05-1

Note 05 Applying the Laws of MotionSections Covered in the Text: Chapter 5

In this note we apply Newton’s Laws to solve somespecific problems. We describe an object in uniformcircular motion and consider the physics of an objectmoving through a viscous medium. We begin withthe concept of friction.

The existence of the force of friction eluded thenatural philosophers such as Aristotle and others wellinto modern times. Indeed, it could be said that theunderstanding of the force of friction paved the wayfor the laws of motion.

Forces of FrictionWhenever two objects in contact are moved oneagainst the other, some resistance to the movement isobserved. This resistance is attributed to a force offriction over the area of the surfaces in contact. Twoforces have been identified, the force of static frictionand the force of kinetic friction. In both cases, the forceis directed opposite to the direction of motion or tothe direction of impending motion (in the case ofobjects not actually moving). Both types of friction arecommonly quantified in terms of a coefficient. Themotion of an object through a viscous medium such asair or a fluid is also resisted by a force of friction.

Static FrictionWe all learn in childhood that in order to make anobject like a block move across a floor we need toapply a force to the block. Careful study reveals that ifthe force applied is not great enough, then the blockdoes not move. If the force is increased steadily, thena condition is eventually reached at which the block“breaks free” from the supporting surface and moves.This effect is interpreted to be the result of a force ofstatic friction between the block and the supportingsurface.

The origin of the force is not difficult to identify. Acritical observation is that the same object can bemoved more easily over a smooth surface than arough one. This means that the underside of the blockand the supporting surface are not infinitely smooth.Microscopic jagged points in both surfaces must catchone another thereby resisting movement. Before theonset of movement, the force of static friction equalsthe applied force (as required by Newton’s third law)and is directed opposite to the applied force (oropposite to the direction of impending motion). Thisforce is found to be proportional to the normal force

the supporting surface exerts on the object (which isthe same as the weight of the object if the surface ishorizontal).

Kinetic FrictionWe soon learn as well that once the block is moving,less force is required to keep the block moving atconstant speed than is required to get the blockmoving in the first place. This implies the existence ofa force of kinetic (moving) friction. If a force is appliedto keep the block moving at constant speed, then theforce of friction equals the applied force (as requiredby the first law). This force, too, is found to be propor-tional to the normal force. All things being equal theforce of kinetic friction is less than the force of staticfriction.

A wealth of experimental evidence reveals that theforces of static and kinetic friction can be written

fs ≤ µsn …[5-1a]

and fk = µkn , …[5-1b]

where µs and µk are the coefficients of static and kineticfriction, respectively, and n is the normal force. Eq[5-1a] gives an upper bound; the maximum force ofstatic friction is

fs.max = µsn ,

which exists just before the block moves. Figure 5-1summarizes these and other observations about staticand kinetic friction.

Figure 5-1. How the forces of static and kinetic friction varywith the force applied.

Note 05

05-2

Unhappily, the forces of static and kinetic friction are“messy” forces in the sense that the correspondingcoefficients are not true physical constants; theydepend on the types of materials in contact, on thesmoothness of the surfaces and on environmentalfactors such as humidity and the existence oflubrication. Let us consider an example.

Example Problem 5-1Determining µs and µk Experimentally

For any two dry materials in contact, the coefficientsof static and kinetic friction can be calculated with thehelp of the simple apparatus sketched in Figure 5-2. Ablock made of the one material is placed on the sur-face of an inclined plane made of the other material.The angle θ of the plane is increased slowly until theblock starts to move. (a) How is the coefficient of staticfriction related to the critical angle θc at which theblock starts to move? (b) Describe how the coefficientof kinetic friction might be subsequently found.

Figure 5-2. Free-body diagram of a block on an inclinedplane.

Solution:For convenience we choose a coordinate system inwhich the x-axis points downwards parallel to theincline and the y-axis points upwards perpendicularto the incline. Three forces act on the block whether itis moving or not: the component of the force ofgravity acting parallel to the incline downwards, thenormal force acting perpendicular to the incline andthe force of friction (static or kinetic depending onwhether the block is stationary or moving) actingupwards parallel to the incline.

(a) Before the block moves, the three forces put theblock in a state of translational equilibrium. ApplyingNewton’s second law in component form to the blockgives the equations

Fx∑ = mgsinθ – fs = 0 , …[5-2a]

and Fy∑ = n – mgcosθ = 0 . …[5-2b]

These equations apply for any angle θ of inclination.But at the critical angle θc at which the block is on theverge of slipping, the force of static friction has itsmaximum magnitude, µsn. Thus eq[5-2a] at that pointgives

mgsinθc = µsn . …[5-3a]

Now rearranging eq[5-2b] we have

mgcosθc = n . …[5-3b]

Dividing eq[5-3a] by eq[5-3b] gives

µs = tanθ c . …[5-4a]

Thus the coefficient of static friction equals the tan-gent of the critical angle. Thus a measurement of θcenables us to calculate µs.

(b) The coefficient of kinetic friction can be found asfollows. Ordinarily, if the angle of inclination wereheld constant after the block starts to move, then theblock would actually accelerate down the incline. Butif immediately after the block starts to move, the anglewere reduced to, say, θ’ so that the block moves downthe incline at constant speed then the first lawrequires that the sum of the x-components of force iszero, i.e.,

Fx∑ = 0 .

Thus from eq[5-2a]fk = mgsinθ ,

and

€

fk = µkn = µkmgcosθ '= mgsinθ' .

Thus

€

µk = tanθ ' . …[5-4b]

Thus the coefficient of kinetic friction equals the tan-gent of the angle θ’. Since θ’ is observed to be less thanθc it follows that µk < µs.

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Example Problem 5-2Calculating Coefficients of Friction

A wooden block of mass 1.00 kg is placed on an in-clined plane made of steel. The angle of inclination ofthe plane is slowly increased, until at 32.0˚ the blockstarts to move. Subsequently, when the angle isreduced to 26.0˚, the block moves at a constant speeddown the incline. (a) Calculate the coefficients of staticand kinetic friction. (b) Calculate the maximum forceof static friction and the force of kinetic friction.

Solution:(a) From eqs[5-4] we have:

µs = tan32.0o = 0.625

and µk = tan26.0o = 0.488 .

(b) The maximum force of static friction is

fs =mgsinθc

= (1.00kg)(9.80m.s –2 )sin32.0o = 5.19 N.

and the force of kinetic friction is

fk = mgsinθ ' ,

= (1.00kg)(9.80m.s –2 )sin26.0o = 4.30 N.

Thus the force required to start the block moving isgreater than the force required to keep the blockmoving at constant speed once it is moving.

Newton’s Second Law Applied to a Particlein Uniform Circular Motion

We have all seen objects being spun in a circle. A ballon the end of a string is one example. With the toolswe have developed we can describe this motion.

Consider a ball modelled as a particle moving clock-wise in a circle at constant speed v (Figure 5-3). Shownare two positions i and f on the particle’s path. Thevectors locating i and f relative to the center of thecircle are also shown, along with the instantaneousvelocity vectors of the object at the two positions. Theinstantaneous velocity vectors point along tangents tothe particle’s path and are therefore perpendicular to

the corresponding position vectors. In the elapsedtime ∆t = tf – ti the position vector sweeps out an angle∆θ.

ri rf

viv f

Δθ

i fΔr

vi

v f

Δθ Δv

Figure 5-3. A particle moving clockwise with uniform circul-ar motion.

We wish (1) to show that the particle is undergoing anacceleration, (2) to calculate the magnitude of thatacceleration and (3) to show that the accelerationvector is directed towards the center of the circle.

The first thing to appreciate about this kind ofmotion is that although the particle is moving withconstant speed and the magnitude of the velocityvector does not change, the direction of the velocityvector does change and changes continuously. Adiagram illustrating the relationship between thevelocity vectors is shown to the right in the figure.From this diagram it can be seen that

Δv = v f – vi .

Since the direction of the particle’s velocity vector ischanging, the particle is, by definition, undergoing anacceleration. According to its definition laid down inNote 03, the average acceleration is the change invelocity divided by the elapsed time:

a = v f – vitf – ti

=ΔvΔt

. …[5-5]

You should be able to see that the triangles formed bythe position and velocity vectors in the figure areisosceles triangles and are similar having the sameinner angle ∆θ . Thus the following relationshipholds:1

1 These ratios define the inner angle in radians. There will be

Note 05

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| Δv |v

=| Δr |r

.

Subsituting for |∆ v| from eq[5-5] we can obtain themagnitude of the average acceleration:

| a | Δtv

=| Δr |r

yields | a |= vr| Δr |Δt

.

The magnitude a of the instantaneous accelerationvector is obtained by taking the limit of the magnitudeof the average acceleration as ∆t → 0:

a = limΔ t→ 0

| a |= vrlimΔt→ 0

| Δr |Δt

=v2

r.

Thus we have the magnitude of the acceleration. Toget the direction of the acceleration vector notice thatas ∆t → 0, so also does ∆θ → 0 and Δv becomesperpendicular to vi or antiparallel to r i . Since it isantiparallel to r i it is automatically pointing towardthe center of the circle. This acceleration is called acentripetal acceleration (centripetal, meaning center-seeking) and is commonly denoted ac:

ac =v2

r. …[5-6]

The time required for the particle to execute one com-plete revolution is called the period, and is commonlydenoted T. The distance travelled by the particle inone revolution is the circumference of the circle, 2πr. Tis the distance travelled divided by the speed:

T =2πrv

.

We can also describe uniform motion in a circle interms of force. Figure 5-4 shows a more real-worldrepresentation of Figure 5-3. A ball on the end of astring is being moved at uniform speed v in a circle ofradius r. To avoid the question of how the ball stays“up” we can assume that the ball is moving on africtionless horizontal table (not shown) and that weare looking down from directly above the table.

more on the radian in Note 10.

We have already seen that since the velocity of theball is continuously changing, the ball is undergoing acentripetal acceleration. Its magnitude is given byeq[5-6] and its direction is towards the center of thecircle. This centripetal acceleration is caused by theforce exerted on the ball; the magnitude of this force isthe tension T in the string. We can therefore write inaccordance with the second law:

Figure 5-4. A body in uniform circular motion. The handexerts a force on the ball whose magnitude is the tension Tin the string.

F∑ =mac = mv2

r .

The magnitude of the centripetal force required tomove the ball in a circle is the tension T in the string.

Example Problem 5-3An Object in Uniform Circular Motion

A ball of mass 0.500 kg is moved in a horizontal circleat a constant speed of 2.00 m.s–1 on a frictionless table(as depicted in Figure 5-4). Calculate the centripetalacceleration and the tension in the string.

Solution:The magnitude of the centripetal acceleration is givenby eq[5-6]:

ac =v2

r=(2.00m.s–1)2

(0.500m)= 8.00 m.s–2 .

Thus T = mac = (0.500kg)(8.00m.s–2 ) = 4.00 N.

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An object in the real world has to move through amedium like air or water. We have already describeda number of examples of objects moving through airand we have chosen to simply ignore the possibleeffect of the air on the object’s motion. In the next sec-tion we examine this special kind of frictional force.

Motion in the Presence of Velocity-Dependent Resistive Forces

If the medium through which an object moves hassufficient density and the object moves with sufficientvelocity, then the object will experience an appreciablefrictional force that depends in a non-obvious way onits velocity relative to the medium. If the velocity islow then the frictional force is observed to be directlyproportional to the velocity. If the velocity is highthen the force is observed to be proportional to thesquare of the velocity. We consider here only theformer type.

Resistive Force Proportional to Object VelocityWe consider this physical situation illustrated inFigure 5-5 because it is at the heart of the experiment“Simple Measurements” that you will be doing in thelab soon. The experiment is a simple one. A marble isreleased at the top of a cylinder of shampoo andallowed to fall slowly to the bottom. Two marks 0.500m apart are etched on the cylinder to enable you tomeasure the marble’s average velocity (which is alsothe average speed in this case). (Average speed equalsdistance travelled divided by elapsed time.) Thenfrom various measurements you can calculate theviscosity of the fluid.2

The expression you will use to calculate theviscosity is derived on the assumption that the marblemoves downwards at a constant velocity called theterminal velocity. Therefore a fundamental questionasked is “does the marble move downwards at avelocity that is truly constant?” This question can beanswered at least in principle by calculating how longthe marble takes to reach a constant, maximumvelocity, and if this time elapses before the timing ofthe marble’s fall is begun when the marble passes thetop mark on the cylinder.

We make the assumption that the resistive force Rexerted on the marble by the fluid is directlyproportional to the object’s velocity and is oppositely 2 The viscosity of a fluid is a measure of what could be called thefluid’s “stickiness”. This topic belongs to the study of fluids (Notes12 & 13). In this first experiment in the lab the major objective is tomaster various kinds of measurements and not just to obtain theviscosity.

directed. Thus it has the form:

Figure 5-5. (a) The free-body diagram of a marble fallingthrough a fluid, and (b) the velocity of the marble as a func-tion of elapsed time. What is meant by the time constant isshown.

R = –bv .

where b is a constant, dependent on factors such asthe viscosity of the fluid and the radius of the marble.3The object is moving downwards so we can apply thesecond law:

Fy∑ = may ,

or Fg + R =ma ,

which gives mg – bv = m dvdt

.

Rearranging we have

dvdt

= g – bmv . …[5-7]

This is a first-order differential equation in v. You should

3 For a falling marble the constant b has the form 6πη(d/2) whered is the diameter of the marble and η is the viscosity.

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be able to show that a solution of the equation is:4

v(t) =mgb1 – e

–btm

, …[5-8]

= vT 1– e−tτ

where we write vT = mg/b and τ = m/b. The factor vThas units of velocity and is called the terminal velocity.This is the velocity the object approaches after asufficiently long time, i.e., as t → ∞. The quantity τ hasunits of time and is called the time constant (it goes byother names in other areas of physics where thevariable has a similar form as eq[5-8]).5 It is the timerequired for the velocity of the object to approach towithin 1/e of the terminal velocity. In the fluid usedin the experiment “Simple Measurements” this time isseen to be of the order of 10 milliseconds. Theconclusion to be drawn is that the marble does indeedreach terminal velocity before timing actually begins.

The Fundamental Forces of NatureAs far as is known to physicists today, there are fourfundamental forces in nature. They are the gravita-tional force, the electromagnetic force, the nuclear forceand the weak force. We can discuss these only briefly.

The Gravitational ForceIsaac Newton was the first to describe the gravita-tional force in mathematical terms in 1686. Accordingto his description any two masses attract each otherwith forces that are directly proportional to theproduct of their masses and inversely proportional tothe square of the distance between their centers. Inother words, for any two masses m1 and m2 a distancer apart the magnitude of the force is

Fg = Gm1m2

r 2.

G is a constant that today is called the universalgravitational constant. From numerous experiments itsaccepted value is:6

4 To show that eq[5-8] is a solution of eq[5-7] it is sufficient toshow that when it and its first derivative are substituted into eq[5-7]the LHS=RHS.5 Time constants appear in other areas of physics, in particularin the experiment “The Capacitor” in the PHYA21 lab.6 Do not confuse the universal gravitational constant G with the

G = 6.67x10–11 N.m2.kg–2,

to 3 significant digits. We shall continue the study ofthis force in later notes.

The Electromagnetic ForceWe have seen that mass is a fundamental property ofmatter. Charge is another. Charges attract or repel eachother with a force called the electromagnetic force . It hasa form similar to the gravitational force (with charge qinstead of mass m and the electric constant k instead ofthe gravitational constant G). We shall discuss thisforce in detail in Note 20.

The Nuclear and Weak ForcesThe nuclear force is the force responsible for keepingprotons and neutrons bound together in the atomicnucleus. If this force did not exist then mutual repul-sion between protons in the nucleus would cause thenucleus to fly apart. This force is a very short-rangeforce and drops to virtually zero much beyond thenucleus itself. The weak force figures in radioactivedecay. Its description belongs in a higher-level coursein physics and is beyond the scope of these notes.

The Gravitational FieldThe fundamental forces of nature are all non-contactforces. When Newton proposed that the force ofgravity was, in essence, a non-contact force the ideawas resisted by the natural philosophers of the day, inspite of the fact that the predictions based on it wereso accurate. Scientists trained on ropes and pulleysfound it hard to believe that two masses, like theEarth and the Moon, could exert forces on one anothereven though separated by many kilometers of emptyspace.

The issue of contact was essentially side-stepped bythe brilliant English experimentalist, Michael Faraday,in his proposal of the concept of the gravitational field.Faraday conjectured that any object by virtue of itsmass sets up an entity in its surroundings called agravitational field. The field is continuous in the senseof having a strength (a magnitude) and a direction atevery point. The field is therefore a vector field. Asecond mass in the region of this field is, of necessity,in contact with the field at that point. The field incontact with the mass then gives rise to the gravita-tional force on the object.7 A field is associated with acceleration due to gravity g.7 Throughout their lifetimes, neither Faraday nor Newton hadany idea as to how this might happen. Suffice us to say that theexistence of the field, either gravitational, electric, nuclear or weak,has now been proven beyond doubt.

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each force of nature, and in each case the field isregarded as the source of the force (and the source ofthe energy transferred).

The gravitational field at the position of a test massmt is defined as the vector

g ≡ Fg

mt

, …[5-9]

where Fg is the gravitational force on mt. In words,the gravitational field at some point in space is thegravitational force per unit mass at that point. In thecase of gravity the gravitational field vector is whatwe know more familiarly as the acceleration due togravity vector g .

The definition, eq[5-9], provides a way of imagininghow a gravitational field might be mapped, or repres-ented graphically. We imagine that a test mass mt canbe moved to various selected (arbitrary) positions in aregion of space and the force per unit mass measured(by means of some instrument) at the chosenpositions. A vector can then be drawn at every chosenposition, with a magnitude equal to the magnitude ofthe gravitational force per unit mass, and with thedirection of the gravitational force vector. The collect-ion of vectors is the field map.

For example, this effort carried out in the neighbor-hood of a spherically symmetric source mass mightresemble Figure 5-6a in 2D space. The gravitationalfield is seen to have a radial geometry. The fieldalways points toward the source mass and themagnitude of the field decreases as r increases (withan inverse square dependency). Of course, this pictureis only a representation.

If this procedure were carried out near the surfaceof the Earth the result might resemble Figure 5-6b in2D space. This field is seen to be uniform. Themagnitude of the field strength at every point is g and

the field vector always points toward the center of theEarth.

Figure 5-6. 2D representations of the gravitational field neara spherically symmetric source mass (a) and near thesurface of the Earth (b)

Having studied a number of applications of Newton’sLaws we are ready to consider the concept of energy.This we will do in Notes 08 and 09.

To Be Mastered

• Definitions: force of static friction, force of kinetic friction, coefficient of static friction, coefficient of kinetic friction• Definitions: centripetal acceleration, uniform circular motion• Definitions: viscous force, time constant• Definitions: field, gravitational field vector

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Typical Quiz/Test/Exam Questions

1. A block of mass 1.0 kg is in contact with a horizontal surface (see the figure). The coefficient of static frictionbetween the surfaces is 0.7 and the coefficient of kinetic friction is 0.6. Three cases are shown: (1) when the netexternal force is zero, (2) when a force of 1.0 N is applied at an angle of 60˚ to the horizontal, and (3) when aforce of 2.0 N is applied at an angle of 60˚ to the horizontal.

block block block

1.0 N 2.0 N60˚ 60˚

(1) (2) (3)

Draw a free body diagram for the block in each case.

2. A ball of mass M is tied to the end of a string of length R and spun in a vertical circle at a constant speed v (seethe figure).

A

B

C

D

T

(a) Draw free body diagrams for the ball at positions A, B, C and D.(b) If M = 1.0 kg, R = 1.0 m and v = 4.00 m.s–1, calculate the tension T in the string, in N, at the four positions.For convenience, take g = 10.0 m.s–2.