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# 2 nd Law - Advanced - PowerPoint PPT Presentation

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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#### Presentation Transcript

“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

• Same velocity

• Same acceleration

• Move the same distance

• Move for the same amount of time

• Linking connector has constant tension

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

F

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

• 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!

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:

N

T

• Complete the force diagrams above

• You should have:

• Weight for both

• Tension for both

• Normal for block #2

T

M2g

M1g

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

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:

• 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

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

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 weight

Both masses have tension

Mass #1:m1g – T = m1a

Mass #2:T–m2g = m2a

m2

m1

T

T

m2g

m1g

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.