Transcript 9. 2nd Law Advanced

2nd Law - Advanced
“This is just like college!”
Presentation 2003 R. McDermott
AP Level – What’s the Difference?
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Two (or more) objects moving
Friction acting
Inclined planes
Hanging objects
No numbers
Linked Objects
• Linked objects move together
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Same velocity
Same acceleration
Move the same distance
Move for the same amount of time
Linking connector has constant tension
Sample #1:
M2
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M1
F
F is acting only on M1 (touching)
Connecting cord acts on both
Assume no friction
Find tension in connecting cord
Find acceleration
Sample cont.
M2
M1
F
• To find two unknowns requires two
equations
• Set up free-body diagrams for both objects
Force Diagram:
N2
T
T
M2g
N1
F
M1 g
• Both objects feel tension from the connecting
cord:
• Both objects feel weight:
• Both objects feel a normal force:
Axis, etc
N2
T
M2g
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T
N1
F
M1 g
The axis system is normal; no components
No vertical acceleration, so:
For object 1,
N1 = M1g
For object 2,
N2 = M2g
This is a trivial result since we usually know the
masses.
Horizontal Results
N2
T
M2g
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T
N1
F
M1 g
Horizontally, we do not have equilibrium
Horizontally then, F = ma
For object 1,
F – T = M1 a
For object 2,
T = M2a
Since F is usually known, and a and T are the
same for each, we can solve for either a or T
Add Friction?
• How would the problem be changed if friction
was involved?
• What are the potential pitfalls in the problem
when friction is present?
• The two frictional effects might be large enough
to balance F so that the objects cannot move!
• Remember that f = N is the maximum potential
friction; f cannot exceed the applied force F!
Sample #2:
M2
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M1 is hanging, M2 is on the table
Connecting cord acts on both
Assume no friction
Find tension in connecting cord
Find acceleration
M1
Force Diagram #2:
N
T
T
M2 g
• Complete the force diagrams above
• You should have:
– Weight for both
– Tension for both
– Normal for block #2
M1 g
Diagram #2:
N
T
T
M2 g
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M1 g
Once you’ve gotten the diagrams above,
You pick a direction to be positive
I’ll elect to make right and down positive
Note that this is consistent for the motion of the
two objects
#2 Equations:
N
T
T
M2 g
• For object #2:
N = m2g
• And also:
T = m2a
• We get for object #1: m1g – T = m1a
• The first equation is trivial, of course
M1 g
Sample #3:
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M1 is hanging, M2 on the incline
Connecting cord acts on both
Assume no friction
Find tension in connecting cord
Find acceleration
Force Diagram #3:
N2
T
T
m2gsin
m2gcos
• Complete the force diagrams above
• You should have:
– Weight (or weight components) for both
– Tension for both
– Normal for block #2
m1g
Diagram #3:
N2
T
T
m2gsin
m2gcos
m1g
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Once you’ve gotten the diagrams above,
You pick a direction to be positive
I’ll elect to make right and down positive
Note that this is consistent for the motion of the
two objects
#3 Equations:
N2
T
T
m2gsin
m2gcos
m1g
N2 = m2gcos
• And also:
T – m2gsin = m2a
• We get for object #1: m1g – T = m1a
• For object #2:
• The first equation is trivial, of course
Atwood Machine
• The last step in the
progression is the diagram to
the right:
• Both masses hanging
• Draw the force diagrams
• Choose a direction to make
positive (I’ll still use over to
right and down)
m2
m1
Atwood Diagram
T
T
• Both masses have weight
• Both masses have tension
m2
m2g
• Mass #1:
• Mass #2:
m1g – T = m1a
T – m2g = m2a
m1
m1g
General Principles:
• Always make a force diagram
• Choosing a direction is arbitrary,
but you must be consistent or
there will be sign problems.
• If your choice of direction is
wrong, the sign on your answers
will indicate that.
• The setup and equations are more
important than the numerical
answers!
Listen up!