2 nd Law - Advanced

1. 2nd Law - Advanced “This is just like college!” Presentation 2003 R. McDermott

2. AP Level – What’s the Difference? • Two (or more) objects moving • Friction acting • Inclined planes • Hanging objects • No numbers

3. 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

4. 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

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

6. 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:

7. 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.

8. 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

9. 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!

10. 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

11. 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

12. 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

13. 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

14.  Sample #3: • M1 is hanging, M2 on the incline • Connecting cord acts on both • Assume no friction • Find tension in connecting cord • Find acceleration

15. 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

16. 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

17. 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

18. 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)

19. 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

20. 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!