Transcript Lecture 6.1

Welcome back to Physics 215
Today’s agenda:
• Weight, elevators, and normal forces
• Static and kinetic friction
• Tension
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Current homework assignment
• HW5:
– Knight textbook Ch.6: 38, 42, 56, 58
– Ch.7: 46, 54
– due Friday, Oct. 3rd in recitation
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Summary
• To solve problems in mechanics, identify all
forces and draw free body diagrams for all
objects
• If more than one object, use Newton’s Third
law to reduce number of independent forces
• Use Newton’s Second law for all components
of net force on each object
• Choose component directions to simplify
equations
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Weight, mass, and acceleration
•
What does a weighing scale
``weigh’’?
•
Does it depend on your frame of
reference?
•
Consider elevators….
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A person is standing on a bathroom scale while riding
an express elevator in a tall office building. When the
elevator is at rest, the scale reads about 160 lbs.
While the elevator is moving, the reading is frequently
changing, with values ranging anywhere from about
120 lbs to about 200 lbs.
At a moment when the scale shows the maximum
reading (i.e., 200 lbs) the elevator
1.
2.
3.
4.
must be going up
must be going down
could be going up or going down
I’m not sure.
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Motion of elevator
Motion of elevator
(if a < 0 )
(if a > 0 )
• Moving upward and
slowing down,
• Moving upward and
speeding up,
OR
OR
• Moving downward
and speeding up.
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• Moving downward
and slowing down.
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A person is standing on a bathroom scale while riding
an express elevator in a tall office building. When the
elevator is at rest, the scale reads about 160 lbs.
While the elevator is accelerating, a different reading is
observed, with values ranging anywhere from about
120 lbs to about 200 lbs.
At a moment when the scale shows the maximum
reading (i.e., 200 lbs) the acceleration of the elevator is
approximately
1.
2.
3.
4.
1 m/s2
2.5 m/s2
5 m/s2
12.5 m/s2
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Conclusions
• Scale reads magnitude of normal force |NPS|
• Reading on scale does not depend on
velocity (principle of relativity again!)
• Depends on acceleration only
* a > 0  normal force bigger
* a < 0  normal force smaller
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Reminder of free-fall experiment
• Objects fall even when there is no atmosphere
(i.e., weight force is not due to air pressure).
• When there is no “air drag” things fall “equally
fast.” i.e. same acceleration
• From Newton’s 2nd law, a = W/m is
independent of m -- means W = mg
• The weight (i.e., the force that makes objects in
free fall accelerate downward) is proportional to
mass.
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Inertial and gravitational mass
• Newton’s second law:
F = mI a
• For an object in “free fall”
W = mG g
• If a independent of mI, must have
mI = mG
 Principle of equivalence
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Forces of friction
• There are two types of situations in which
frictional forces occur:
– Two objects “stick to each other” while at rest
relative to one another (static friction).
– Two objects “rub against each other” while moving
relative to each other (kinetic friction).
• We will use a macroscopic description of
friction that was obtained by experiment.
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Friction demo
• Static friction: depends on surface and
normal force for pulled block
• Kinetic friction: generally less than
maximal static friction
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The maximum magnitude of the force of static friction
between two objects
• depends on the type of surfaces of the objects
• depends on the normal force that the objects exert on
each other
• does not depend on the surface area where the two
objects are touching
f
static
B,T
£m
static
NB,T
The actual magnitude of the force of static friction is
generally less than the maximum value.
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A 2.4-kg block of wood is at rest on a concrete floor.
(Using g = 10 m/s2, its weight force is about 24 N.)
No other object is in contact with the block. If the
coefficient of static friction is ms = 0.5, the frictional force
on the block is:
1.
0N
3.
12 N
2.
8N
4.
24 N
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A 2.4-kg block of wood is at rest on a concrete floor.
(Using g = 10 m/s2, its weight force is about 24 N.)
Somebody is pulling on a rope that is attached to the
block, such that the rope is exerting a horizontal force of
8 N on the block. If the coefficient of static friction is
ms = 0.5, the frictional force on the block is:
1.
0N
3.
12 N
2.
8N
4.
24 N
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Having no choice, you have parked your old
heavy car on an icy hill, but you are worried that
it will start to slide down the hill.
Would a lighter car be less likely to slide when
you park it on that icy hill?
1.
2.
3.
No, the lighter car would start sliding at a
less steep incline.
It doesn’t matter. The lighter car would
start sliding at an incline of the same angle.
Yes, you could park a lighter car on a
steeper hill without sliding.
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Block on incline revisited
F
N
W
q
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Initially at rest
• What is the largest angle before the
block slips?
• Resolve perpendicular to plane
 N = Wcosq
• Resolve parallel  F = Wsinq
• Since F ≤ msN, we have
Wsinq ≤ msWcosq i.e.
tanq ≤ ms
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What if q >
-1
tan ms
?
The magnitude of the force of kinetic friction between
two objects
• depends on the type of surfaces of the objects
• depends on the normal force that the objects exert on each
other
• does not depend on the surface area where the two objects are
touching
• does not depend on the speed with which one object is moving
relative to the other
f
kinetic
B,T
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=m
kinetic
NB,T
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What if q >
-1
tan ms
?
• Block begins to slide
• Resolve along plane:
Wsinq - mKWcosq = ma
• Or:
a = g(sinq - mKcosq)
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Summary of friction
• 2 laws of friction: static and kinetic
• Static friction tends to oppose motion and is
governed by inequality
Fs ≤ msN
• Kinetic friction is given by equality FK = mKN
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Tension
• For an ideal string or rope connecting two objects:
• does not stretch  inextensible
• has zero mass
• Let’s look at an example of a cart connected
to a falling mass by an ideal string...
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What if the mass of the rope/string is not zero?
Two blocks are connected by a heavy rope. A
hand pulls block A in such a way that the blocks
move upward at increasing speed.
The (downward) tension force on the upper
block by the rope is
1.
2.
3.
less than
equal to
greater than
the (upward) tension force on the lower block by the rope.
4.
Answer depends on which block is heavier.
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Hand pulls block A so blocks move up at increasing speed.
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Notice: for mR = 0, the tension forces
exerted at either end are the same.
The term “tension in the string” is
therefore often used as a short-hand for
the tension forces exerted on or by
the string at either end.
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Blocks A and C are initially held in place as
shown. After the blocks are released, block A
will accelerate up and block C will accelerate
down.
The magnitudes of their accelerations are the
same.
Will the tension in the string be
1.
equal to 1.0 N (i.e. the weight of A),
2.
between 1.0 N and 1.5 N,
3.
equal to 1.5 N (i.e. the weight of C), or
4.
equal to 2.5 N (i.e. the sum of their weights)?
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Reading assignment
• Tension, pulleys
• Chapters 6, 7 in textbook
• Start circular motion
• Ch. 8 in textbook
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