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Ch 4 Force and Motion
Chapter 4
Newton’s 1st Law and 2nd law
Newton’s 3rd Law
Internal and External forces
Common Forces
Applications
•
•
•
•
•
Force and Newton’s
Laws of Motion
(Continuation)
Newton’s 1st and 2nd Laws
Newton’s Third Law
• For each action there is a reaction! r
r
| FAB |=| FBA |
• They are equal in magnitude.
r
r
• They are opposite in direction.
FAB = −FBA
• Newton’s 1st law: an object will
• stay at rest, or
• maintain its motion at a constant velocity and in
a straight line
as long as
• no force is exerted on the object, or
• all forces cancel each other (Fnet=0)
• Newton’s 2nd law:
• They are applied in different bodies.
r
FBA
r
r
Fnet = ma
r
FAB
• Galilei Relativity: Laws of Physics take the
same form in all inertial frames
Third-law Force Pair
A System of Bodies
• Definition: two or more bodies that are
considered as one entity.
• Forces acting between bodies internal to the
system are internal forces
• If an external body acts on a body of the
system, the force is called an external force
• When applying Newton’s second law to a system,
we consider only the external forces (why?).
A
F1
• When: interaction between any two bodies with mass
(apple + Earth).
• Direction: towards the center of the planet (downwards)
• Application point: at the center of mass of the body.
• Magnitude: A free falling body (neglect air) drops with a
constant acceleration g pointing towards the center of the
Earth.
y
r
r
r
)
W = mg = − mgy
F1, F2are internal for the AB-system
F3 is external for the AB-system
B F2
F3
C
r
W
Gravitational Force
•
g
g
r
W
is the free-fall acceleration
Earth
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Gravitation
Applications of Universal Gravitation
•Gravitational force on
Earth surface
W =G
mmE
RE2
mg = G
mmE
RE2
m
g = G E2
RE
mE =
Why can we take g=const. at earth’s
surface?
So it must be that
GM E
≠ const.
r2
≈
g (r ) = G
v
r
When: supporting surface
Direction: normal to the supporting surface
Application point: in the pressing object.
Magnitude: reactive
r
N
N = W = mg
• A 0.2 kg block is sliding
along a frictionless surface
that is inclined at an angle
r
θ=30o. Find the normal FN
force.
y
v
N
Fnet,y = may
N −W = 0
y
r
N
x
v
W
ME
, r ≥ RE
r2
y
Example
• Force:
r
r
Fnet = ma
gRE2
G
GM E
2
RE
The Normal Force N or FN
•
•
•
•
⇒ mE = 6 ×10 24 kg
•Acceleration due to
gravity g will vary with
altitude
F=mg
g=
RE = 6380 km
Applications of Universal Gravitation
Newton says force on mass m a
distance r from center of earth is
But earlier we said
take g = 9.8 m / s 2
v
W
x
θ
r
r
Fnet = ma
r
N
r
W
Fnet,y = may , ay = 0
Fnet , y = N − W cos θ
x
θ
r
W
N − W cos θ = 0
N = mg cos θ = 0.2 ⋅10 cos 30 = 1.41
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r
f
Friction
Properties of Friction
• When: when in contact with a rough surface.
• Static Friction
• Varies in magnitude
• Along contact surface
• Depends on other forces (reactive)
• Has maximum value:
• Slippery surface - no friction!
• Application point: along the surface of contact.
• Direction: opposite to the attempted slide.
• Magnitude: see next slide
r
f
f s ,max = µ s FN
• If sum of other forces exceeds
• Body starts to move
• Kinetic friction
r
υ
r
f
r
f
f s ,max
r
υ
r
f
r
f
Properties of Friction
Tension
• When: when pulling an object with a
inelastic cord or string.
• Application point: the point of attachment
• Direction: along the string and away from
the body
• Magnitude: it is equal to the force with
which we pull. The magnitude of the tension
is the same at all points of the cord (if the
cord is massless).
r
r
r
T T
T
r
r
T
r
T
T
• Kinetic Friction
• Once the body starts sliding, the friction force
rapidly decreases
• Parallel to contact surface
• Magnitude depends on normal force
r
f
f k = µ k FN
• Friction coefficients depend on both contact
surfaces
µ <µ
•
k
s
• dimensionless
• Independent of velocity, acceleration
Example
Applying Newton’s Laws-Example
• Two blocks with masses mA, and mB
are connected with a cord that is
wrapped around a massless
frictionless pulley. Find the
acceleration, a of the blocks.
r
r
FA,net = mA a
r
r
FB,net = mB a
T − WA = m A a
T − WB = −mB a
T = WA + m A a T = WB − mB a
WA + mA a = WB − mB a
a = g (mB − m A ) (m A + m B )
r
T
y
r
T
A
r
WA
r
T
B
r
WB
• A block with mass m=1 kg is pulled
up an inclined surface with an
acceleration a=1 m/s2. The tension
in the cord should not exceed 6 N.
What is the maximum friction that
is allowed? (θ=30o)
r
FN
y
r
f
θ
v
W
r
FN
y
x
x
r
θ f
r
T
r
T
r
a
v
r
r
W
Fnet = ma
r r
r r
r
W + FN + f + T = ma
T − W sin θ − f = ma
(x)
− W cos θ + FN = 0
(y)
f = −ma + T − W sin θ = 0.1N
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