2 nd law advanced
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2 nd Law - Advanced. “This is just like college!”. Presentation 2003 R. McDermott. AP Level – What’s the Difference?. Two (or more) objects moving Friction acting Inclined planes Hanging objects No numbers. Linked Objects. Linked objects move together Same velocity Same acceleration

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2 nd Law - Advanced

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2 nd law advanced

2nd Law - Advanced

“This is just like college!”

Presentation 2003 R. McDermott


Ap level what s the difference

AP Level – What’s the Difference?

  • Two (or more) objects moving

  • Friction acting

  • Inclined planes

  • Hanging objects

  • No numbers


Linked objects

Linked Objects

  • Linked objects move together

    • Same velocity

    • Same acceleration

    • Move the same distance

    • Move for the same amount of time

    • Linking connector has constant tension


Sample 1

F

M1

M2

Sample #1:

  • F is acting only on M1 (touching)

  • Connecting cord acts on both

  • Assume no friction

  • Find tension in connecting cord

  • Find acceleration


Sample cont

F

M1

M2

Sample cont.

  • To find two unknowns requires two equations

  • Set up free-body diagrams for both objects


Force diagram

Force Diagram:

N1

N2

  • Both objects feel tension from the connecting cord:

T

T

F

M2g

M1g

  • Both objects feel weight:

  • Both objects feel a normal force:


Axis etc

N1

N2

T

T

F

M2g

M1g

Axis, etc

  • 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

N1

N2

T

T

F

M2g

M1g

Horizontal Results

  • Horizontally, we do not have equilibrium

  • Horizontally then, F = ma

  • For object 1,F – T = M1a

  • 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

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

M1

Sample #2:

  • M1 is hanging, M2 is on the table

  • Connecting cord acts on both

  • Assume no friction

  • Find tension in connecting cord

  • Find acceleration


Force diagram 2

Force Diagram #2:

N

T

  • Complete the force diagrams above

  • You should have:

    • Weight for both

    • Tension for both

    • Normal for block #2

T

M2g

M1g


Diagram 2

N

T

T

M2g

M1g

Diagram #2:

  • 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

M2g

M1g

#2 Equations:

  • For object #2:N = m2g

  • And also:T = m2a

  • We get for object #1:m1g – T = m1a

  • The firstequation is trivial, of course


Sample 3

Sample #3:

  • 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

Force Diagram #3:

N2

T

T

  • Complete the force diagrams above

  • You should have:

    • Weight (or weight components) for both

    • Tension for both

    • Normal for block #2

m2gsin

m2gcos

m1g


Diagram 3

N2

T

T

m2gsin

m2gcos

m1g

Diagram #3:

  • 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

T

N2

T

m2gsin

m2gcos

m1g

#3 Equations:

  • For object #2:N2 = m2gcos

  • And also:T – m2gsin = m2a

  • We get for object #1:m1g – T = m1a

  • The firstequation is trivial, of course


Atwood machine

m2

m1

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)


Atwood diagram

Both masses have weight

Both masses have tension

Mass #1:m1g – T = m1a

Mass #2:T–m2g = m2a

m2

m1

Atwood Diagram

T

T

m2g

m1g


General principles

Listen up!

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!


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