Normal Force Example: Incline
m1
α
The angle of the frictionless incline is α = 30°. Mass m1
slides down the incline, starting from rest. What is the
speed of the mass after it slid 10 meters downhill?
[use g = 10 m/s2]
PHY2053, Fall 2013, Lecture 7 – Gravity, Contact Forces, Tension
Newton’s Third Law
LawIII:
Lex
III:Actioni
To every
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semper
is always
et æqualem
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and
Law
3: Inreaction:
ansive
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opposite
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or the between
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oftwo
two
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seeach
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other.
Thesein
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other
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et in
directed
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opposite
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dirigi.
directions.
m1
PHY2053, Fall 2013, Lecture 7 – Gravity, Contact Forces, Tension
m2
Example: Two Blocks
m1
m2
In the picture above, the two blocks are placed on a
frictionless surface, in contact with each other. The
masses of the two blocks are m1 = 10 kg and m2 = 15 kg.
A force F1 = 100 N is being applied to mass m1 from the
left, and force F2 = 200 N is being applied to mass m2
from the right. Compute the force F2,1 that mass m1 is
exerting on mass m2. Contact Forces, Tension
PHY2053, Fall 2013, Lecture 7 – Gravity,
Gravitational force
● Newton’s Law of Gravitation:
● where m1 and m2 are the masses of the two bodies
● r is the distance between them, measured from the
center of the (spherical) object
● G is the universal gravitational constant
G = 6.674 × 10-11 N m2 / kg2
● provides excellent description of planetary motion
PHY2053, Fall 2013, Lecture 7 – Gravity, Contact Forces, Tension
4
Example: Cranial Attraction
● The heads of neighboring students in the classroom
are separated by ~ 1 m. A typical human head weighs
about 5 kg. Compute the attractive force (due to
gravitation) between the heads of two students.
m1 m2
F =G 2
r
● m1 = m2 = 5 kg, r = 1 m, G = 6.674 × 10-11 N m2/kg2
● use in formula → obtain F = 1.67 × 10-9 N
PHY2053, Fall 2013, Lecture 7 – Gravity, Contact Forces, Tension
5
Example: Weight change while
airplane is in flight
● An airplane is cruising at the altitude of 10 km above
sea level. What factor less does a passenger weigh at
that altitude compared to their weight at sea level?
The radius of the Earth is 6370 km and
its mass is ME = 6 × 1024 kg.
m1 m2
F =G 2
r
● m1 = ME, m2 = m (unknown)
● Compare F(r) for r = RE to F(r) for r = (RE + 10 km)
PHY2053, Fall 2013, Lecture 7 – Gravity, Contact Forces, Tension
6
Example: “g” on surface of Mars
m1 m2
F =G 2
r
Declare this “g” when computed at the
surface of a specific planet.
Formula then simplifies to F = m2 g
● For all objects on the surface of a planet, one of the
masses (mass of the planet) and r are the same
● formula can then be simplified to F = m g, except g
is different on each planet (depends on M and R)
● for Mars, MMars = 1/9 ME , RMars = 0.533 RE
● we can then compute gMars in terms of gEarth
PHY2053, Fall 2013, Lecture 7 – Gravity, Contact Forces, Tension
7
Tension Force
● “Ideal String” concept:
● massless, infinitely thin, fixed length
● withstands any force without breaking
“String with tension T”
m1
PHY2053, Fall 2013, Lecture 7 – Gravity, Contact Forces, Tension
m2
Tension Example: Pulley + Weights
The masses in the depicted
Atwood machine are
m1 = 5 kg and m2 = 3 kg.
● What is the acceleration
of mass m1?
● What is the tension T2 of
the cable holding the
massless pulley?
PHY2053, Fall 2013, Lecture 7 – Gravity, Contact Forces, Tension
T2
m1
m2
Friction
● Force due to imperfections of surfaces in contact
● For a pair of surfaces, the frictional force depends
(only) on the normal force between the surfaces:
Frictional
force
Normal force
Coefficient of friction
● To a good approximation, does not depend on size or
shape of contact surface, only the materials in contact
● Always opposite to the direction of motion
● Static friction – objects not moving w.r.t. each other
● Kinetic friction – one object is being dragged across
Fall
2013,
Lecture
7 – Gravity,
Contact
Forces,
theTension
(surface of the) other
PHY2053,
Problem: Pushing a crate at an angle
A crate of mass 100 kg is
being pushed across a
horizontal surface. The
coefficient of friction between
the crate and the surface is
μ = 0.2. The force is being
applied downward on the
crate at a 30º angle with
respect to the vertical axis.
What is the minimum force at
which the crate will move
from its prone position?
[use g = 10 m/s2]
PHY2053, Fall 2013, Lecture 6 – Newton’s Laws
∘
30
F