Physics 2211: Lecture 10 Todays Agenda

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Physics 2211 Lecture 10Todays Agenda

• Dynamics of many-body systems
• Atwoods machine
• Inclined Plane
• General case of two attached blocks on inclined
  planes
• Some interesting problems
• Friction
• WA06 due Monday evening
• Quizzes available outside N018 (down the hallway)

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Many-body Dynamics

• Systems made up of more than one object
• Objects are typically connected
• By ropes pulleys today
• By rods, springs, etc. later on

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Problem Two strings Two Masses onhorizontal
frictionless floor

• Given T1, m1 and m2, what are a and T2?
• T1 - T2 m1a (a)
• T2 m2a (b)
• Add (a) (b) T1 (m1 m2)a
  a

Plugging solution into (b)
a
i
m1
m2
T2
T1
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Lecture 10, Act 1Two-body dynamics

• Three blocks of mass 3m, 2m, and m are connected
  by strings and pulled with constant acceleration
  a. What is the relationship between the tension
  in each of the strings?

(a) T1 gt T2 gt T3 (b) T3 gt T2 gt T1
(c) T1 T2 T3
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Lecture 10, Act 1Solution

• Draw free body diagrams!!

T1 gt T2 gt T3
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Lecture 10, Act 1 Solution

• Alternative solution

T1 gt T2 gt T3
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Atwoods Machine
Masses m1 and m2 are attached to an ideal
massless string and hung as shown around an ideal
massless pulley.
Fixed Pulley

• Find the accelerations, a1 and a2, of the masses.
• What is the tension in the string T ?

j
T1
T2
m1
a1
m2
a2
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Atwoods Machine...

• Draw free body diagrams for each object
• Applying Newtons Second Law ( j -components)
• T1 - m1g m1a1
• T2 - m2g m2a2
• But T1 T2 T since pulley is ideal
• and a1 -a2 -a.since the masses are
  connected by the string

Free Body Diagrams
T1
T2
j
a1
a2
m2g
m1g
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Atwoods Machine...

• T - m1g -m1 a (a)
• T - m2g m2 a (b)
• Two equations two unknowns
• we can solve for both unknowns (T and a).
• subtract (b) - (a)
• g(m1 - m2 ) a(m1 m2 )
• a
• add (b) (a)
• 2T - g(m1 m2 ) -a(m1 - m2 )
• T 2gm1m2 / (m1 m2 )

-
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Atwoods Machine...

• So we find

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Is the result reasonable? Check limiting
cases!

• Special cases
• i.) m1 m2 m a 0 and T mg. OK!
• ii.) m2 or m1 0 a g and T 0.
  OK!
• Atwoods machine can be used to determine g (by
  measuring the acceleration a for given masses).

-
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Lecture 10, Act 2Two-body dynamics

• In which case does block m experience a larger
  acceleration? In (1) there is a 10 kg mass
  hanging from a rope. In (2) a hand is providing
  a constant downward force of 98.1 N. In both
  cases the ropes and pulleys are massless.

m
a
F 98.1 N
Case (1)
Case (2)
(a) Case (1) (b) Case (2) (c)
same
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Lecture 10, Act 2 Solution
Add (a) and (b)
98.1 N (m 10kg)a
Note
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Lecture 10, Act 2 Solution
T 98.1 N ma
For case (2)
m
m
a
a
10kg
F 98.1 N
Case (1)
Case (2)
The answer is (b) Case (2)
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Problem Inclined plane

• A block of mass m slides down a frictionless ramp
  that makes angle ? with respect to the
  horizontal. What is its acceleration a ?

m
a
?
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Inclined plane...

• Define convenient axes parallel and perpendicular
  to plane
• Acceleration a is in x direction only.

m
a
?
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Inclined plane...

• Consider x and y components separately
• i mg sin ? ma. a g sin ?
• j N - mg cos ? 0. N mg cos ?

ma
mg sin ?
N
?
mg cos ?
mg
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Angles of an Inclined plane
The triangles are similar, so the angles are the
same!
N
?
mg
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Inclined plane...

• Vectors

m
N
?
mg
a g sin ???i N mg cos ???j
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Attached bodies on two inclined planes
smooth peg
m2
m1
?1
?2
All surfaces frictionless
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How will the bodies move?
From the free body diagrams for each body, and
the chosen coordinate system for each block, we
can apply Newtons Second Law
x
y
Taking x components 1) T1 - m1g sin ?1 m1
a1X 2) T2 - m2g sin ?2 m2 a2X? But T1 T2
T and a1X -a2X -a (constraints)
x
y
N
T1
T2
m2
N
m1
?2
?1
m2g
m1g
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Solving the equations
Using the constraints, solve the equations. T -
m1gsin ?1 -m1 a (a) T - m2gsin ?2 m2
a (b) Subtracting (a) from (b) gives
m1gsin ?1 - m2gsin ?2 (m1m2 )a So
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Special Case 1
Boring
m2
m1
If ?1 0 and ?2 0, a 0.
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Special Case 2
T
Atwoods Machine
T
m1
m2
If ?1 90 and ?2 90,
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Special Case 3
m1
Lab configuration
m2
-
If ?1 0 and ?2 90,
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New Topic Friction

• What does it do?
• It opposes motion!
• How do we characterize this in terms we have
  learned?
• Friction results in a force in the direction
  opposite to the direction of motion!

j
N
FAPPLIED
i
ma
fFRICTION
W
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Friction...

• Friction is caused by the microscopic
  interactions between the two surfaces

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Friction...

• Force of friction acts to oppose motion
• Parallel to surface.
• Perpendicular to Normal force.

j
N
F
i
ma
fF
W
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Model for Sliding Friction

• The direction of the frictional force vector is
  perpendicular to the normal force vector N.
• The magnitude of the frictional force vector fF
  is proportional to the magnitude of the normal
  force N .
• fF ?K N ( ?K??W in the previous
  example)
• The heavier something is, the greater the
  friction will be...makes sense!
• The constant ?K is called the coefficient of
  kinetic friction.

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Model...

• Dynamics
• i F ? ?KN ma
• j N mg
• so F ???Kmg ma

j
N
F
i
ma
?K mg
W W mg
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Inclined Plane with Friction

• Draw free-body diagram

ma
?KN
j
N
?
mg
?
i
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Inclined plane...

• Consider i and j components of FNET ma

?KN
ma
j
N
?
a / g sin ?????Kcos ?
?
mg
mg cos ??
i
mg sin ??
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Static Friction...

• So far we have considered friction acting when
  something moves.
• We also know that it acts in un-moving static
  systems
• In these cases, the force provided by friction
  will depend on the forces applied on the system.

j
N
F
i
fF
W W mg
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Static Friction...

• Just like in the sliding case except a 0.
• i F ??fF 0
• j N mg

While the block is static fF ??F
j
N
F
i
fF
W W mg
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Static Friction...

• The maximum possible force that the friction
  between two objects can provide is fMAX ?SN,
  where ?s is the coefficient of static friction.
• So fF ? ?S N.
• As one increases F, fF gets bigger until fF ?SN
  and the object starts to move.

j
N
F
i
fF
W W mg
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Static Friction...

• ?S is discovered by increasing F until the block
  starts to slide
• i FMAX ???SN 0
• j N mg
• ?S ??FMAX / mg

j
N
FMAX
i
?Smg
W W mg
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Static Friction

• We can also consider ?S on an inclined plane.
• In this case, the force provided by friction will
  depend on the angle ? of the plane.

?
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Static Friction...

• The force provided by friction, fF , depends on ?.

fF
ma 0 (block is not moving)
mg sin ????ff ???
N
?
(Newtons 2nd Law along x-axis)
mg
?
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Static Friction...

• We can find ?s by increasing the ramp angle until
  the block slides

mg sin ????ff????
In this case
?ff????SN ? ??Smg cos ?M
?SN
mg sin ?M????Smg cos ?M????
N
mg
??M
?S???tan ?M?
?
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Additional comments on Friction

• Since fF ?N , the force of friction does not
  depend on the area of the surfaces in contact.
• By definition, it must be true that ?S gt ?K
  for any system (think about it...).

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Recap of todays lecture

• Dynamics of many-body systems
• Atwoods machine.
• General case of two attached blocks on inclined
  planes.
• Some interesting special cases.
• Friction