Physics 211: Lecture 8 Todays Agenda

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Physics 211 Lecture 8Todays Agenda

• Friction Recap
• Drag Forces
• Terminal speed
• A special very cool demo.
• Dynamics of many-body systems
• Atwoods machine
• General case of two attached blocks on inclined
  planes
• Some interesting problems

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Model for Surface Friction

• The direction of the frictional force vector fF
  is perpendicular to the normal force vector N, in
  the direction opposing relative motion of the two
  surfaces.
• Kinetic (sliding) The magnitude of the
  frictional force vector is proportional to the
  magnitude of the normal force N.
• fF ?KN
• It moves, but it heats up the surface it moves
  on!
• Static The frictional force balances the net
  applied forces such that the object doesnt move.
  The maximum possible static frictional force is
  proportional to N.
• fF ? ?SN and as long as this is true, then fF
  fA in opposite direction
• It doesnt move!

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Lecture 8, Act 1friction dynamics

• A block of mass m, when placed on a rough
  inclined plane (m gt 0) and given a brief push,
  keeps moving down the plane with constant speed.
• If a similar block (same m) of mass 2m were
  placed on the same incline and given a brief
  push, it would

(a) stop (b) accelerate (c)
move with constant speed
m
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Lecture 8, Act 1Solution

• Since the velocity is constant, its just broken
  free from S. F.
• Net force down ramp is essentially zero
• Draw FBD and find the total force in the
  x-direction

FNET,X mg sin q - mKmg cos q
ma 0 (first case)
Doubling the mass will simplydouble both
termsnet forcewill still be zero! Speed will
still be constant! Increase the friction and the
downhill force by the same factor ? nothing
changes!
q
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Friction in Fluids Drag Forces

• When an object moves through a viscous medium,
  like air or water, the medium exerts a drag or
  retarding force that opposes the motion of the
  object relative to the medium.

FDRAG
j
Fg mg
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Drag Forces
Parachute

• This drag force is proportional to the speed v of
  the object raised to some power, n. This will
  result in falling at a maximum (terminal) speed.

FD bvn

j
feels like n1
v
Fg mg
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Terminal Speed

• Suppose FD bv. Sally jumps out of a plane.
  After falling for a while her downward speed is a
  constant vTERMINAL.
• What is FD after she reaches this terminal speed?
• What is the terminal speed vTERMINAL?
• FTOT FD - mg ma 0. ? why??
• FD mg when??
• Since FD bv
• bv mg

FD bv
j

v
Fg mg
8
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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Atwoods Machine
Masses m1 and m2 are attached to an ideal
massless string and hung as shown around an ideal
massless pulley.
Fixed Pulley

• What are the tensions in the string T1 and T2 ?
• Find the accelerations, a1 and a2, of the masses.
• Use FBD
• Solve for motion

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
• Yikes! 2 eqn, but 4 unk???
• But T1 T2 T
• since pulley is ideal
• and a1 -a2 -a.
• since the masses are
• connected by the string

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Atwoods Machine...

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

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Atwoods Machine...
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).

If m1 is almost m2, then a will be small. You
can measure motion for a long time. More
accurate.
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A related situationAttached bodies on two
inclined planes
smooth peg
m2
m1
?1
?2
all surfaces frictionless peg is 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
m2
m1
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Solving the equations
Using the constraints, we get 2 equations and 2
unknowns, 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
Air-track
m1
Lab configuration
m2
-
If ?1 0 and ?2 90,
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Lecture 8, 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.

Case (1)
Case (2)
(a) Case (1) (b) Case (2) (c)
same
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Lecture 8, Act 2 Solution

• For case (1) draw FBD and write FNET ma for
  each block

(a)
T ma (a) mWg -T mWa (b)
mW10kg
(b)
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Lecture 8, Act 2 Solution

• The answer is (b) - Case (2) In this case the
  block experiences a larger acceleration

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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)

i
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Lecture 8, Act 3Two-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 8, Act 3Solution

• Draw free body diagrams!!

T1 gt T2 gt T3
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Lecture 8, Act 3 Solution

• Alternative solution

T1 gt T2 gt T3
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Problem Rotating puck weight.

• A mass m1 slides in a circular path with speed v
  on a horizontal frictionless table. It is held
  at a radius R by a string threaded through a
  frictionless hole at the center of the table. At
  the other end of the string hangs a second mass
  m2.
• What is the tension (T) in the string?
• What is the speed (v) of the sliding mass?

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Problem Rotating puck weight...
T

• Draw FBD of hanging mass
• Since R is constant, a 0.
• so T m2g

m2
m2g
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Problem Rotating puck weight...
T m2g
Puck
N
T m2g

• Draw FBD of sliding mass

m1
Use F T m1a where a v2 / R
m1g
m2g m1v2 / R
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Recap of todays lecture

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