Transcript Document

Newton’s Laws (II)
• Free-body diagrams
• Normal Force
• Friction, ropes and pulleys
Serway and Jewett : 5.7, 5.8
Physics 1D03 - Lecture 7
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Free-Body Diagrams
• Pick one object (the “body”).
• Draw all external forces which act directly on that body
(gravity, contact, electromagnetic).
Imagine cutting around the body to separate it from its surroundings.
Replace each external object with a force applied at the point of
contact.
• Indicate the direction of the acceleration of the object
beside the diagram; but remember, ma is not a force
on the diagram.
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Example: free-body diagram
m
A block is pulled up a frictionless ramp:
Note :
Forces on Block

a

FT

mg
• title, to indicate the chosen
object (use m or mA etc)
• contact forces, to replace the
rope and the ramp
• gravity doesn’t require contact

n
• a may be indicated for
reference, but is not a force
Physics 1D03 - Lecture 7
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Ropes and Pulleys
• A rope attached to something
exerts a force parallel to the rope
• The magnitude of the force is
called the tension in the rope
rope
Force
• Tension is uniform in a rope of negligible mass
• The tension is not changed if the rope passes over
an ideal pulley (frictionless and massless)
• Tension has units of force (newtons)
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Quiz
Since the elevator is broken, a student
rigs up a bucket-and-pulley system to
avoid climbing stairs.
The student
weighs 750 N, and the bucket weighs
250 N. How hard (with what force) must
the student pull on the rope to go up
with the bucket?
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
Fg
Contact Forces
Example: We try to push a
block across a table; the table
pushes back.
FA= applied force

f
FA

n

Ftable
Divide the contact force from
the table into two components:
-
normal force, perpendicular to the surfaces
-
friction , parallel to the surface
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If we look closely, the normal force
arises from the table being bent :
as the table tries to straighten, it
pushes back.
This is really an elastic force; the table
behaves like a spring.
At the atomic level, contact forces are due
to electromagnetic interactions.
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Friction
Friction is the force which resists sliding of two surfaces across
each other.
We distinguish between static and kinetic friction:
Static Friction :
- there is NO relative motion
- fs prevents sliding
Kinetic Friction :
- the block is sliding
- fk is opposite to v
FA

F  0

fs
v
fk
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Friction is complicated. A useful empirical model was
presented by Charles Coulomb in 1781:
1.
The force of static friction has a maximum value; if you
push too hard, the block moves. This maximum value is
proportional to the normal force the surfaces exert on each
other.
2.
Once the object is sliding, kinetic friction is approximately
independent of velocity, and usually smaller than the
maximum static friction force. The force of kinetic friction is
also proportional to the normal force.
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Define two pure numbers (no units):
s (“coefficient of static friction”)
k (“coefficient of kinetic friction”)
(“” is a Greek letter, pronounced “mu”)
Then Coulomb’s rules are:

FA

mg

f

n
f s  s n
f k  k n
Question : would
 itbe correct to write these as vector
equations, f  n ?
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Copper on steel
Aluminum on aluminum
Teflon on Teflon
s
0.53
1.5
0.04
k
0.36
1.1
0.04
• Values depend on smoothness, temperature, etc. and
are approximate
• Usually  < 1, but not always
• Usually, k is less than s , and never larger
• The coefficients depend on the materials, but not on
the surface areas, contact pressure, etc.
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Concept Quiz
Each block weighs 100 N, and the coefficient of
static friction between each pair of surfaces is 0.50.
What minimum force F is needed to pull the lower
block out?
a) 50 N
b) 100 N
c) 150 N
F
100 N

100 N
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Equilibrium

• A special case : a  0 (object doesn’t move, or moves
at constant velocity)


• Newton’s second law gives  F  ma  0
The vector sum of forces acting
on a body in equilibrium is zero
• This is equivalent to three independent component
equations:  Fx  0,  Fy  0,  Fz  0
• We can solve for 3 unknowns (or 2, in 2-D problems)
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Example
A block is in equilibrium on a
frictionless ramp. What is the
tension in the rope?
m
f
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Quiz
The block has weight mg and is in equilibrium on the ramp.
If s = 0.9, what is the frictional force?
37o
A)
B)
C)
D)
0.90 mg
0.72 mg
0.60 mg
0.54 mg
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A Heavy Rope
B
A
2
1
1  60
 2  45


mg  200 N
Find the tension in the rope at A and
at B. Should we assume TA = TB ?
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Example 3 (solutions)
Free-body diagram of rope
TA
2
1
mg
TB
TB
TA
1
2
mg

• weight mg should be applied at the “center of mass”,
which we will discuss later.
• For summing forces, the force locations don’t matter,
only directions.
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Example
For what angle θ is the system in equilibrium?
(assume frictionless surfaces, ideal pulleys, etc.)
M
m

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Summary
• Free-body diagrams (Text section 5.7)


• Equilibrium: if a  0,  F  0.
• Force Laws:
- ropes and pulleys
- normal force
• Static and kinetic friction, f=μN
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