Download Lecture 5 - Purdue Physics

PHYSICS 149: Lecture 5
• Chapter 2
– 2.5 Newton’s Third Law
– 2.6 Gravitational Forces
– 2.7 Contact Forces: Normal Force and Friction
Lecture 5
Purdue University, Physics 149
1
Newton’s Third Law
• All forces come in pairs
• Third law forces involve TWO OBJECTS.
OBJECTS
• The two forces are:
– th
the force
f
object
bj t one exerts
t on object
bj t ttwo
– the force object two exerts on object one
• Three
Th
ways tto state
t t the
th 3rd
3 d law:
l
– Forces on each other are equal and opposite
– For every action there is an equal and opposite reaction
– You can’t push on something without it pushing back on
y
you
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Newton’s Third Law of Motion
• In an interaction between two objects, each
object exerts a force on the other
other. These two
forces are equal in magnitude and opposite in
direction.
– To every action, there is always opposed an equal
reaction.
– Forces always come in equal but opposite actionreaction pair.
• Note that these two forces act on different
objects; they do not cancel in any way.
Don’tt forget that forces always exist in pairs
pairs.
• Don
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Purdue University, Physics 149
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Free Body Diagram (FBD)
• A simplified sketch of a single object with force
vectors drawn to represent every force acting
“on” that object. (It must not include any forces
that act on other objects.)
j
)
• FBD is useful to find the net force acting on an
object.
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Internal and External Forces
• Internal Forces: Forces which act on one part of an object
by another part of the same object
• External Forces: Forces which act on an object by some
other object.
• Net force on a system = vector sum of internal forces +
vector sum of external forces
• But,
B t vector
t sum off internal
i t
l forces
f
is
i zero because,
b
from
f
Newton’s third law, internal forces will occur in equal and
opposite pairs and so they contribute nothing to the sum.
Th never influence
They
i fl
the
h system’s
’ motion.
i
• Eventually, net force on a system = vector sum of external
y We need to consider external forces onlyy in
forces only.
order to describe the motion of the system.
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Examples
External
• Net force on a baseball =
forces:
interaction with the Earth (gravity)
what we
need to
+ interaction with a bat
consider
+ interaction
i t
ti with
ith th
the air
i
Internal Forces:
+ interactions among protons,
their vector sum
neutrons, and
d electrons
l
iin iit
is zero
• I am hit by myself (internal forces) and other
person (external forces). I am pushed due to
external forces only
only. Internal forces do not make
any contribution.
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Pushing a Stalled Car
Two people are pushing a stalled car. The mass of
the car is 2000kg. One person applied a force of 300
N, the other 400N. Friction opposes this motion with a
force of 600N. What is the acceleration of the car:
y
Fc,man1
Fc,man2
ΣF = +600 N − 300 N − 400 N
ΣF = −100 N
Lecture 5
a=
Fcg=f
x
ΣF
100 N
m
=−
= −0.05 2
m
2000kg
s
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A FBD for Every Situation
A skydiver is
d
descending
di
with a constant
velocity
e oc ty.
A force is applied to the right
to drag a sled across
l
loosely-packed
l
k d snow with
ith a
rightward acceleration.
A car is coasting to the
right and slowing down
Lecture 5
Purdue University, Physics 149
A football is
moving
upwards
towards its
peak after
having been
booted by
the punter.
8
ILQ 1
When a car accelerates from rest, what
force causes the acceleration of the car?
A) The rotating engine on the drive shaft
B) The
Th force
f
off the
th axell on th
the ti
tires
C) The friction force of the road on the tires
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ILQ 2
A person is standing on a bathroom scale. Which
of these is not a force exerted on the scale?
A)
B)
C)
D)
Lecture 5
a contact force due to the feet of the person
the weight of the person
a contact force due to the floor
the weight of the scale
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Newton’s Law of Universal Gravitation
F12
m1
F21
m2
r
• The magnitude of gravitational force is:
(= F12 = F21)
where G = 6.674 × 10-11 Nm2/kg2 (universal gravitational constant)
Note: m1 and m2 need to be in kg, and r needs to be in m.
• The
Th direction
di ti off gravitational
it ti
l force
f
is:
i
– each object is pulled toward the other’s center (attractive force)
– on line connecting
g the masses;; always
y attractive
• very weak, but this holds the universe together!
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Comparison with EM Force
q2
F1,2
F2,1
r12
F1,2 = force on q1 due to q2 =
= F2,1 = force on q2 due to q1
q1
kq 1 q 2
2
r12
Direction: on line connecting the masses; can be
attractive or repulsive
k = universal constant = 8.99 x 109 N-m2/c2
Gm e2
−43
=
= 2.4 × 10
2
E electric (2 electrons ) kqe
Fgravity (2 electrons )
q e = 1.6 × 10 − 9 C
Lecture 5
m e = 9.11 × 10 − 31 K g
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Weight
• Weight is the force of gravity on an object with mass
• Units of weight are Newtons or Pounds
S
Same
mass b
butt Different
Diff
t weight!
i ht!
F =G
mM pplanet
r2
r
On earth W=mg where g=9.8 m/s2
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Weight on Earth
• Your weight on Earth is the magnitude of Earth’s
gravitational force exerted on you (m)
(m).
GM E m
⎛ GM E ⎞
W=
= m⎜ 2 ⎟
2
R
⎝ R ⎠
where R is the distance
between you and Earth’s
Earth s center
• The weight of an object of mass m “near” Earth’s
surface is:
where
(g is called the gravitational field strength)
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Weight on Other Planets
• The weight of an object of mass m “near” a
planet’s
planet
s surface is:
⎛ GM Planet
GM Planet m
= m⎜⎜ 2
Wat Planet =
2
RPlanet
Pl t
Pl t
⎝ RPlanet
GM Planet
where g Planet =
2
RPlanet
⎞
⎟⎟ = mg Planet
⎠
• For example, gMoon = 1.62 N/kg ≈ 1/6 gEarth.
– Let
Let’s
s say there is a man whose mass is 100 kg
kg.
• At the surface of Earth, his mass and weight are 100 kg and
980 N (=m·gEarth), respectively.
• At the
th surface
f
off M
Moon, hi
his mass and
d weight
i ht are 100 kkg and
d
162 N (=m·gMoon), respectively.
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Weight on Earth and on Moon
• How far above the surface of the Earth does an object
have to be to have the same weight as it would have
on the
th surface
f
off the
th moon?
? Neglect
N l t effects
ff t from
f
the
th
Earth’s gravity on the Moon’s surface and vice versa
ME
MM
FE = Gm 2 = Gm 2 = FM
r
rM
r2 =
ME 2
rM
MM
5.97 × 1024 Kg
× 1
r=
1.744 × 103 km
k
22
7.35 × 10 Kg
(
)
2
= 1.57
1 × 104 kkm
Height over surface r-rE = 1.57 × 104 km − 6.371 × 103 km = 9.3 × 103 km
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Things are different on the Moon
Surface g
gravity
y
compared to Earth
Your mass
Energy to stop a 1 Kg
ball moving at 90
km/hour
How much can you lift
How high
g can yyou jjump
p
How far can you kick a
ball
Lecture 5
Earth
Moon
1
0.17
40 Kg
625 Joules
40 Kg
625 Joules
10 kg
20 cm
20 m
60 kg
120 cm
120 m
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Force of Gravity
For objects on the surface of the earth:
2) = mg
(
g
• F = GMm/R2 = m(GM/R
• g = GM/R2 = 9.8 N/kg = 9.8 m/s2
What about at the top of Mount Everest? (h=8850m or
GME m
29,035 feet
F =W =
r2
−2
WEverest
1
2
⎛ hEverest ⎞
=
r
=
1+
=
⎜
⎟
2( )
Wsurface ( r + hEverest )
r ⎠
⎝
hEverest ⎞
⎛
−3
1
−
2
=
1
−
2.76
×
10
= 1 − 0.00278 = 0.99722
⎜
⎟
r ⎠
⎝
The approximation
works well since:
Lecture 5
hEverest 8.850×10
−3
=
=
1
1.389
389
×
10
r
6.371×106
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r2
Lecture 5
r1
Your weight
g
decreases as your
altitude g
goes up.
p
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Weight
The weight (W) of an object is
equal
q
to the magnitude
g
of the
gravitational force acting on a
body of mass m
W = mg
Dropping
D
i an object
bj t causes it
to accelerate at free-fall
acceleration g
Fg = mg
W = Fg
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ILQ: Gravitation
Does a man weigh more
A)) at the top
p of Mt. Everest or
B) at the base of the mountain?
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Which Forces Enter in a FBD?
Several force must be taken into account:
• Gravity
• Normal Force
• Friction
• Push or Pull
• Tension
W
Gravity: if the sled has a mass m the force
d to gravity
due
i iis W = mg
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Which Forces Enter in a FBD?
•
•
•
N
Normal force: always perpendicular
th surface
the
f
with
ith which
hi h a b
body
d iis iin
contact.
Friction: the frictional force is parallel
to the surface and it always opposes
the direction of motion.
P h or pullll
Push
W
N
f
P
W
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Normal (= Perpendicular) Force
• The normal force is a contact force
perpendicular to the contact surfaces that
prevents two objects from passing through one
another.
• Normal force is a vector
vector.
– Direction: always perpendicular to the “contact
surface” ((rather than the horizon))
– Magnitude: depends on the weight of the object
(see different cases on next pages)
• Type: contact force (not long-range force)
• Normal force is usually denoted by N.
N
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What Causes Normal Force?
• Atoms inside solid objects are
inter-connected by molecular
bonds which act like springs.
• When you place an object on top
off a table,
t bl the
th table
t bl d
deforms
f
slightly. This bend is usually not
visible to the eye
eye.
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Normal Force: Case 1
• If the table’s surface (contact surface) is
horizontal,
horizontal
– Direction of the normal force is perpendicular to the
“contact surface.” In this case, vertically upward.
– Magnitude of the normal force is
the book’s weight, according to
Newton’s First Law of Motion.
N = W (= mg)
according to
ΣFy = 0 for an object in equilibrium
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Normal Force: Case 2
• If the contact surface is horizontal and there is
another vertical force acting on the book
book,
– Direction of the normal force is perpendicular to the
“contact surface.” In this case, vertically upward.
– Magnitude of the normal force is
the book’s weight plus the magnitude
of the additional force, according to
Newton’s First Law of Motion.
N = W (= mg) + F
according to
ΣFy = 0 for an object in equilibrium
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Normal Force: Case 3
• If the contact surface is not horizontal (with an
inclination angle θ),
θ)
– Direction of the normal force is perpendicular to the
“contact surface.” In this case, it is not vertical.
– Magnitude of the normal force is
+y
the book’s weight times cosθ,
according to Newton’s
First Law of Motion.
θ
N = W cosθ (=
( mg cosθ)
+x
W θ
Wcosθ
according to
Wsinθ
ΣFy = 0 for an object in equilibrium
θ
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Purdue University, Physics 149
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