General Physics I Spring 2011 - University of Northern...
Transcript of General Physics I Spring 2011 - University of Northern...
General Physics I
Spring 2011
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Applying Newton’s Laws
Friction• When you push horizontally on a
heavy box at rest on a horizontal
floor with a steadily increasing
force, the box will remain at rest
initially, i.e., remain in
equilibrium. Since the box is in
equilibrium, the net force in the
horizontal direction is zero, i.e.,
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horizontal direction is zero, i.e.,
there must be a force opposing
your push. This force is the force
of static friction. Static friction is
friction that acts when surfaces
in contact are at rest relative to
each other.
Static Friction
• As you continue to increase the applied force, the
magnitude of the static friction (fs) rises to match the
magnitude of the applied force and the box remains at
rest. Eventually, the box will begin to move relative to
the floor when the applied force exceeds the
magnitude of the maximum static friction force (fs,max)
for the two surfaces. Experimentally, it has been found
that f is approximately proportional to the
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that fs,max is approximately proportional to the
magnitude of the normal force acting on each surface:
fs,max = µsn,
where µs is the coefficient of static friction for the two
surfaces.
Kinetic Friction
• When there is relative motion between the surfaces, the
friction force becomes kinetic friction (fk). Kinetic friction is in
the opposite direction to the motion of one surface relative
to the other.
• It has also been found experimentally that the magnitude of
the kinetic friction force is approximately proportional to the
magnitude of the normal force:
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magnitude of the normal force:
fk = µkn,
where µk is the coefficient of kinetic friction.
• For the same two surfaces, usually µk < µs.
• Note that both surfaces experience the friction force,
according to Newton’s third law.
Kinetic Friction
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Kinetic friction is independent of speed and area of contact.
Static and Kinetic Friction
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Graph of friction force vs. time as the box is pushed.
Friction and Driving
• Under normal circumstances, the force that propels your car forward is static friction between the tires are the road. The bottom surface of a tire does not slip relative to the road. Note that friction is in the same direction as the motion of the car. That is why it is important to note that friction opposes relative motion between surfaces. (The tire pushes back on the road; friction opposes this attempted backward relative motion and so acts in the forward direction.)
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motion and so acts in the forward direction.)
• When you are braking, it is also better that static friction slows you down. If the brakes lock, the tires slip and the friction will be kinetic friction, which is usually less than the maximum static friction force. Anti-lock brakes prevent this from happening.
• When the roads are icy, the coefficients of static and kinetic friction are much smaller. The tires slip much more easily and the friction forces are much reduced, making driving more difficult.
Workbook: Chapter 5, Question 21
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Interacting Objects• As we saw before, when
objects interact, they experience forces that come in action/reaction pairs due to Newton’s third law. Each force that constitutes a pair acts on a different object. These forces have to be treated carefully when solving
An�
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carefully when solving problems involving interacting objects. Note that the forces of a pair have the same magnitude and this is a useful piece of information in solving problems involving interacting objects. The figure and free-body diagrams illustrate these ideas.
A
A
Aw�
handF�
BonAF�
B
Bn�
Bw�
AonBF�
AonB BonAF F=� �
Interacting Objects: Accelerations
• In many cases, the accelerations of
two interacting bodies will be
related. For example, for two
blocks being pushed while they are
in contact, their accelerations are
equal. Otherwise, the blocks would
separate. Applying Newton’s 2nd
law to each block yields:A
An�
handF�
BonAF�
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law to each block yields:A
Aw�
B
Bn�
Bw�
AonBF�
,
Block A. ---- (1)
Block B. ---- (2)
Since we can substitute for
in Eq. (1) to get or,
.hand
hand
x BonA Ahand
x BAonB
BonA AonB
BBonA A
BA
F
F
F F F m a
F F m a
F F
F m a m a
m m a
−
= − =
= =
=
=
= +
∑
∑
x
x
Interacting Objects: Accelerations• For two objects connected by a
rope or string that doesn’t
stretch or shrink, the
magnitudes of the accelerations
are the same. In the example to
the right, though the directions
of the accelerations are
different, their magnitudes are
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different, their magnitudes are
the same. Thus,
• In terms of components,
(Note that the acceleration of B
is in the negative y direction and
so is actually negative.)
.BA
a a=� �
.ByAx
a a=−
Bya
Ropes and Pulleys• In this course, we will
assume that ropes or strings
are massless and pulleys are
massless and frictionless.
• The tension in a massless
rope is the same everywhere
in the rope, including at both
ends (which are connected to
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ends (which are connected to
other objects). The rope
exerts a force on a connected
object equal to the tension.
(The object of course exerts
a reaction force on the rope
having the same magnitude
as the tension.)
Ropes and Pulleys• When a rope or string passes
over a pulley, the magnitude of
the tension in the rope is the
same on both sides of the
pulley.
T�
Frictionless surface
y
y
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A
An�
Aw�
AT�
B
BT�
Bw�
BAT T=� �
x
Workbook: Chapter 5, Question 26, 29
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Textbook: Chapter 5, Problem 74
B
BT�
Bw�
mB = 100 kgA
AT�
Aw�
y
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212
First, we have to find the acceleration of the 100-kg block.
Use: ( ) ( ) .
Knowns: 1 m, 0 m, 6.0 s, ( ) 0. Unknown: .
2 ( )Solving for gives:
(
yi i yBf
yi i yBf
yi if
yB yB
y y v t a t
y y t v a
y y v ta a
= + ∆ + ∆
=− = ∆ = =
− − ∆=
∆2
2 2
2( 1 m 0 0) 0.0556 m/s .) (6.0 s)t
− − −= =−
Problem 74 (continued)
2 2
Apply Newton's 2nd law to the 100-kg block.
.
So, ( ).
100 kg[9.8 m/s ( 0.0556 m/s )] 974 N.
Apply Newton's 2nd law to block of unknown mass.
y B B B yB
B B B yB B B yB B yB
B
y A A A y
F T w m a
T w m a m g m a m g a
T
F T w m a
= − =
= + = + = +
= + − =
= − =
∑
∑ . So,
( ). Thus,
A
= + = + = +
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2
2 2
( ). Thus,
.( )
Now, 974 N. Further, 0.0556 m/s .
974 NHence, 99 kg.(9.8 m/s 0.0556 m/s )
A A A yA A A yA A yA
AA
yA
B yBA yA
A
T w m a m g m a m g a
Tm
g a
T T a a
m
= + = + = +
=+
= = =− =
= =+