# Chapter 5 Applications of Newton's Laws

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```					Chapter 5       Applications of Newton’s Laws

We shall apply Newton’s laws of motion to a variety
of situations in which several types of forces are
involved:
• Gravitational force
• Tension
• Contact Forces (including friction)

(5-1)
Gravitational force Fg
Also known as the “weight” of
m                Fg     an object
g

earth

All objects near the surface of the earth are uniformly
accelerated towards the center of the earth. The
magnitude of the acceleration g = 9.8 m/s 2
From Newton’s second law we get that Fg = ma = mg
(5-2)
A practical method
of measuring Fg Use
a spring scale as
shown in the figure.                    y-axis
The spring has a
pointer attached to it
so that we can
determine its
elongation

System = grapes. There are two forces acting on the grapes:
The gravitational force mg and the spring force Fs which is
given by: Fs = k|y|. Here y is the elongation of the spring
The net force Fynet = Fs - mg = may = 0 because the grapes are
stationary. Thus mg = Fs = k|y|                (5.3)
T                    Tension
crate     rope
It is the force exerted by a rope or a
floor
cable

Tension has the following characteristics:
• Tension always pulls (never pushes)
• The direction of T is along the rope
pulley
• The magnitude |T| is constant                        T
even when T changes direction            T

(as in the case of the pulley)      crate                  pull
(5-4)
Normal Force (contact force)
Characteristics of the normal force FN
• FN is perpendicular to the contact surface
• FN always pushes away from the contact surface
(never pulls)

FN
crate
Fg
floor
(5-5)
The normal force and Newton’s third law
FN = Fcf     System = crate
crate                      The normal force on the crate
exerted by the floor is labeled Fcf
floor
System = floor
crate
The normal force on the floor
exerted by the floor is labeled F fc
floor
Ffc

Newton’s third law: Fcf + F fc = 0 → Ffc = - Fcf = - FN
The crate exerts a force F fc on the floor which is equal and
opposite to FN                          (5-6)
Q: How can we
measure F N ?
A: With a bathroom
scale!
The bathroom scale
measures the normal
force (and not
necessarily mg unless
the object on the scale
does not accelerate)

(5-7)
Fas is the normal force exerted
on the apple by the scale.
apple            FN = Fas    Fsa is the normal force exerted
on the scale by the apple
scale
These two forces are equal and
FN = Fsa   opposite by virtue of Newton’s
third law
y-axis                       System = apple
Fynet = FN - mg
FN          Fynet = may = 0       →

mg          FN - mg = 0       →
(5-8)               Indeed: FN = mg
What does the scale reads when the
y-axis
elevator   elevator accelerates upwards ?
System = man
..              Fynet = FN - mg
Fynet = ma   (Newton’s second law)

FN     →    FN - mg = ma
→    FN = mg + ma

a                 mg          FN > mg
The scale in this case gives a reading
which is higher than the actual weight
scale
mg                         (5-9)
What does the scale reads when the
y-axis
elevator   elevator accelerates downwards ?
System = man
..              Fynet = FN - mg
Fynet = - ma   (Newton’s second law)

FN     →    FN - mg = - ma
→    FN = mg - ma

a                 mg          FN < mg
The scale in this case gives a reading
which is lower than the actual weight
scale
mg                         (5-10)
Atwood’s Machine
a
It consists of two masses m1 and
a   m2 (m2 > m 1) connected with a
y
rope that goes around a pulley.
The rope and the pulley have
m2         negligible masses.
m1

The two masses are released and move under gravity which
acts downwards. The heavier mass m 2 will move down and
the lighter mass m1 will move up. Determine the
acceleration a of the two masses
(5-11)
y                                          a
a                                                           T
T    a
a
y                                        y

m1                            m2
m1g                             m2g
m2
m1
System = m1                        System = m2
Fynet = T − m1 g = m1a →           Fynet = T − m2 g = − m2a →
T = m1g + m1a     eqs.1            T = m2 g − m2a       eqs.2

We combine eqs.1 and eqs.2 → m1 g + m1a = m2 g − m2 a Solve for a
( m2 − m1 )
→                   a=g
( m2 + m1 )                             (5-12)
Friction   Friction opposes motion or impending motion

(5-13)
(5-14)
motion               Friction was first studied by
FN          Leonardo da Vinci
crate               T
He discovered that friction has the
f              rope   following characteristics:
floor    mg
• Friction opposes motion or impending motion
• Static and kinetic friction are independent of contact area
• The static and kinetic friction are proportional to FN
Static friction fs                  0 < f S < µ S FN
µs is a constant known as “coefficient of static friction”
Kinetic friction fk                  f k = µ k FN
µk is a constant known as “coefficient of kinetic friction”
Leonardo da Vinci
(self portrait)
(1452-1519)
motion                 Consider the following experiment:
FN
T      We start pulling a heavy crate using
crate                     a rope with a force T that increases
f            rope
with time. At some instant tm the
floor     mg           crate will start to move

f             µk < µs
We then plot the frictional force f
as function of time. The static                      µ s FN
friction increases with time. At
tm the static friction becomes                                    µ k FN
maximum (µs FN) and the crate             Static kinetic
friction friction
starts to move. The kinetic                    tm
friction has a smaller constant                Motion         t
value of µk FN .           (5-15)              starts
Example (5-6),page 121
y
A mass m is at rest on a
x       horizontal ramp. The ramp is
slowly raised. Find the critical
angle θc at which the mass will
start to slide

Fynet = FN - mg cos θ = ma y = 0 → FN = mg cos θ (eqs.1)
Fxnet = mg sin θ - f S = max = 0 → mg sin θ = f S = µ S FN (eqs.2)
From eqs.1 substitute FN into eqs.2 →
mg sin θ = µ S mg cos θ       →      tan θ C = µ S

(5-16)
Example (5-8) page 123. An automobile has a mass
m = 1000 kg. The static friction with the road is µs = 0.8
Determine the maximum forward acceleration (without
spinning the wheels)
Fynet = FN - mg = may = 0    →     FN = mg     (eqs.1)
Fxnet = fs = ma   →    a = fs /m
amax = (fs)max /m = µs FN /m (eqs.2)
From eqs.1 substitute FN into eqs.2          → amax = µsmg /m
amax = µs g = 0.8×9.8 = 7.8 m/s2
When the rope breaks
the book does not stay
on its circular path but
tangent to the orbit

Conclusion: For an object to be able to move on a circular
orbit a force that points towards the center of the circle is
required. This force is known as the “centripetal” force.
When the centripetal force stops acting the object cannot stay
on its circular orbit
(5-18)
(5-19)
m                    Consider an object of mass m
m        moving on a circle of radius r
with uniform speed v. The
acceleration points towards the
center of the orbit and has
magnitude a = v2/r = ω2r
m
From Newton’s second law a force F = ma must act on the
object . Thus the centripetal force F is given by the
expression:
mv 2
F=          or     F = mω 2r
r
The magnitude of F is constant, its direction is not. It
always points towards the center of the circular orbit.
Recipe for problems that involve
uniform circular motion of an
C   .   y
R        object of mass m on a circular
orbit of radius R with speed v

x
m
v
• Draw the force diagram for the object
• Choose one of the coordinate axes (the y-axis in this
diagram) to point towards the orbit center C
• Determine Fynet
• Set Fynet = mv2/R or    Fynet = mω2R        (5-20)
Example (5-11) page 128
A small dice sits 0.15 m from
the center of a turntable. If µs
between the turntable and the
(5-21)                          dice is 0.55 what is the largest
y
possible angular speed such
x
that the dice will not slide off

Fynet = FN - mg = may = 0 →          FN = mg     (eqs.1)
Fxnet = fs = mω2r (eqs.2) → ω2 = fs /mr
ω2max = (fs )max /mr       From eqs.1 we have:    (fs)max = µs FN
Thus    ω2max = µs mg/mr → ωmax = [µs g/r]1/2 = 6 rad/s
Example (5-12) page 129
A space station in the form of a hollow
circular tube rotates around its axis. If
the distance from the station outer wall
to the center is 50 m what should be
y             the speed v of the outer wall so that a
earth
Fynet = FN = mv2/R      (eqs.1)
Also       FN = mg      (eqs.2)
If we combine the two equations we get: mv2/R = mg
→      v = [Rg]1/2 = [50×9.8]1/2 = 22.1 m/s
(5-22)

```
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