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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
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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 =
A%%L&N( N)WT"N'* T+)) LAW*
←
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
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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.
APPLYING NEWTON'S THREE LAWS
←%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:
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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 .
A%%L&N( N)WT"N'* T+)) LAW*
←
rictional orces
→
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.
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• 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.
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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$. Though friction applies in so
many situations, it is often ignored in order to simplify the situation. =nless friction is e4plicitly stated
to be present in a given problem, in can be ignored. That said, friction remains one of the most
widely used applications of Newton's Laws.
APPLYING NEWTON'S THREE LAWS
←%roblems
→
Pro0lem 1
The coefficient of static friction, μ s , between a given bloc6 of weight 92N and a surface has a value
of .9. A hori;ontal force is applied to the bloc6. +ow much force must be applied for the bloc6 to
move>
When e4actly enough force is applied for the bloc6 to move, the static frictional force will be e!ual to
its ma4imum possible value. Thus:
F s 1 F s ma4 1 μ s F N
3ecause the bloc6 rests on a hori;ontal plane, the normal force is simply e!ual to the weight of the
bloc6: F N 1 92N . Thus F s 1 #92$#.9$ 1 B.9N . ecall that the static frictional force always cancels
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e4actly with the force applied to the bloc6. Thus the minimum necessary force to move the bloc6 is
simply B.9N .
A%%L&N( N)WT"N'* T+)) LAW*
←
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
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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.
APPLYING NEWTON'S THREE LAWS
←
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%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.
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"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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