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

Sample #1: F is acting only on M1 (touching) Connecting cord acts on both Assume no friction Find tension in connecting cord Find acceleration M2 M1 F

Sample cont. To find two unknowns requires two equations Set up free-body diagrams for both objects M2 M1 F

Force Diagram: Both objects feel tension from the connecting cord: F M1gM1g N1N1 T M2gM2g T N2N2 Both objects feel weight: Both objects feel a normal force:

Axis, etc The axis system is normal; no components No vertical acceleration, so: For object 1,N 1 = M 1 g For object 2,N 2 = M 2 g This is a trivial result since we usually know the masses. F M1gM1g N1N1 T M2gM2g T N2N2

Horizontal Results Horizontally, we do not have equilibrium Horizontally then,  F = ma For object 1,F – T = M 1 a For object 2,T = M 2 a Since F is usually known, and a and T are the same for each, we can solve for either a or T F M1gM1g N1N1 T M2gM2g T N2N2

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: M1 is hanging, M2 is on the table Connecting cord acts on both Assume no friction Find tension in connecting cord Find acceleration M1 M2

Force Diagram #2: Complete the force diagrams above You should have: –Weight for both –Tension for both –Normal for block #2 M1gM1g T M2gM2g N T

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 M1gM1g T M2gM2g N T

#2 Equations: For object #2: N = m 2 g And also: T = m 2 a We get for object #1: m 1 g – T = m 1 a The first equation is trivial, of course M1gM1g T M2gM2g N T

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: Complete the force diagrams above You should have: –Weight (or weight components) for both –Tension for both –Normal for block #2 m1gm1g T T m 2 gcos  m 2 gsin  N2N2

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 m1gm1g T T m 2 gcos  m 2 gsin  N2N2

#3 Equations: For object #2: N 2 = m 2 gcos  And also: T – m 2 gsin  = m 2 a We get for object #1: m 1 g – T = m 1 a The first equation is trivial, of course m1gm1g T T m 2 gcos  m 2 gsin  N2N2

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) m1m1 m2m2

Atwood Diagram Both masses have weight Both masses have tension Mass #1:m 1 g – T = m 1 a Mass #2:T – m 2 g = m 2 a m1m1 m2m2 m1gm1gm2gm2g T T

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!