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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
- Move the same distance
- Move for the same amount of time
- Linking connector has constant tension

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

M1

M2

Sample cont.- To find two unknowns requires two equations
- Set up free-body diagrams for both objects

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:

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.

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?

- 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!

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:

N

T

- Complete the force diagrams above
- You should have:
- Weight for both
- Tension for both
- Normal for block #2

T

M2g

M1g

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

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:

- 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

- Complete the force diagrams above
- You should have:
- Weight (or weight components) for both
- Tension for both
- Normal for block #2

m2gsin

m2gcos

m1g

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

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

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)

Both masses have tension

Mass #1: m1g – T = m1a

Mass #2: T–m2g = m2a

m2

m1

Atwood DiagramT

T

m2g

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!

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