Lecture 3 Chapter 6 Force and Motion II

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Lecture 3Chapter 6Force and Motion II
Friction More Circular motion Apparent
Weight Drag Forces Numerical integration Derivativ
es/Integration (Lecture 1) Tension between
Blocks (Lecture 2) Misconceptions
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Friction
You are standing still, then begin to walk. What
was the external forced that caused you to
accelerate?
Hint It is very hard to start walking if you are
standing on ice.
What force causes a car to accelerate when a
traffic light turns green?
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Friction
N
Fixed block
F
fs
mg
Static Friction Push with a force F and block
does not move because fs F. The force of
friction varies from 0 up to some maximum. The
maximum value equals fs msN, where N is the
normal force. Above we would have fs msmg. The
coefficient of static friction ranges from 0 to
1.2
Kinetic Friction If we increase F until the
block starts to move, the friction force
decreases to fk mkN and remains constant
throughout the motion.
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Frictional Forces
Friction is an attractive force between two
surfaces that is a result of the vector sum of
many electrical forces between the surface atoms
of the two different bodies. Only about 10-4 of
the surface atoms actually contribute.
Model of dry friction 3 or 4 asperites support
top block. Temporarily weld together
Models of friction See Chabay and Sherwood Matter
and Ineractions Volume 1 ISBN 0-471-35491-0
The Friction and Lubrication of Solids F. P.
Bowden and D. Tabor, Oxford University Press 1964
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Problem Solving with Newtons 2nd Lawinvolving
friction

• Vector sum of external forces in x direction
  max
• Vector sum of external forces in y direction
  may
• If no acceleration, then set sum equal to 0

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Mass on Inclined plane
Question How high can we tilt the book before
the coin slides down the block?
Apply Newton's 2nd law in the x and y directions
and using fs ms N
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Mass on Inclined plane
Question How high can we tilt the book before
the coin slides down the block?
Apply Newton's 2nd law in the x and y directions
and using fs ms N
Take x direction along the plane
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Mass On Inclined Plane
F
F
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Free Body Diagram of an accelerating system
Atwoods machine with friction.Find a and T.
Frictionless pulley
T
fk
-----
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Problem 13 Chapter 6 edition 6 and 7
Question What is the minimum magnitude force
required to start the crate moving?
y components
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Problem 13 continued
Newton's 2nd law
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Inertial Drag Force and Terminal Velocity
Drag force Whenever you have a body like a ball
moving through a medium that behaves like a
fluid, there will be a drag force opposing the
motion.
Imagine a falling ball slowed down due to elastic
collisions with air molecules. Simply pushing the
air out of the way.
Hand waving argument
Inertial drag
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Terminal speeds in air
Using Newtons 2nd law,
where m is the mass of the falling ball
Stokes-Napier Law
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TERMINAL SPEEDS IN AIR
Object Speed (m/s) Speed (mph)
Feather 0.4 0.9 Snowflake 1 2.2 BB 9 20 Mou
se 13 29 Tennis ball 31 66 Baseball 42 86 S
ky diver 60 -120 134 -268 Cannonball 250 560
Show demo of falling feather in vacuum
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How to solve this equation?Two ways
One way is to use a spread sheet in Exel.
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Use Newtons 2nd Law
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Newtons 2nd Law
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Go to Excel Spread Sheet
631 Website Lecture 3 Materials
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C16(g-(b_1/m_1)C16C16)delta_t
D161/2(C16C17)delta_t
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We can also solve the equation to get the
velocity as a function of time before it reaches
terminal velocity. Let b 1/2CrA
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Solving equation continued
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This can be integrated, and fixing the constant
of integration by the requirement that the
velocity be zero at t 0 , which is the case for
free fall we find
Now show comparison of this solution with
numerical integration with Excel.
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Comparison
The curve modeled by velocity squared for
terminal velocity Differs from the true equation
due to a large delta t. When delta t becomes
small enough the two curves are
Indistinguishable.
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Water Resistance and Drag Forces
Whenever you have a body moving through a liquid
there will be a drag force opposing the motion.
Here the drag force is proportional to - kv.
Viscous drag. Stokes Law terminal velocity is
proportional to mass
A 1000 km boat in the water shuts off its engine
at 90 km/hr. Find the time required to slow down
to 45 km/hr due to a water drag force equal to
-70v, where v is the speed of the boat. Let k
70.
ma - kv
v/v0 45/90 1/2
t m/k ln 21000/70 ln 2 9.9 s
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Uniform circular motion
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VERTICAL CIRCULAR MOTION
Down is negative, Up is positive
Minimum v for N 0 (apparent weightlessness)
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VERTICAL CIRCULAR MOTION
Down is negative, Up is positive
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What do we mean by Fictitious ForcesFf - ma
(the fictitious force always acts in the opposite
direction of acceleration)
T
ma
T
T
mg
T
mg
q
ma
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Example of fictitious force (Ff - ma)
In a vertically accelerated reference frame, eg.
an elevator, what is your apparent weight?
Apparent weight N
Upwards is positive Downwards is negative
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Tug-of-war demo illustrates how a small sideways
force can produce a large horizontal force
Problem 13-10
Suppose two guys in the tug of war are at 4.5
meters apart and I pull the rope out 0.15 meters.
Then f 4 degrees
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ConcepTest 4.6 Force and Two Masses
1) 3/4 a1 2) 3/2 a1 3) 1/2 a1 4) 4/3
a1 5) 2/3 a1

• A force F acts on mass m1 giving acceleration
  a1. The same force acts on a different mass m2
  giving acceleration a2 2a1. If m1 and m2
  are glued together and the same force F acts on
  this combination, what is the resulting
  acceleration?

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ConcepTest 4.6 Force and Two Masses
1) 3/4 a1 2) 3/2 a1 3) 1/2 a1 4) 4/3
a1 5) 2/3 a1

• A force F acts on mass m1 giving acceleration
  a1. The same force acts on a different mass m2
  giving acceleration a2 2a1. If m1 and m2
  are glued together and the same force F acts on
  this combination, what is the resulting
  acceleration?

Mass m2 must be (1/2)m1 because its acceleration
was 2a1 with the same force. Adding the two
masses together gives (3/2)m1, leading to an
acceleration of (2/3)a1 for the same applied
force.
F m2 a2 (1/2 m1 )(2a1 )
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ConcepTest 4.7 Climbing the Rope
1) this slows your initial velocity which is
already upward 2) you dont go up, youre too
heavy 3) youre not really pulling down it
just seems that way 4) the rope actually pulls
you up 5) you are pulling the ceiling down

• When you climb up a rope, the first thing you do
  is pull down on the rope. How do you manage to
  go up the rope by doing that??

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ConcepTest 4.7 Climbing the Rope
1) this slows your initial velocity which is
already upward 2) you dont go up, youre too
heavy 3) youre not really pulling down it
just seems that way 4) the rope actually pulls
you up 5) you are pulling the ceiling down

• When you climb up a rope, the first thing you do
  is pull down on the rope. How do you manage to
  go up the rope by doing that??

When you pull down on the rope, the rope pulls
up on you!! It is actually this upward force by
the rope that makes you move up! This is the
reaction force (by the rope on you) to the
force that you exerted on the rope. And voilá,
this is Newtons 3rd Law.
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ConcepTest 4.8a Bowling vs. Ping-Pong I
1) The bowling ball exerts a greater force on
the ping-pong ball 2) The ping-pong ball exerts
a greater force on the bowling ball 3) The
forces are equal 4) The forces are zero because
they cancel out 5) There are actually no forces
at all

• In outer space, a bowling ball and a ping-pong
  ball attract each other due to gravitational
  forces. How do the magnitudes of these
  attractive forces compare?

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ConcepTest 4.8a Bowling vs. Ping-Pong I
1) The bowling ball exerts a greater force on
the ping-pong ball 2) The ping-pong ball exerts
a greater force on the bowling ball 3) The
forces are equal 4) The forces are zero because
they cancel out 5) There are actually no forces
at all

• In outer space, a bowling ball and a ping-pong
  ball attract each other due to gravitational
  forces. How do the magnitudes of these
  attractive forces compare?

The forces are equal and opposite by Newtons
3rd Law!
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ConcepTest 4.10a Contact Force I

• If you push with force F on either the heavy box
  (m1) or the light box (m2), in which of the two
  cases below is the contact force between the two
  boxes larger?

1) case A 2) case B 3) same in both cases
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ConcepTest 4.10a Contact Force I

• If you push with force F on either the heavy box
  (m1) or the light box (m2), in which of the two
  cases below is the contact force between the two
  boxes larger?

1) case A 2) case B 3) same in both cases
The acceleration of both masses together is the
same in either case. But the contact force is
the only force that accelerates m1 in case A (or
m2 in case B). Since m1 is the larger mass, it
requires the larger contact force to achieve the
same acceleration.
Follow-up What is the accel. of each mass?
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ConcepTest 5.3b Tension II
1) 0 N 2) 50 N 3) 100 N 4) 150 N 5) 200
N

• Two tug-of-war opponents each pull with a force
  of 100 N on opposite ends of a rope. What is the
  tension in the rope?

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ConcepTest 5.3b Tension II
1) 0 N 2) 50 N 3) 100 N 4) 150 N 5) 200
N

• Two tug-of-war opponents each pull with a force
  of 100 N on opposite ends of a rope. What is the
  tension in the rope?

This is literally the identical situation to the
previous question. The tension is not 200 N !!
Whether the other end of the rope is pulled by a
person, or pulled by a tree, the tension in the
rope is still 100 N !!
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ConcepTest 5.4 Three Blocks

• 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?

1) T1 gt T2 gt T3 2) T1 lt T2 lt T3 3) T1
T2 T3 4) all tensions are zero 5)
tensions are random
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ConcepTest 5.4 Three Blocks

• 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?

1) T1 gt T2 gt T3 2) T1 lt T2 lt T3 3) T1
T2 T3 4) all tensions are zero 5)
tensions are random
T1 pulls the whole set of blocks along, so it
must be the largest. T2 pulls the last two
masses, but T3 only pulls the last mass.
Follow-up What is T1 in terms of m and a?
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ConcepTest 5.5 Over the Edge
1) case 1 2) acceleration is zero 3) both
cases are the same 4) depends on value of m 5)
case 2

• In which case does block m experience a larger
  acceleration? In (1) there is a 10 kg mass
  hanging from a rope and falling. In (2) a hand
  is providing a constant downward force of 98 N.
  Assume massless ropes.

m
a
10kg
Case (1)
Case (2)
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ConcepTest 5.5 Over the Edge
1) case 1 2) acceleration is zero 3) both
cases are the same 4) depends on value of m 5)
case 2

• In which case does block m experience a larger
  acceleration? In (1) there is a 10 kg mass
  hanging from a rope and falling. In (2) a hand
  is providing a constant downward force of 98 N.
  Assume massless ropes.

In (2) the tension is 98 N due to the hand. In
(1) the tension is less than 98 N because the
block is accelerating down. Only if the block
were at rest would the tension be equal to 98 N.
m
a
10kg
Case (1)
Case (2)
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ConcepTest 5.12 Will it Budge?
1) moves to the left 2) moves to the right 3)
moves up 4) moves down 5) the box does not
move

• A box of weight 100 N is at rest on a floor
  where ms 0.5. A rope is attached to the box
  and pulled horizontally with tension T 30 N.
  Which way does the box move?

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ConcepTest 5.12 Will it Budge?
1) moves to the left 2) moves to the right 3)
moves up 4) moves down 5) the box does not
move

• A box of weight 100 N is at rest on a floor
  where ?s 0.4. A rope is attached to the box
  and pulled horizontally with tension T 30 N.
  Which way does the box move?

The static friction force has a maximum of msN
40 N. The tension in the rope is only 30 N.
So the pulling force is not big enough to
overcome friction.
Follow-up What happens if the tension is 35 N?
What about 45 N?
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ConcepTest 5.19c Going in Circles III
1) Fc T mg 2) Fc T N mg 3) Fc
T mg 4) Fc T 5) Fc mg

• You swing a ball at the end of string in a
  vertical circle. Since the ball is in circular
  motion there has to be a centripetal force. At
  the top of the balls path, what is Fc equal to?

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ConcepTest 5.19c Going in Circles III

• You swing a ball at the end of string in a
  vertical circle. Since the ball is in circular
  motion there has to be a centripetal force. At
  the top of the balls path, what is Fc equal to?

1) Fc T mg 2) Fc T N mg 3) Fc
T mg 4) Fc T 5) Fc mg
Fc points toward the center of the circle, i.e.
downward in this case. The weight vector points
down and the tension (exerted by the string)
also points down. The magnitude of the net
force, therefore, is Fc T mg
v
T
mg
R
Follow-up What is Fc at the bottom of the
balls path?