Download Chapter 5 Applications of Newton`s Laws

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
m
Fg
Also known as the “weight” of
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.
The spring has a
pointer attached to it
so that we can
determine its
elongation
y-axis
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)
Tension
T
crate
floor
rope
It is the force exerted by a rope or a
cable
Tension has the following characteristics:
• Tension always pulls (never pushes)
• The direction of T is along the rope
• The magnitude |T| is constant
pulley
even when T changes direction
T
(as in the case of the pulley)
(5-4)
T
crate
pull
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
crate
System = crate
The normal force on the crate
exerted by the floor is labeled Fcf
floor
System = floor
crate
floor
The normal force on the floor
exerted by the floor is labeled F fc
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
scale
FN = Fsa
Fsa is the normal force exerted
on the scale by the apple
These two forces are equal and
opposite by virtue of Newton’s
third law
System = apple
y-axis
Fynet = FN - mg
(5-8)
FN
Fynet = may = 0
mg
FN - mg = 0
→
Indeed: FN = mg
→
y-axis
elevator
What does the scale reads when the
elevator accelerates upwards ?
System = man
..
Fynet = FN - mg
Fynet = ma
FN
mg
a
scale
(Newton’s second law)
→
FN - mg = ma
→
FN = mg + ma
FN > mg
The scale in this case gives a reading
which is higher than the actual weight
mg
(5-9)
y-axis
elevator
What does the scale reads when the
elevator accelerates downwards ?
System = man
..
Fynet = FN - mg
Fynet = - ma
FN
mg
a
scale
(Newton’s second law)
→
FN - mg = - ma
→
FN = mg - ma
FN < mg
The scale in this case gives a reading
which is lower than the actual weight
mg
(5-10)
Atwood’s Machine
a
m2
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
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
T
a
T
a
a
y
y
m1g
m2
m1
m2g
m2
m1
System = m1
System = m2
Fynet = T − m1 g = m1a →
Fynet = T − m2 g = − m2a →
T = m1g + m1a
T = m2 g − m2a
eqs.1
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
FN
T
crate
f
floor
rope
Friction was first studied by
Leonardo da Vinci
He discovered that friction has the
following characteristics:
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
0 < f S < µ S FN
Static friction fs
µs is a constant known as “coefficient of static friction”
f k = µ k FN
Kinetic friction fk
µk is a constant known as “coefficient of kinetic friction”
Leonardo da Vinci
(self portrait)
(1452-1519)
motion
Consider the following experiment:
FN
T
crate
f
floor
rope
mg
We start pulling a heavy crate using
a rope with a force T that increases
with time. At some instant tm the
crate will start to move
µk < µs
f
We then plot the frictional force f
as function of time. The static
friction increases with time. At
tm the static friction becomes
maximum (µs FN) and the crate
starts to move. The kinetic
friction has a smaller constant
value of µk FN .
(5-15)
µ s FN
µ k FN
Static kinetic
friction friction
tm
Motion
starts
t
Example (5-6),page 121
y
A mass m is at rest on a
horizontal ramp. The ramp is
slowly raised. Find the critical
angle θc at which the mass will
start to slide
x
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
Fxnet = fs = ma
→
→
FN = mg
(eqs.1)
a = fs /m
amax = (fs)max /m = µs FN /m (eqs.2)
From eqs.1 substitute FN into eqs.2
amax = µs g = 0.8×9.8 = 7.8 m/s2
→ amax = µsmg /m
When the rope breaks
the book does not stay
on its circular path but
instead flies along the
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
m
Consider an object of mass 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=
r
or
F = mω 2r
The magnitude of F is constant, its direction is not. It
always points towards the center of the circular orbit.
C
.
Recipe for problems that involve
uniform circular motion of an
object of mass m on a circular
orbit of radius R with speed v
R
y
m
x
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
(5-21)
A small dice sits 0.15 m from
the center of a turntable. If µs
between the turntable and the
dice is 0.55 what is the largest
possible angular speed such
that the dice will not slide off
y
x
Fynet = FN - mg = may = 0 →
FN = mg
(eqs.1)
Fxnet = fs = mω2r (eqs.2) → ω2 = fs /mr
ω2max = (fs )max /mr
Thus
From eqs.1 we have:
(fs)max = µs FN
ω2max = µs mg/mr → ωmax = [µs g/r]1/2 = 6 rad/s
Example (5-12) page 129
y
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
the speed v of the outer wall so that a
scale will read what it would read on
earth
Fynet = FN = mv2/R
(eqs.1)
Also
(eqs.2)
FN = mg
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)