(a + µg)

The forces acting on the hanging block include that of gravity −m2g andthe upward tension force, by Newton’s third law also of magnitude T. Thusthe second law statement for the hanging block is

m2a = T −m2g

T = m2(a + g)

We have intentionally used the same a as with the tabletop block since thetaut string implies that they have the same magnitude of acceleration. Weequate our two expressions for T and find

m1(a + µg) = m2(a + g)

(m1 −m2)a = g(m2 − µm1)

a =m2 − µm1

m1 −m2g.

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3. NEWTON’S LAWS

This was not a difficult example by any means, but it did reiterate somekey points regarding the application of Newton’s laws. First, if two objectsare constrained to move together, such as those connected by a taut string,we can often find a useful relationship between their accelerations. In thiscase, it was a very simple equality. Second, if there is a force whose mag-nitude we do not know, we might try to eliminate it from our problem bysolving for it in two (or more ways) as we did above. �

m2

Fm1


Example 3.5 A block of mass m2 is free to slide on a frictionless floor while ablock of mass m1 slides on top of the first block. There is a coefficient of staticfriction µ between the two blocks. If you apply a force F to the bottom blockas shown (Figure 3.5), was is the magnitude of the friction force f that actsbetween the two blocks, assuming that the top block does not slip? What isthe minimum µ such that the top block does not slip?

We first draw a free-body diagram for the top block. Clearly, the frictionforce f acts in the same direction as the force F. Furthermore, f depends onthe normal force N acting upward on the block, which is equal to the block’sweight m2g.

By Newton’s third law, the same normal force N acts on the bottomblock, except in the opposite direction. Similarly, the friction force actingon the bottom block is − f . (The bottom block also experiences a normalforce from the ground, but it is canceled out by gravity and N and is not

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relevant.) Newton’s second law gives

F− f = m1a

f = m2a,

where a is the (common) acceleration of the two blocks. Dividing the topequation by the bottom one gives

F− ff

=m1

m2

m2F−m2 f = m1 f

f =m2

m1 + m2F.

Since f ≤ µm2g,

µm2g ≥ m2

m1 + m2F

µ ≥ F(m1 + m2)g

. �

Example 3.6 A person of mass m stands on a balance in an elevator thathas a vertical acceleration a. What is the reading on the balance? (Assume|a| < g.)

Drawing a free-body diagram for the person, we find that there are twoforces acting on the person: a normal force N (exerted by the balance onthe man) pointing upward and gravity pointing downward. Since |a| < g,the person always stays in contact with the elevator (why?) and his or heracceleration is a. Therefore

N −mg = ma

N = m(a + g). (3.7)

By Newton’s third law, the force exerted by the balance on the man is equalto the normal force N exerted by the man on the balance, so the reading onthe balance is simply m(a + g).

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3. NEWTON’S LAWS

m

a b

T1 T2

W


Our result, albeit simple, shows us a couple things. First, when a > 0,the person’s apparent weight, that is, the reading on the balance, is greaterthan mg. In other words, the person feels heavier than his or her normalweight. This happens at the beginning of an upward elevator ride and atthe end of a downward ride. Second, when a < 0, the person’s apparentweight is less than mg, and the person feels lighter than normal.

While riding in a decently fast elevator, you can more easily notice yourdecreased apparent weight during an upward elevator ride if you jump inthe air at the very end of your trip, where a is negative (at the beginning fora downward elevator ride, though the timing is harder, at least for me).11 �

Example 3.7 Two strings hold up a box of mass m as drawn in Figure 3.6.The string on the left makes an angle α with the horizontal, while the one onthe right makes an angle β. What are the tension forces ~T1 and ~T2, the tensionforce holding up the box from the left and right strings, respectively?

We know that the sum of the forces on the box ~F is equal to the (vector)sum of the two tension forces and the weight. But since the box remains at

11I assume no responsibility if you get hurt while doing this.

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rest, we know that the acceleration of the box is zero, in which case ~F = 0:

~T1 + ~T2 + ~W = 0.

We know that ~W = (0,−mg). Now, we find the tension forces in terms oftheir magnitudes and the angles α and β. Note that since the tension forcesact in the direction of their respective strings, it must be that

~T1 = (−|~T1| cos α, |~T1| sin α)

and~T2 = (|~T2| cos β, |~T2| sin β).

Putting all the forces together, we have

(−|~T1| cos α + |~T2| cos β, |~T1| sin α + |~T2| sin β−mg) = 0.

Now we break up the vector equation into a system of two linear equationsto solve for |~T1| and |~T2|:

−|~T1| cos α + |~T2| cos β = 0

|~T1| sin α + |~T2| sin β−mg = 0.

The top equation tells us that |~T2| =cos α

cos β|~T1|, which we substitute into the

bottom equation:

|~T1| sin α +cos α

cos β|~T1| sin β−mg = 0,

which gives

|~T1| =

1

sin α +cos α

cos βsin β

mg

=

(cos β

sin α cos β + cos α sin β

)mg

=

(cos β

sin(α + β)

)mg,

where we used the angle sum formula

sin(α + β) = sin α cos β + cos α sin β. (3.8)

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3. NEWTON’S LAWS

Therefore, by substitution (or by symmetry—why?)

|~T2| =(

cos α

sin(α + β)

)mg.

Thus~T1 =

(−(

cos α cos β

sin(α + β)

)mg,

(sin α cos β

sin(α + β)

)mg)

(3.9)

and~T2 =

((cos α cos β

sin(α + β)

)mg,

(sin β cos α

sin(α + β)

)mg)

. (3.10)�

Example 3.8 A towel of length L hangs over a horizontal bar of negligi-ble diameter. If the coefficient of friction between the bar and the towel isµ, what is the maximum overhang such that the towel remains stationary?(The overhang is defined as the difference in length between the two halvesof the towel.)

Intuitively, we might simply claim that the bar must provide a normalforce equal to the weight of the bar. Using this, we can find the largest pos-sible frictional force that can be generated. Since the towel slips because ofthe difference in weights of the hanging parts on the left and right sides, wecan simple equate the frictional force with the weight difference to find themaximum overhang. These are correct assertions, but let’s be a little morecareful to make sure we’re right.

Let the portion of the towel to the left of the bar be of length l and theportion to the right be of length L − l. Without loss of generality, assume.L− l ≤ l. We divide the towel up into three portions: the part of the towelto the left of the bar, the part to the right, and the small piece hanging overthe top of the bar. If the entire towel has mass M, then the mass of the leftportion is Ml/L, and the mass of the right portion is M(L− l)/L; the centerpiece has negligible mass.

Since the tension forces exerted by the center piece on the other twopieces are the only forces holding them up, by Newton’s third law it must bethe case that the tension forces T1 and T2 exerted by the left and right pieces,

98


respectively, on the center piece are given by

T1 =Mgl

L

T2 =Mg(L− l)

L.

Drawing a free-body diagram of the center piece reveals that the normalforce provided by the bar must be of magnitude T1 + T2 = Mg, as we rea-soned earlier. Because the center piece is essentially massless, the net forceacting on the center piece must be zero, and we have

T1 − T2 = f .

Since f ≤ µMg,

MglL− Mg(L− l)

L≤ µMg

2l − LL≤ µ

l ≤ (1 + µ)L2

.

Taking l = (1 + µ)L/2, we find that the maximum overhang is

l − (L− l) = 2l − L = (1 + µ)L− L = µL. �

Dealing with constraints

A few of the problems at the end of this section illustrate the techniqueof finding kinematic constraints. After we’ve drawn our free-body dia-gram for a problem and applied Newton’s second law, sometimes we’llfind that we simply don’t have enough equations to solve for all the un-knowns. This may be either because the problem does not have a uniquesolution (we won’t pull any of those on you in this book!) or becausethere are pieces of information we have forgotten to take into account.

The most common thing to forget are our kinematic constraints: oftenthere are implicit restrictions on the position or velocity of an object. Forinstance, if we know an object slides on a circular track, it must have the

99

3. NEWTON’S LAWS

specific acceleration and, by extension, net force that allows it to move ina circular path. As we’ll see in the next section, knowing that an objectmoves in a circular path of radius r presents a very powerful kinematicconstraint—the acceleration of the object must be a = v2/r, where v isthe object’s speed at any given point.

A more complicated case arises when the positions of two differentobjects are related. Even if we know for what we’re looking, these typesof constraints are often tricky to find. (In particular, see Example 3.13.)To find the constraint, a couple of notable examples below require us torelate the accelerations of several different objects.

Atwood’s machine

The Atwood’s machine, a simple pulley system with two masses as shownin Figure 3.7, was first developed in the latter half of the eighteenth cen-tury as a means to verify mechanics principles, though it later evolvedinto a device to torture physics students.12 For the moment, we considerthe machine to consist of two point masses m1 and m2, a massless stringconnecting the two, as well as a pulley over which the string runs. (Laterin this book, we will consider what happens when we no longer considerthe pulley to be massless.)

Problems involving pulleys and strings are particularly good ones notonly for learning how to applying the different types of forces we men-tioned earlier but also for understanding how to place kinematic con-straints on various systems. It’s important to remember that while wecan find ~F to get ~a, sometimes we instead know ~a and have to find ~F.Often, some combination of both—determining some of the forces actingon various objects in a system, as well as finding some constraints on thepaths or rates of their motion—is necessary.

Example 3.9 Given the Atwood’s machine in Figure 3.7, find the accelera-tion of each of the two masses m1 and m2 and find the tension force in thestring connecting the two masses, as well as in the string connecting thepulley to the ceiling.

12Intellectually, not physically.

100


m1

m2

Figure 3.7 A simple Atwood’s machine with two masses. Here we consider thepulleys and string to be massless.

The net force acting on each mass consists of a downward force due togravity and an upward force due to the tension from the string. By Newton’sthird law, the tension force T must be the same for each mass, so

m1a1 = T −m1g

m2a2 = T −m2g.

Here we have three unknowns and only two equations, so there must besome additional piece of information we haven’t accounted for yet. Indeed,we know that the constraint a1 = −a2 must hold if the string is taut.

Subtracting the bottom equation from the top one gives

m1a1 −m2a2 = −m1g + m2g

(m1 + m2)a1 = (m2 −m1)g

a1 =m2 −m1

m1 + m2g. (3.11)

Since m1 and m2 are symmetric with respect to each other (we can’t distin-guish between m1 and m2 in either the physical situation or in the algebra),we could simply have solved for a2 first and should have gotten the samesolution, except with the subscripts swapped. Another way of looking atthings is to say that a system in which naming the variables yields different

101

3. NEWTON’S LAWS

solutions is completely unphysical; it shouldn’t matter which mass we callm1 and which we call m2.13 Thus we simply swap the subscripts to get

a2 =m1 −m2

m1 + m2g. (3.12)

Notice that our results check out in various commonsense cases. First of all,when m1 = m2, both masses stand still and a1 = a2 = 0. Second, when themasses are different, the lighter mass should come up and the heavier massshould fall down. Accordingly, when m1 > m2, it follows that a1 < 0 anda2 > 0; when m2 > m1, we have a1 > 0 and a2 < 0. Finally, it is reasonableto have |a1|, |a2| < g.

Now we solve for T:

T = m1(a1 + g)

= m1

(m2 −m1

m1 + m2g +

m1 + m2

m1 + m2g)

=2m1m2

m1 + m2g. (3.13)

To find the tension in the string connected to the ceiling, we draw a free-body diagram for the pulley. Since the pulley is massless and does not ac-celerate, the tension force upward must exactly cancel the downward totalforce of −2T. Therefore, the upward tension force is 2T. �

Now we consider the case of a double Atwood’s machine, that is, a com-pound Atwood’s machine with two pulleys (Figure 3.8).

Example 3.10 Given the Atwood’s machine in Figure 3.8, find the accelera-tion of each of the three masses m1, m2, and m3.

Let the strings be as numbered as in Figure 3.10: rope 1 is the portion ofthe upper string to the left of the upper pulley, rope 2 is the portion of the

13Be careful, however, since in this simple situation, we only have one solution for a1and a2, so the one solution must be symmetric with respect to a1 and a2. If there weretwo solutions, the solutions might be symmetric with respect to each other, but eachsolution itself might not be symmetric. Later on, we’ll see more examples of symmetry,some of which are less trivial than this one.

102


m1

m2

m3

1

4

23

Figure 3.8 A double Atwood’s machine with three masses. Again, we considerthe pulleys and strings to be massless.

bottom string to the left of the bottom pulley, and rope 3 is the portion to theright. Rope 4 is the portion of the upper string between the two pulleys.

We first examine the lower pulley, which we approach similarly to thecase of the simple Atwood’s machine (Example 3.9). If the upward tension inrope 2 is T, then the tension in rope 3 is also T upwards. Furthermore, as weshowed for the simple Atwood’s, the tension in rope 4 must be 2T, whichimplies that the tension in rope 1 must also be 2T by Newton’s third law.Therefore, applying Newton’s second law to each of the masses producesthe following equations:

2T −m1g = m1a1 (3.14)

T −m2g = m2a2 (3.15)

T −m3g = m3a3. (3.16)

However, we might notice that, similar to before, we have more unknownsthan equations, which necessitates an additional equation if we are to finda unique solution. This time, however, finding the constraint equation forthe strings is not so simple. Some may find it via some trial and error or bymentally moving around the strings and masses, but we use the following

103

3. NEWTON’S LAWS

systematic and surefire method. It may seem somewhat tedious for thisexample, but for more complicated problems it is invaluable.

Let the length of rope i be denoted by li. The length of rope 1 determinesthe position of m1, and as the rope gets longer, m1 drops, so we have

d2

dt2 l1 = −a1.

Since rope 4 is part of the same string as rope 1, it must be the case that l1 + l4is constant, d2/dt2(l1 + l4) = 0 and

d2

dt2 l4 = − d2

dt2 l1 = a1. (3.17)

Applying the same thinking to the lower pulley gives

d2

dt2 (l2 + l4) = −a2

d2

dt2 (l3 + l4) = −a3.

Summing these two equations gives

d2

dt2 (l2 + l3 + 2l4) = −(a2 + a3).

Since l2 + l3 is constant, d2/dt2(l2 + l3) = 0, in which case our equationreduces to

d2

dt2 l4 = − a2 + a3

2. (3.18)

Equating Equations 3.17 and 3.18 yields

a1 = − a2 + a3

2. (3.19)

We probably could have written down Equations 3.17 and 3.18 without pro-ceeding as deliberately as we did, but again, we purposely tried to be sys-tematic here.

Equations 3.14–3.16 and 3.18 are now sufficient to determine the accel-erations of the three masses. Solving them is straightforward but somewhatmessy and is left to the reader as an exercise (you may wish to use a com-puter). We simply give the final results:

T =4m1g

4 + m1m2

+ m1m3

=4m1m2m3

4m2m3 + m1(m2 + m3)(3.20)

104


h

q

m

Figure 3.9 A block of mass m sits on an incline of height h and angle θ with thehorizontal. Except in Example 3.13, the incline is fixed.

and

a1 =4m2m3 −m1(m2 + m3)

4m2m3 + m1(m2 + m3)g (3.21)

a2 =3m1(m3 −m2)− 4m2m3

4m2m3 + m1(m2 + m3)g (3.22)

a3 =3m1(m2 −m3)− 4m2m3

4m2m3 + m1(m2 + m3)g. (3.23)

�

Forces along an incline

One of the most common types of introductory physics problems in-volves a block sliding down an inclined plane—a wedge, roughly speaking—either with or without friction. Let’s try a few.

Example 3.11 Suppose we have a fixed, frictionless inclined plane of heighth and acute angle θ with the horizontal (Figure 3.9). If a box of mass m isreleased from rest at the top of the incline, what is the box’s speed at thebottom of the incline?

A free-body diagram of the block (Figure 3.10) reveals that gravity actsstraight downwards on the block, while the normal force acts perpendicular

105

3. NEWTON’S LAWS

q

N

mg cos q

mg sin qm

mg

Figure 3.10 A free-body diagram of a block on an inclined plane of inclination θrelative to the horizontal.

to the plane. One valid way to do the problem is to break normal force intovertical and horizontal components (relative to the ground), and then applythe kinematic constraint ay/ax = tan θ, where the acceleration of the block is~a = (ax, ay). However, as we saw in Example 2.16, it is much easier simpleto rotate our coordinates such that the x-axis lines up along the plane.

As shown in Figure 3.10, we can break gravity into components paral-lel and perpendicular to the plane, which lets us avoid having to find thekinematic constraints. This essentially turns our two-dimensional probleminto a one-dimensional one since the normal force N and the perpendicularcomponent of gravity −mg cos θ must sum to zero (why?). Thus the block’snet force, which is in the direction of the plane, is the parallel componentmg sin θ.

It follows that the acceleration of the block is g sin θ. Because the blocktravels a total distance h/ sin θ, we can use Equation 2.10 to find the block’sfinal speed v at the bottom of the incline:

v2 = 2(g sin θ)

(h

sin θ

)v =

√2gh. (3.24)

�

106


Example 3.12 There is a heavy package of mass m on a flatbed cart. What isthe largest angle θ with the horizontal at which the cart can be tipped suchthat the package stays on the cart if the package and cart surfaces have acoefficient of static friction µ?

We rotate our coordinates such that the x-axis lies along the incline. Asin Example 3.11, the normal force is of magnitude mg cos θ, in which casethe magnitude of the static friction force f is bounded by f ≤ µmg cos θ. Ifthe package stays on the cart, the block’s parallel as well as perpendicularacceleration is zero, and Newton’s second law in the x direction yields

mg sin θ − f = 0

mg sin θ ≤ µmg cos θ

tan θ ≤ µ. �

Example 3.13 A free, frictionless inclined plane of mass M and acute angleθ with the horizontal sits on a frictionless horizontal surface (Figure 3.9).If a box of mass m is released from rest at the top of the incline, what arethe accelerations of the block and plane while the block travels down theincline?

As usual, the only two forces acting on the block are gravity and thenormal force, of magnitude N. However, not as usual it need not be thatN = mg cos θ. The inclined plane is sliding, so it is not completely clearexactly how much normal force is needed. All we can do is decompose thenormal force into horizontal and vertical components, respectively N sin θ

and N cos θ, and express the acceleration of the block ~a = (ax, ay) in termsof N:

ax =N sin θ

m(3.25)

ay =N cos θ

m− g. (3.26)

By Newton’s third law, the normal force exerted on the block by theplane is equal and opposite the normal force exerted on the plane by theblock. Since the vertical component of this force is canceled by the normal

107

3. NEWTON’S LAWS

force from the horizontal surface, we quickly find that the (horizontal) ac-celeration ~A = (A, 0) of the inclined plane is

A = −N sin θ

M. (3.27)

The negative sign indicates that the incline moves to the left.We thus have four unknowns but only three equations. Clearly, the way

out of this quandary is some sort of kinematic constraint; we have not yetspecified how the motion of the inclined plane affects the path of the block.To find the relationship between the two objects, we note that the distancethe block has fallen divided divided by the horizontal distance the block hasmoved relative to the inclined plane must be a constant, namely tan θ, if theblock is to remain on the plane (Figure 3.11). Where (x, y) is the positionof the block and (X, Y) is the position of the top left corner of the inclinedplane relative to their starting points, we have

tan θ =Y− yx− X

(x− X) tan θ = Y− y.

Taking two derivatives gives

(ax − A) tan θ = −ay (3.28)

since the vertical height of the plane is constant.We have found our fourth equation, and now solving the problem is

simply a matter of algebra. As much as we might like to save this as anexercise for the reader, we will see this problem, at least, through to the end.We substitute Equations 3.25–3.27 into Equation 3.28, yielding(

N sin θ

m+

N sin θ

M

)tan θ = −N cos θ

m+ g

N(

sin θ tan θ

m+

sin θ tan θ

M+

cos θ

m

)= g

N((m + M) sin θ tan θ + M cos θ

mM

)= g.

ThereforeN =

Mmg(m + M) sin θ tan θ + M cos θ

108


Y

qy

O X

x

Y – y

X – x

Figure 3.11 The point represents the position of the block, with lengths mea-sured from the origin O defined as shown.

and

ax =Mg sin θ

(m + M) sin θ tan θ + M cos θ

ay =Mg cos θ − g[(m + M) sin θ tan θ + M cos θ]


= − (m + M)g sin θ tan θ


A = − mg sin θ

(m + M) sin θ tan θ + M cos θ. �

Notice that as M becomes much larger than m, the problem reduces to thefixed-incline case (Example 3.11).

As you can start to see from the previous example, identifying thecorrect forces and kinematic constraints can be quite a hairy task. Ingeneral, as problems start becoming more and more complicated, wetend to start using approximations, if appropriate, or we turn to a dif-ferent method entirely. Lagrangian and Hamiltonian mechanics are moresystematic but completely equivalent formulations of classical mechan-ics. In many cases they offer distinct advantages over directly workingwith Newton’s laws. One of these is greater ease in choosing coordi-

109

3. NEWTON’S LAWS

nates, which is often quite cumbersome with Newtonian mechanics. Inthis book we won’t need to use either the Lagrangian or Hamiltonianmethods, which can be found in more advanced mechanics texts.

3.4 Circular motion

Usually we begin by analyzing all the forces acting on an object, findingthe net force, and then determining the acceleration and, ultimately, po-sition of the object in question. However, often times we wish to go inthe opposite direction: we might ask ourselves what net force is requiredto sustain a given trajectory. This is essentially what we do when we dealwith kinematic constraints.

Circular motion is no different. Many objects move in a circular tra-jectory: a roller-coaster traveling on a circular track, a car driving in aloop on a banked track, and a swirling weight tied to a string. To analyzeeach of these situations, we first examine the kinematic constraints posedby circular motion. Luckily, it turns out that the force acting on each ofthese objects is governed by a simple relationship.

Uniform circular motion

To give us a feel for circular motion, we begin with the simplest of allcases. Imagine that an object begins at the point (1, 0) and moves coun-terclockwise at a constant speed of 1 in a unit circle centered at the ori-gin. As you might have seen from doing the problems from the previ-ous chapter, we can express the object’s position using the trigonometricfunction

~r(t) = (cos t, sin t),

where t is time. We can then find the first and second derivatives todetermine the velocity and acceleration, respectively:

~v(t) =~r ′(t) = (− sin t, cos t)

and~a(t) = ~v ′(t) = (− cos t,− sin t).

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3.4. Circular motion

As expected, we note that the magnitude of each of the functions is 1since

√sin2 t + cos2 t = 1. Now, to find the net force, we simply multiply

by the object’s mass m:

~F(t) = m~a(t) = (−m cos t,−m sin t).

Let’s take a look at the direction of each of the vectors. As we con-structed, the position vector always points outward from the origin tothe object. The velocity vector, on the other hand, is always perpendicu-lar to the position vector since

~r(t) ·~v(t) = − cos t sin t + sin t cos t = 0.

Thus, the velocity must be tangent to the circular path.Similarly, the acceleration is perpendicular to the velocity:

~a(t) ·~v(t) = cos t sin t− sin t cos t = 0.

However, the acceleration is not parallel but antiparallel to the positionbecause ~a(t) = −~r(t). Therefore, we find that the acceleration pointsin toward the center of the circle in uniform circular motion. The force,likewise, points inward in the same direction as the acceleration.

We can now generalize our results. How do we modify our previousequations to describe the motion of an object moving at some speed v in acircle of radius r? We might start by adding a coefficient r to the positionto change the radius:

~r(t) = (r cos t, r sin t).

Accounting for the speed v is a little trickier. Recall that the period T(the time for the object to make a full revolution) of the sine and cosinefunctions can be found by dividing 2π by the coefficient of t (Section 1.6).Since the object moves at a constant speed, we can simply divide thedistance traveled in one revolution, the circumference of the circle, bythe period to find the speed. Thus, for

~r(t) = (r cos(ct), r sin(ct)),

111

3. NEWTON’S LAWS

we havev =

2πrT

=2πr2πc

= cr,

so c = v/r and~r(t) =

(r cos

(vr

t)

, r sin(v

rt))

. (3.29)

We can confirm that we’ve indeed found the correct position function bycalculating the velocity vector:

~v(t) =~r ′(t) =(−v sin

(vr

t)

, v cos(v

rt))

, (3.30)

where we used the chain rule to calculate the derivative of each compo-nent. As desired,

|~v(t)| =√

v2 sin2(v

rt)+ v2 cos

(vr

t)= v.

We can now find the acceleration and force:

~a(t) = ~v ′(t) =(−v2

rcos

(vr

t)

,−v2

rsin(v

rt))

(3.31)

and

~F(t) = m~a(t) =(−mv2

rcos

(vr

t)

,−mv2

rsin(v

rt))

. (3.32)

Like before, the velocity vector is perpendicular to the position and accel-eration vectors, which you can verify, and the acceleration vector pointsinward toward the center of the circle.

Finally, we’ve arrived at our most important result. If an object movesin a circular path at a constant speed, a situation known as uniform cir-cular motion, the net force acting on the object must point inward andhave magnitude

|~F(t)| =√(−mv2

r

)2

cos2(v

rt)+

(−mv2

r

)2

sin2(v

rt)

,

or

|~F(t)| = mv2

r. (3.33)

112


A net force that results in circular motion is commonly called the cen-tripetal force. The result expressed by Equation 3.33 is important enoughthat we state it in words:

If an object of mass m undergoes uniform circular motion atspeed v and radius r, the magnitude of the centripetal force ismv2/r.

One very important thing to remember is that the centripetal force,in and of itself, is not the source of motion; it is merely the net force weknow must be acting on an object if the object is to undergo circular mo-tion. As with all other net forces, the centripetal force is simply the forceor the sum of all the forces acting on the object, such as a tension forcefrom a string or a normal force from a curved track. Therefore, just as wewouldn’t draw the net force on a free-body diagram, it’s usually not agood idea to label centripetal force on a free-body diagram, either. (Youdefinitely don’t want to “double count” any forces.) It merely is a wayto account for the net force—and subsequently determine the individ-ual forces acting on the object—based on the kinematic requirements ofa circular path.

In hindsight, notice that we could have come by the correct positionfunction by another means—the answer lies in integration and the chainrule. When we take the derivative of a function, the inside coefficient oft always ends up “outside” the function. Since we need a coefficient ofsine and cosine to be r in the position function and but v in the velocityfunction, we can figure out that we need an “inside” coefficient (of t) ofv/r to cancel out the r and introduce a v when we take the derivative ofthe position.

Example 3.14 You swirl around in a circle a small ball connected to a taut,massless string of length L (Figure 3.12). If the string makes an angle θ withthe vertical, what is the speed v of the ball, assuming it is constant?

Assume the ball has mass m. The only forces acting on the ball are grav-ity and the tension force. Breaking up the tension force ~T into vertical andradial components reveals that the net vertical force must be zero, in which

113

3. NEWTON’S LAWS

q

m

L


case

|~T| cos θ −mg = 0

|~T| = mgcos θ

.

Newton’s second law in the radial direction gives

|~T| sin θ =mv2

L sin θ.

Therefore

v2 =|~T|L sin2 θ

m

=mg

cos θ· L sin2 θ

m,

and

v =

√gL sin2 θ

cos θ.

The speed can be more neatly expressed in terms of the angular speed ω (theGreek letter omega), which is the number of radians traveled per second andis given by ω = v/(L sin θ):

ω =

√g

L cos θ. (3.34)

�

114


Example 3.15 The International Space Station (ISS) orbits the earth in a cir-cular orbit of radius R. We’ll find in Section 6.2 that the acceleration due tothe earth’s gravity here is not well approximated by g = 9.81 m/s2 but isinstead given by GM/R2, where G is the gravitational constant (see the ap-pendix) and M is the mass of the earth. At what speed must the ISS travelfor the astronauts inside to feel weightless?

An astronaut of mass m standing in the ISS feels two forces: a normalforce from the floor (or ceiling, depending on the speed) and a force due togravity. As we saw in Example 3.6, weightlessness occurs when the normalforce equals zero. Since the net force on the astronaut must be mv2/R, byNewton’s second law we have

mv2

R= m

(GMR2

)v =

√GM

R. (3.35)

We’re done, but we might as well check to see that our answer is reasonablegiven real-world values. The ISS orbits the earth at about 385 km above thesurface of the earth. We add this to the radius of the earth (in the appendix)to obtain R = 6.76 × 106 m. Using the values for M and G given in theappendix, we obtain v = 7700 m/s. This matches the actual average speedof the ISS to two significant figures. �

Non-uniform circular motion

Now consider the case in which our object is still constrained to a circularpath but does not have to move at a constant speed. We call such motionnon-uniform circular motion. For reasons that will soon become clear,we replace the speed v in our position function for the uniform case byd(t), a function of distance traveled in terms of t:

~r(t) =(

r cos(

d(t)r

t)

, r sin(

d(t)r

t))

. (3.36)

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3. NEWTON’S LAWS

Then the chain rule gives

~v(t) =(−d′(t) sin

(d(t)

rt)

, d′(t) cos(

d(t)r

t))

. (3.37)

Notice that we chose to place d(t) inside the sine and cosine expressionsin the position function so that differentiation would yield the result|~v(t)| = d′(t), which is simply the instantaneous speed.

Differentiating again, we can now find the acceleration:

~a(t) =(−d′′(t) sin

(d(t)

rt)− [d′(t)]2

rcos

(d(t)

rt)

,

d′′(t) cos(

d(t)r

t)− [d′(t)]2

rsin(

d(t)r

t))

.

We see two distinct parts to each component: one with a coefficient of−d′′(t) and one with a coefficient of [d′(t)]2/r. Suppose we split~a(t) upinto two distinct vector components, which we’ll call~aT(t) and~aR(t):

~a(t) =~aT(t) +~aR(t), (3.38)

where

~aT(t) =(−d′′(t) sin

(d(t)

rt)

, d′′(t) cos(

d(t)r

t))

(3.39)

and

~aR(t) =(− [d′(t)]2

rcos

(d(t)

rt)

,− [d′(t)]2

rsin(

d(t)r

t))

. (3.40)

We’ve named our components~aT(t) and~aR(t) because the former istangential to the object’s circular path (perpendicular to the position vec-tor and parallel or antiparallel to the velocity) and the latter points in theradial direction (antiparallel to the position vector and perpendicular tothe velocity). You’ll be asked to prove this in the problems.

Notice that the size of the tangential component~aT(t) equals the changein speed d′′(t), which makes sense. Thus the tangential acceleration(and the tangential force ~FT(t) = m~aT(t)) is the factor that changes thespeed of the object. The radial component, on the other hand, serves the

116


same purpose as the acceleration did in the case of uniform circular mo-tion. The magnitude of the radial acceleration ~aR(t) is simply equal to[d′(t)]2/r, the square of the speed divided by the radius, as it did before.Therefore we can write

|~FR(t)| = m|~aR(t)|

=m[d′(t)]2

r,

so

|~FR(t)| =m[v(t)]2

r. (3.41)

This result is completely analogous to Equation 3.33. If our derivationshave confused you a bit, just ignore the mathematics for now and keepin mind the one key idea of this section:

The magnitude of the radial centripetal force is always mv2/r,where v is the speed, even if v is changing. The magnitudeof the tangential force is dv/dt, which reduces to zero if v isconstant.

Again, we repeat: you should never mark centripetal force on a free-body diagram. Remember that centripetal force is simply the net forcethat must result for an object to undergo circular motion.

Example 3.16 A bead of mass m is constrained to slide on a fixed hoop ofradius R as shown (Figure 3.13). If the bead begins at the top of the hoop,where it is given a negligible initial speed, find the normal force exertedby the hoop on the bead as a function of θ, the angle measured from thehorizontal. For convenience, let angles above the horizontal be positive andbelow be negative (imagine placing the hoop onto a unit circle). Where doesthe force vanish?

Drawing a free-body diagram for the bead shows that the normal forceand gravity both act on the bead. However, only the component of gravityperpendicular to the hoop mg sin θ is relevant to the centripetal motion ofthe bead; the parallel component mg cos θ simply serves as the tangentialacceleration. (The normal force N, by definition, only acts in the direction

117

3. NEWTON’S LAWS

perpendicular to the hoop, so it is always relevant.) Taking the directioninward (toward the center of the hoop) to be negative and outward (awayfrom the hoop) positive, we find that the second-law statement in the direc-tion perpendicular to the hoop reads

−mv2

R= −mg sin θ + N,

where v is the speed of the bead. You can should confirm that this equationholds for negative θ. (Technically, here we are using polar coordinates, whichwe described at the end of Section 2.4.)

Using methods from the next chapter, it can easily be proven that thespeed is a function of θ as follows:

v =√

2gR(1− sin θ).

(See the problems if you’ve unsure why we can’t find v using our currenttools.)

Therefore we have

N = −mv2

R+ mg sin θ

= −2mg(1− sin θ + mg sin θ

= mg(3 sin θ − 2). (3.42)

We can make a couple commonsense checks here. When θ = π/2, thebead is at the top of the hoop and has zero speed. Therefore the net forceshould be zero and the normal force should be exactly enough to balance thebead’s weight. Accordingly, we find N = mg using Equation 3.42. Whenθ = −π/2, the bead is at the bottom of the hoop and has speed

√4gR,

therefore, we expect a net force of 4mg. Again, we find a consistent result ofN = −5mg.

The normal force vanishes when

mg(3 sin θ − 2) = 0 (3.43)

sin θ =23

, (3.44)

for which θ is about 0.73 radians or 42◦. Since N is strictly decreasing asθ decreases, we see that the normal force decreases as the bead falls down.About halfway between the perpendicular and parallel, the normal force

118


vanishes, at which point it starts to increase in magnitude, this time in theinward (negative) direction.

We’ve described the normal force completely via our expression for theforce in the radial direction, but we can do one last thing, which could beuseful if we want to study the vertical and horizontal components of thenormal force. The normal force in Cartesian coordinates ~N = (Nx, Ny) isgiven by

~N = N(cos θ, sin θ) = mg(3 sin θ − 2)(cos θ, sin θ). (3.45)�

qR

N

mg sin qmg cos q

Figure 3.13 Diagram for Example 3.16. A bead of mass m is constrained to slideon a fixed hoop of radius R. Shown are the forces acting on the beadat the instant it makes an angle θ with the horizontal.

Example 3.17 Similar to Example 3.16, two beads, each of mass m, begin atthe top of a circular hoop of radius R, where they are given negligible initialvelocities in opposite directions. The hoop is of mass M and is not fixedto the ground. Show that given certain conditions, the hoop will lift offthe ground at some point during the beads’ motion. Given conditions suchthat the hoop lifts off the ground, determine the angle θ, the angle betweenthe bead on the right and the horizontal, at which the hoop first lifts off theground.

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3. NEWTON’S LAWS

Three forces act on the hoop: gravity, the normal force from the floor,and the normal force provided by the two beads; only the last of these isnon-trivial. Since the beads travel in opposite directions, the horizontal com-ponents of the normal force cancel. Thus, only the vertical components arerelevant. By Newton’s third law, the total normal force exerted on the hoopNM is the opposite of twice the vertical component Ny of the normal forceexerted on each of the beads:

NM = −2Ny.

In Example 3.16, we showed that Ny was given by

Ny = mg sin θ(3 sin θ − 2),

in which caseNM = 2mg sin θ(2− 3 sin θ). (3.46)

We can see that NM is possibly positive only for 0 < sin θ < 2/3, so if thereexists some θ at which the hoop rises, it must be subject to these bounds.

The hoop lifts of the ground the instant the normal force exerted on thehoop is zero. The initial acceleration at this instant is zero. Thus Newton’ssecond law yields

2mg sin θ(2− 3 sin θ)−Mg = 0

6m sin2 θ − 4m sin θ + M = 0, (3.47)

for which the quadratic formula gives the following (putative) solutions forsin θ:

sin θ =4m±

√16m2 − 24mM

12m

=13±√

4m− 6M6√

m. (3.48)

Real solutions to this equation exist only if 4m − 6M is nonnegative; thehoop lifts off the ground only if 4m− 6M is positive, or

m >32

M. (3.49)

Note that we can safely discard the smaller of the two roots since the hoopwill first rise at the larger angle.

120


We’re not quite done yet, however, since we need to show that this yieldsa solution for θ and not just sin θ, which is bounded between −1 and 1. Wecan easily show that

√4m− 6M/(6

√m) < 1/3 as long as Equation 3.49

is satisfied, in which case 1/3 < sin θ < 2/3 and a solution for θ existsif and only if Equation 3.49 is satisfied. In general, as sin θ increases from1/3 to 2/3, the ratio m/M increases and is bounded asymptotically at 2/3(why?). �

qR

mm

M

Figure 3.14 Diagram for Example 3.17. The setup is the same as in the singlebead case (Example 3.16), but there are two beads, and the hoop isfree to move.

Example 3.18 A block of negligible size slides off a frictionless circular con-vex14 ramp of radius R as shown in Figure 3.15. At what angle θ, measuredfrom the horizontal, does the block lose contact with the ramp?

The problem is essentially equivalent to Example 3.16, except the ramp,unlike the hoop, cannot exert an inward force. Therefore, the block trav-els along the ramp with the same normal force N we found before (Equa-tion 3.42) until the normal force becomes zero, at which point the block losescontact with the ramp.

Thus the solution is identical to that of our earlier example (Equa-tion 3.44), where the normal force vanished for

sin θ =23

,

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3. NEWTON’S LAWS

for which θ is about 0.73 radians or 42◦. �

qR


3.5 Drag

In Section 2.3, we ignored the effects of drag, the resistive force that op-poses an objects motion as it travels through a fluid. Now we examinesome basic models for drag.15

Drag at low speeds

Using a simple direct proportion, sometimes called Stokes’ law, we caneffectively model the resistive force due drag for slow-moving objects.16

Where ~D is the drag force, ~v is the velocity of the object, and k is some

14You may be familiar with the terms “concave up” and “concave down” instead.These terms are perfectly well defined, but in general mathematicians instead prefer“concave” and “convex,” which mean the same thing, respectively.

15Air resistance is one type of drag. More generally, the same principles apply tomany types of fluids, which includes liquids in addition to gases.

16For instance, we could accurately describe the motion of a stone falling through alake.

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3.5. Drag

constant that depends on the viscosity of the fluid and shape and size ofthe object,

~D = −k~v. (3.50)

Note that Equation 3.50 is a vector statement. It implicitly says that thedirection of the drag force is opposite that of the direction of motion,which is what we would expect.

We first consider drag in a simple one-dimensional case: suppose aball of mass m is thrown straight downward at an initial velocity v0. Ifthe drag force is given by Equation 3.50, let’s find the ball’s velocity v as afunction of time t. (We take the upward direction to be positive.) Wherea is the ball’s acceleration, Newton’s second law gives us

ma = −kv−mg

−mdvdt

= kv + mg (3.51)

Try as we might, we won’t be able to solve the preceding equation,which has both a variable (v) and its derivative (dv/dt), with our currentmathematical toolbox. (Taking derivatives or integrals of both sides onlychanges the order17 of the equation but fails to change its essential char-acter.) Here we have a first-order differential equation. That is, we havean equation that contains a first derivative of the function for which weare trying to solve.18 (Don’t worry if you haven’t seen this before; youwon’t be asked to solve a differential equation in the problems.) We cansolve the equation at hand using a method called separation of variables.We’ll save the mathematical details of why this works to the problems,but we’re essentially allowed to treat dv/dt as if it were a fraction.

17The order of a derivative is the number of repeated derivatives taken. For instance,d3x/dt3 is third-order term. The order of a differential equation (an equation that con-tains derivatives), then, is the order of the highest-ordered derivative that appears inthe equation.

18If we want to solve for position, Newton’s second law is strictly speaking also adifferential equation—a second-order one, to be exact—but a trivial one at that, so wenever explicitly refer to it as such.

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3. NEWTON’S LAWS

We move dt to the right side and integrate to obtain∫ dvkv + mg

= −∫ dt

m1k

ln(kv + mg) = − tm

+ C

ln(kv + mg) =km

t + kC,

where C is some constant. To get rid of the logarithm, we exponentiateboth sides:

kv + mg = ekCe−km t

v = Ae−km t − mg

k,

where A = ekC/k is some constant.To determine A, we see that when t = 0, v = v0. Thus

v0 = A− mgk

A = v0 +mgk

,

and finallyv(t) =

(v0 +

mgk

)e−

km t − mg

k. (3.52)

As a simple check, it can be shown that as k approaches zero, our ex-pression tends toward v0 + gt, which is the velocity function in the setupwithout drag.

Our exponential function shows some marked differences comparedto the the simple linear function v0 + gt in the case without drag. Whereasthe linear function is unbounded as t grows very large, in our case, thevelocity function with drag has a horizontal asymptote. As t approachesinfinity, e−

km t approaches zero, and v gets very close to, but does not ex-

ceed, −mg/k. We call this limiting value v∞ the terminal velocity19 ofthe object:

v∞ =mgk

. (3.53)19Despite the fact that the velocity points downward, by convention we really mean

terminal “speed,” hence the lack of a negative sign.

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3.5. Drag

In many situations, an object gets close to its terminal velocity quitequickly, so it essentially travels at terminal velocity for the entire dura-tion of its motion.

Notice that an object traveling right around terminal velocity experi-ences approximately zero net force, which must be the case if the object’svelocity is to remain constant, since

~F = −kv∞ −mg = k(mg

k

)−mg = 0.

Example 3.19 A projectile of mass m is fired from the origin with an initialvelocity ~v0 = (v0x, v0y). Determine its position~r(t) as a function of time t.Assume that the effects of air resistance can be described by Stokes’ law withconstant k.

We take advantage of the vector form of Equation 3.50. Where the dragforce is ~D = −k~v = (−kvx,−kvy) and the weight is ~W = (0,−mg), New-ton’s second law gives

m~a = ~D + ~W(m

dvx

dt, m

dvy

dt

)= (−kvx,−kvy −mg),

from which we obtain the two equations

mdvx

dt= −kvx (3.54)

and

mdvy

dt= −kvy −mg. (3.55)

Owing to the independence of the x and y directions, our problem has re-duced to a problem very similar to the one-dimensional case. In fact, wealready solved Equation 3.55 before; it has solution

vy =(

v0y +mgk

)e−

km t − mg

k. (3.56)

Save the constant term −mg, Equation 3.54 is identical to Equation 3.55 andthus has solution

vx = v0xe−km t, (3.57)

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3. NEWTON’S LAWS

where we simply replaced the mg terms by zero. Now we integrate to getthe components of the position function:

x =∫

vx dt

=∫

v0xe−km t dt

= −mk

v0xe−km t + C.

x(0) = 0, so C = (m/k)v0x and

x(t) = −mk

v0xe−km t +

mk

v0x. (3.58)

Likewise,

y =∫

vy dt

=∫ [(

v0y +mgk

)e−

km t − mg

k

]dt

= −mk

(v0y +

mgk

)e−

km t − mg

kt + C.

y(0) = 0, so C = (m/k)(v0y + mg/k) and

y(t) = −mk

(v0y +

mgk

)e−

km t − mg

kt +

mk

(v0y +

mgk

). (3.59)

We could additionally eliminate the parameter t (as we did before toprove that, neglecting drag, projectiles follow a parabolic trajectory) to de-termine the dependence of y on x directly. However, we end up with a messyexpression that gives us relatively little insight. We simply note that theshape of the trajectory contains a term linear in x, as well as a term logarith-mic in x, which serves to make the trajectory “drop off” near the end. Thetrajectory is shown for several sample values of m, k, and ~v (Figure 3.16). �

126

3.5. Drag

0

5

10

15

0 5 10 15x (m)

y (m)

k = 0.05

k = 0.01

k = 0

Figure 3.16 For the projectile described in Example 3.19, several sample trajec-tories are given. Here m = 0.1 kg, v0x = 5 m/s, and v0y = 15 m/s,with k as given in the graph.

Drag at high speeds

At higher speeds, a more accurate model is the Rayleigh equation, whichrelates the magnitude of the drag force to the square of the object’s speed:

D = −kv2, (3.60)

where k is again some constant (though not necessarily the same con-stant in Stokes’ law). Like before, the force points opposite the object’svelocity.20

Strictly speaking, most everyday projectiles are moving quickly enoughto require the application of the Rayleigh equation as opposed to Stokes’law, so we should have used Equation 3.60 for our previous examples.Nonetheless, the steps for solving the problem in the one-dimensionalcase are the same, regardless of which equation we use. In the generalmulti-dimensional case, however, it is a little more difficult to use theRayleigh equation since the differential equation for the y-component ofthe velocity depends on the x-component and vice versa.

20The vector form of the equation reads ~D = −k|~v|~v.

127

3. NEWTON’S LAWS

For drag governed by the Rayleigh equation, we can determine ter-minal velocity for an object in free fall via a net force argument (we couldhave done the same for Stokes’ law as well). The speed v∞ for whichthe the net force is zero can be found by equating the forces provided bydrag and by gravity:

kv2∞ = mg

v∞ =

√mgk

. (3.61)

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3.6. Problems

3.6 Problems

1. a) Analogous to Example 3.3, suppose that we chop up a springof spring constant k into n identical pieces. What is the springconstant of each of the small pieces?

b) Suppose that we connect one end of a spring of spring constantk1 to one end of a spring of spring constant k2. What is theeffective spring constant k1,2 of the connected spring?

2. Suppose you have blocks numbered 1 through n from left to right.The ith block has mass mi, and each adjacent pair of blocks is con-nected by a string. In other words, block i is connected to blocki + 1. If a force F pulls on block 1 toward the left, find the tensionforce Ti between the ith block and the (i + 1)th block. What doesthe expression simplify to if mi = m for all i?

m1

Fm2

... mn–1 mn

T1 T2 Tn–1Tn–2

Figure 3.17 Diagram for Problem 2.

3. A constant force to the right of magnitude F is exerted on a freeinclined plane of mass M, which moves along a frictionless hori-zontal surface (Figure 3.9). Held up by static friction of coefficientµ, a block of of mass m rests on the incline without slipping down.Determine the minimum possible value of F.

4. Given the Atwood’s machine in Figure 3.18, find the acceleration ofeach of the two masses m1 and m2 and find the tension force in thestrings connected to each of the two masses, as well as in the stringsconnecting each of the pulleys to the ceiling.

5. a) Given the same setup as in Example 3.14, replace the stringby a rigid, massless rod. You twirl the ball at a known speedv (you may choose instead to work with the angular speed

129

3. NEWTON’S LAWS

m2

m1


ω) and find that the the rod quickly settles at an angle θ withrespect to the vertical. Determine θ as a function of v or ω.(Hint: Be careful. Note that cos θ ≤ 1.)

b) Explain why it is not possible to twirl the ball in a circular pathparallel to the ground such that the ball lies above the string.

6. Prove that ~aT(t)⊥~aR(t) and ~v(t)⊥~aR(t). Why must ~aR(t) alwayspoint inward toward the center of the circular path? What happensif d′′(t) = 0?

7. A car cannot make a turn on a flat road without friction. However,if the road is banked at an angle θ as shown in Figure 3.19, no fric-tion is required.21 What is the minimum θ needed for a car to drivein a banked circular path of radius R at a maximum constant speedv if the coefficient of friction is µ? What is the result as µ vanishes?(Hint: You will want to obtain an equation in terms of tan θ.)

8. In Example 3.16, we claimed that we needed the methods of thenext chapter to determine v, the speed of the bead as a function of θ.

21Although we focus on circular tracks of radius R in this problem, in general theresult applies to any turn whose (instantaneous) radius of curvature is R.

130

3.6. Problems

q

R


Why couldn’t we just have done something like v(θ) =∫

a(θ) dθ =∫g cos θ dθ? Find a more correct approach and explain why the re-

sulting integral is difficult to evaluate. (Hint: The given expressionis not dimensionally correct.)

9. A ball of mass m is thrown straight downward at an initial veloc-ity v0. If the drag force is given by the Rayleigh equation (Equa-tion 3.60), find the ball’s velocity v as a function of time t. Withoutusing a net force argument, establish that the ball has a terminal

velocity given by v∞ =√

mgk . You will need to use the integral∫ dx

x2 − a2 = −1a

tanh−1 xa+ C,

where C is a constant and tanh−1 is the inverse of the function

tanh x =e2x − 1e2x + 1

.

10. (The following problem is more of an exercise in calculus than inphysics—skip it if you wish.) Given a first-order differential equa-tion of the form

g(y)dydx

= f (x),

separation of variables essentially makes the claim that∫

g(y) dy =∫f (x) dx, which you will justify in this problem. If the integrals are

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3. NEWTON’S LAWS

doable, we end up with an equation containing y on the left and xon the right, which can then be solved for y in terms of x.

a) Where G is the antiderivative of g (i.e., dG/dy = g), find thederivative of G with respect to x.

b) Use your result to prove that∫

g(y) dy =∫

f (x) dx.

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