Applying Newton

7/23/2019 Applying Newton

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APPLYING NEWTON'S THREE LAWS

← Applications of Newton's Laws

→Now that we have established the fundamentals of dynamics through Newton's Laws, we can apply

them to a variety of physical situations to see their full use. As stated before, Newton's Laws apply in

almost every physical situation. There are, however, a number of forces that most commonly arise.

We will limit our application to these forces: normal force, frictional force and tension force. rom

these three !uite common forces we will be able to predict the outcome of a variety of physical

situations.

These three forces, normal, tension and friction, are present in a surprising number of physical

situations. "ften the three forces are all present, as we shall see in the problems. Though Newton's

Laws apply in some sense to everything from !uantum mechanics to electricity, these forces #along

with gravity$ form the basis for applications within classic mechanics.

A%%L&N( N)WT"N'* T+)) LAW*

←

Terms and ormulae

→

Terms

Force - A force is defined as a push or a pull.

Inertia - The tendency of an obect to remain at constant velocity.

Inertial reference frame - Any frame in which Newton's Laws are valid.

Ma - The amount of matter in a given body.

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Ne!ton - The name given to a unit of force. "ne Newton is enough force to cause a / 0( body to

accelerate at a rate of one meter per second per second.

Ne!ton' T"ree La! -

irst Law: f F 1 2 then a 1 2 and v 1 constant*econd Law: F 1 ma

Third Law: F A3 1 - F 3A

Wei#"t - The gravitational force e4erted on a given mass.

Free $o%& ia#ram - A diagram of all forces acting upon a given obect.

Normal Force - The force caused by two bodies in direct contact that is perpendicular to the plane

of contact.

Frictional Force - The force caused by the electrical interaction between two bodies in direct

contact that is parallel to the plane of contact and in the opposite direction of the motion of one

obect relative to the other.

Tenion Force - The force felt by a rope or cable that transmits another force.

Static Frictional Force - The frictional force on two bodies at rest.

(oefficient of Static Friction - 5efines the proportionality between F N and F sfor two given

materials.

)inetic Frictional Force - The frictional force on two bodies in motion relative to one another.

(oefficient of )inetic Friction - 5efines the proportionality between F N and F 6for two given

materials.

(entri*etal Acceleration - The acceleration, directed toward the center of a circle, which causes

uniform circular motion.

(entri*etal Force - The force, directed toward the center of a circle, which causes uniform circular

motion.

Formulae

Ne!ton' Secon% La! F = ma



Ne!ton' T"ir% La! F AB = - F BA

Form+la for ma,im+m tatic

frictional force- F s

max = μ s F N

Form+la for .inetic frictional force- F k = μ k F N

E/+ation for centri*etal

acceleration- a =

E/+ation for centri*etal force- F =


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The Normal orce

→

"f all physical forces in everyday life, perhaps the most common is the normal force. The normal

force comes into play any time two bodies are in direct contact with one another, and always acts

perpendicular to the body that applies the force. The simplest e4ample of the normal force can be

seen in the situation of a man standing on a platform. 7learly a gravitational force acts on the man,pulling him down, perpendicular to the platform8 but since the man is not moving, another force must

act to counteract the gravitational force. This force is applied by the platform, and is called the

normal force, and is referred to as F N .

The normal force can also be seen as a direct conse!uence of Newton's Third Law. 7ontinuing with

the e4ample of the man on the platform, his weight, due to the gravitational force, pushes down on


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the platform. Newton's third law predicts that this force on the platform must be accompanied with an

e!ual and opposite force applied to the man by the platform. This force is precisely the normal force.

*ince the normal force is a reactive force, its magnitude is independent of the nature of the force

causing it. The most common normal force is caused by gravity, as seen in the man on the platform.

+owever, there can be additional forces that also cause a normal force.

7onsider a bloc6 on a platform with weight /2 N. n addition, someone pushes downward on the

bloc6 with an additional force of /9 N. The platform thus e4periences a total force of 9 N, and reacts

with a normal force of 9 N, 6eeping the bloc6 in e!uilibrium. Thus, in the situation of a hori;ontal

obect, the normal force is simple: it is merely e!ual in magnitude and opposite in direction to all

forces applied to the surface.

The Normal Force on an Inclined Plane

The normal force becomes more comple4, however, in situations where forces are not perpendicular

to the plane. 7onsider the case of a bloc6 resting on an inclined plane, or a ramp. n this instance,

the gravitational force on the bloc6 isnot perpendicular to the plane. n order to calculate the normal

force for this situation we must find the component of the gravitational force that is perpendicular to

the plane. We do so by brea6ing down the force vector into two components #see <ectors,

+eading $: one parallel to the plane and one perpendicular to the plane. The normal force thus has

e!ual magnitude and opposite direction of the component of the gravitational force that is

perpendicular to the inclined plane. =sing a free body diagram, all of these forces can be displayed,

and the resultant motion can be predicted:

Figure %: Free Body Diagram of an Inclined Plane



What does our free body diagram predict> To find out we analy;e all forces acting upon the obect.

The perpendicular gravitational force # F cosθ $ cancels e4actly with the normal force # F N $, as we

e4pected, and we are left with one force, the parallel gravitational force # F sinθ $, which points down

the plane. Thus the bloc6 will accelerate down the incline. *uch a prediction seems to fit with our

intuition: a bloc6 placed on an inclined plane will simply slide down the plane.

The normal force thus applies to a variety of situations. Though most commonly used with flat and

inclined planes, the normal force applies in any situation in which a force is e4erted on an obect by

direct contact from another obect.


←%roblems

→

Pro0lem 1

A bloc6 of /2N rests on a plane inclined ?9 o . n addition a hori;ontal force of /2N is applied to the

bloc6. What is the normal force applied by the inclined plane>

We solve the problem by drawing a free body diagram, and resolving all force vectors into

components parallel and perpendicular to the plane:







Solution 1

The component of the gravitational force perpendicular to the plane is given by:

F Gâä¥ = F Gsin 45 o = 10 sin 45 o = 7.07 N

*imilarly, the component of the applied force perpendicular to the plane is:

F âä¥ = F sin 45 o = 10 sin 45 o = 7.07 N

Thus the normal force on the bloc6 is simply the sum of the two perpendicular forces, or /?./?N .


←

rictional orces

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Another !uite common force is frictional force. Li6e the normal force, it is caused by direct contactbetween surfaces. +owever, while the normal force is always perpendicular to the surface, the

frictional force is always parallel to the surface. To fully describe the cause of friction re!uires

6nowledge beyond the scope of classical mechanics. or our purposes, it is enough to 6now that

friction is caused by electrical interactions between the two surfaces on a microscopic level. These


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interactions always serve to resist motion, and differ in nature according to whether or not the

surfaces are moving relative to each other. We shall e4amine each of these cases separately.

Static Frictional Forces

7onsider the e4ample of two bloc6s, one resting on top of the other. f friction is present, a certain

minimum hori;ontal force is re!uired to move the top bloc6. f a hori;ontal force less than this

minimum force is applied to the top bloc6, a force must act to counter the applied force and 6eep the

bloc6 at rest. This force is called the static frictional force, and it varies according to the amount of

force applied to the bloc6. f no force is applied, clearly there is no static frictional force. As more

force is applied, the static frictional force increases until it reaches a certain ma4imum value8 once

the hori;ontal force e4ceeds the ma4imum frictional force the bloc6 begins to move. The frictional

force, defined as F s ma4 , is conveniently proportional to the normal force between the two surfaces:

F s max = μ s F N

The constant of proportionality, μ s is called the coefficient of static friction, and

is a property of the materials that are interacting #i.e. two interacting rough

materials will have a higher value of μ s than two smooth materials$.

This e!uation for ma4imum static frictional force contains a lot of information, and a few remar6smust be made for clarification.

• The e!uation seems to be relating two vectors, F s ma4 and F N . This relation is valid only for

the magnitudes of the vectors, not the direction. n fact, the two vectors will always be

perpendicular.

• The e!uation introduces the concept of the coefficient of static friction. This constant varies

from material to material, but does not depend on the orientation of the material on the

surface. or e4ample, if a bloc6 of wood is set on a concrete platform, μ s is the same whether

the bloc6 is on its side, its front, or its top. n other words, the coefficient does not change

according to the surface area of contact.

• *ince the e!uation does not specify a direction for the frictional force, it must be stated and

understood that the frictional force always acts in the opposite direction as the force applied to

the obect.



• t is vitally important to remember that this e!uation only gives the maximumstatic frictional

force, which corresponds to the ma4imum force that can be applied to a body before it moves.

f a lesser force is applied to the body, a frictional force less than the ma4imum force

counteracts the original force.

Though it is rather surprising that frictional force and normal force are related in such a simple

manner, physical intuition tells us that they should be directly related. 7onsider again a bloc6 of

wood on a concrete platform. The normal force is given by the weight of the wood. f an additional

downward force is applied to the wood #producing a greater normal force$ the surfaces are actually

in closer contact than they were before, and the resulting electrical interactions are stronger. Thus,

intuitively, a greater normal force yields a greater frictional force. "ur intuition agrees with the

e!uation.

Kinetic Frictional Forces

"nce a force is applied to an obect that e4ceeds F s ma4 , the obect begins to move, and static

frictional forces no longer apply. The moving obect does still e4perience a frictional force, but of a

different nature. We call this force the 6inetic frictional force. The 6inetic frictional force always

counteracts the motion of the obect, and is independent of speed. No matter the speed of the obect

#as long asv @ 2 $ it e4periences the same frictional force. Also, for the same reasons as e4plained

with static friction, the 6inetic frictional force is proportional to the normal force:

F k = μ

k F N

This e!uation is of the same form as that for ma4imum static frictional force, and defines the

coefficient of 6inetic friction, μ 6 , which has the same properties as μ s , but a different value. μ 6 is a

property of the interacting materials, and, li6e μ s , is independent of orientation of the obects. The

only significant difference between the two friction e!uations is that the first measures the friction

between two stationary obects and its value is dependent on the force applied to one, while second

measures a frictional force that only e4ists when one of the obects is moving and which is not

depend on the force applied to the bloc6. inally, when comparing static with 6inetic friction, it must

be noted that μ s is always greater in value than μ 6 . *imply stated, this means that it ta6es less force

to 6eep a bloc6 moving than to start its motion.



These two types of friction, li6e the normal force, arise whenever two obects are in direct contact.

"ften both 6inetic and static friction apply to a given situation, as an obect might start at rest #when

static friction applies$ then begin to move #when 6inetic friction applies$. Though friction applies in so

many situations, it is often ignored in order to simplify the situation. =nless friction is e4plicitly stated

to be present in a given problem, in can be ignored. That said, friction remains one of the most

widely used applications of Newton's Laws.


←%roblems

→

Pro0lem 1

The coefficient of static friction, μ s , between a given bloc6 of weight 92N and a surface has a value

of .9. A hori;ontal force is applied to the bloc6. +ow much force must be applied for the bloc6 to

move>

When e4actly enough force is applied for the bloc6 to move, the static frictional force will be e!ual to

its ma4imum possible value. Thus:

F s 1 F s ma4 1 μ s F N

3ecause the bloc6 rests on a hori;ontal plane, the normal force is simply e!ual to the weight of the

bloc6: F N 1 92N . Thus F s 1 #92$#.9$ 1 B.9N . ecall that the static frictional force always cancels







e4actly with the force applied to the bloc6. Thus the minimum necessary force to move the bloc6 is

simply B.9N .


←

Tension orce

→

The final common application of Newton's Laws deals with tension. Tension usually arises in the use

of ropes or cables to transmit a force. 7onsider a bloc6 being pulled by a rope. The person doing the

pulling at one end of the rope is not in contact with the bloc6, and cannot e4ert a direct force on the

bloc6. ather a force is e4erted on the rope, which transmits that force to the bloc6. The force

e4perienced by the bloc6 from the rope is called the tension force.

Almost all situations you will be presented with in classical mechanics deal with massless ropes or

cables. f a rope is massless, it perfectly transmits the force from one end to the other: if a man pulls

on a massless rope with a force of /2 N the bloc6 will also e4perience a force of /2 N. An important

property of massless ropes is that the total force on the rope must be ;ero at all times. To prove this,

we go bac6 to Newton's *econd Law. f a net force acts upon a massless rope, it would cause

infinite acceleration, as a 1 F Cm , and the mass of a massless rope is 2. *uch a situation is physically

impossible and, conse!uently, a massless rope can never e4perience a net force. Thus all massless

ropes always e4perience toe!ual and opposite tension forces. n the case of a man pulling a bloc6

with a rope, the rope e4periences a tension in one direction from the pull of the man, and a tension

in the other direction from the reactive force of the bloc6:

Figure %: !"e !en#ion in a $a##le## ope







Tension and Pulleys

The dynamics of a single rope used to transmit force is clearly !uite simple: the rope ust transmits

an applied force. When pulleys are used in addition to ropes, however, more complicated situations

can arise. n a dynamical sense, pulleys simply act to change the direction of the rope8 theydo not change the magnitude of the forces on the rope. Dust as we assumed the ropes to be

massless, we will similarly assume that the pulleys we wor6 with are massless and frictionless,

unless told otherwise. The simplest case involving a pulley involves a bloc6 being lifted by another

bloc6 connected to a rope:

Figure %: !"e !en#ion in a ope and Pulley Sy#tem

This diagram represents a small bloc6 on the left in the act of being lifted by a

larger bloc6 on the right. Notice the forces T and -T: even when used inaddition to a pulley, the rope must still e4perience two e!ual and opposite

tension forces. rom the figure it may seem that the rope actually e4periences

two forces in the same direction, ma6ing the situation impossible. The

presence of the pulley, however, changes the situation to ma6e it physically

tenable. When analy;ing a rope and pulley situation it is useful to define a

direction not in terms of up or down, but in terms of the shape of the rope. n

the situation above, we can define the positive direction on the rope as

pointing upward on the left side of the pulley, and pointing downward on the

right side. When we define direction in this way the rope does actually

e4perience two e!ual and opposite forces.


←





%roblems

→

Pro0lem 1

A 9&g picture frame is held up by two ropes, each inclined ?9 o below vertical, as shown below. What

is the tension in each of the ropes>

3ecause the picture frame is at rest, the tension in the two ropes must e4actly counteract the

gravitational force on the picture frame. 5rawing a free body diagram we can calculate the vertical

components of the tension in the ropes:

7learly the hori;ontal components of the tension in the two ropes cancel e4actly. n addition, the

vertical components are e!ual in magnitude. *ince F 1 2 , then the vertical components of the

tension in the two ropes must cancel e4actly with the gravitational force: ! y 1 mg EFG! sin ?9 o 1 #9$

#H.I$ 1 ?HN . Thus: ! 1 1 B?.JN . The total tension on each rope is thus B?.JN .

Pro0lem 1

7onsider a /2&g bloc6 resting on a frictionless plane inclined B2

o

connected by a rope through apulley to a /2&g bloc6 hanging free, as seen in the figure below. What is the direction and magnitude

of the resulting acceleration of the -bloc6 system>

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Though this problem seems !uite comple4, it can be solved by simply drawing a free body diagram

for each bloc6. *ince the resulting acceleration of each bloc6 must be of the same magnitude, we

will get a set of two e!uations with two un6nowns, T and a. irst we draw the free body diagram:

"n bloc6 /, there are B forces acting: normal force, gravitational force and tension. The gravitationalforce, in terms of parallel and perpendicular components, and the normal force can be easily

calculated:

F G

= (10kg )

(9.8)= 98 N

F Gâä¥= F Gcos 30o= 84.9 N F G || = F Gsin 30o = 49 N

The normal force is simply a reaction to the perpendicular component of the gravitational force.

Thus F N 1 F (EK 1 I?.HN . F N and F (EK thus cancel, and the bloc6 is left with a force of ?HN down the

ramp, and the tension, T, up the ramp.



"n bloc6 , there only two forces, the gravitational force and the tension. We 6now that F ( 1 HIN ,

and we denote the tension by T. =sing Newton's *econd Law to combine the forces on bloc6 / and

bloc6 , we have e!uations and un6nowns, a and T:

F = ma

10a 1 = T - 4910a 2 = 98 - T

+owever, we 6now that a / and a are the same, because the two bloc6s are bound together by the

rope. Thus we can simply e!uate the right side of the two e!uations:

! - ?H 1 HI - ! Thus ! 1 /? and ! 1 B.9N

With a defined value for T, we can now plug into one of the two e!uations to solve for the

acceleration of the system:

/2a 1 B.9 - ?H 1 ?.9

Thus a 1 .?9mC# . nterpreting our answer physically, we see that bloc6 / accelerates up the

incline, while bloc6 falls, both with the same acceleration of.?9mC# .

Pro0lem 1

Two /2&g bloc6s are connected by a rope and pulley system, as in the last problem. +owever, there

is now friction between the bloc6 and the incline, given by μ s 1 .9 and μ 6 1 .9 . 5escribe the

resulting acceleration.

We 6now from the last problem that bloc6 / e4periences a net force up the incline of ?.9 N. *ince

friction is present, however, there will be a static frictional force counteracting this

motion. F s ma4 1 μ s F n 1 #.9$#I?.H$ 1 ?.9N . 3ecause this ma4imum value for the frictional force

e4ceeds the net force of ?.9 N, the frictional force will counteract the motion of the bloc6s, and the

bloc6 system will not move. Thus a 1 2 and neither bloc6 will move.

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Applying Newton