1. SECOND QUARTER! Its second quarter time!
Lets continue on with Forces, but first

2. EXAMS (seats next class)
Your exams (look at, not keep yet)

3. AN EXAMPLE
A 500kg rocket is going up at 5m/s2. How much force is the engine pushing it up.
Explain why doing F = ma to get (500)(5) = 2500N is WRONG

4. AN EXAMPLE
A 500kg rocket is going up at 5m/s2. How much force is the engine pushing it up.

6. LETS GRAB A VOTING DEVICE
Draw this situation in your notebook, call it: NET FORCE EXAMPLE.
A 3kg block is moving right, and you pull it at a 30 degree angle with 20N of force. There is friction.
It slows at a rate of 2m/s2.

7. A) There is an imbalance left, due to the acceleration direction
VOTE ON:
Which is true?
A) There is an imbalance left, due to the acceleration direction
B) There is an imbalance right, due to the force of your pull
C) There is a balance of forces because the acceleration is opposite the motion

8. A) There is an imbalance left, due to the acceleration direction
VOTE ON:
Which is true?
A) There is an imbalance left, due to the acceleration direction
B) There is an imbalance right, due to the force of your pull
C) There is a balance of forces because the acceleration is opposite the motion

9. How do you find the net force?
VOTE ON:
How do you find the net force?
A) Determine the amount of friction first, then 20 minus that
B) Determine the amount of imbalance , subtract 20 from it
C) Determine the amount of imbalance from ma

10. How do you find the net force?
VOTE ON:
How do you find the net force?
A) Determine the amount of friction first, then 20 minus that
B) Determine the amount of imbalance , subtract 20 from it
C) Determine the amount of imbalance from ma

11. Lets Do A FBD

12. Should the normal force be LARGER or SMALLER or EQUAL than the weight?
A) Larger because there is an upward force in addition to it
B) Smaller because there is an upward force in addition to it
C) Equal, since the normal force opposes the weight

13. Which is true: (Set + as direction of accel)
A) max = Ff – X F) 0 = Y20 + FN -W
B) -max = Ff – X G) 0 = Y20 - FN +W
C) -max = Y20 + FN -W
D) max = Y20 + FN -W
E) 0 = Ff – X20

14. Which is true: (Set + as direction of accel)
A) max = Ff – X F) 0 = Y20 + FN -W
B) -max = Ff – X G) 0 = Y20 - FN +W
C) -max = Y20 + FN -W
D) max = Y20 + FN -W
E) 0 = Ff – X20

15. Solve for the friction now

16. How do you determine the coefficient of friction μ?
VOTE ON:
How do you determine the coefficient of friction μ?
A) Since friction = μ*FN, use FN = mg since it’s the opposite of weight.
B) Since friction = μ*FN, use FN = Y20 because they are equal from 3rd law.
C) Since friction = μ*FN, use the earlier setups to get the FN

17. How do you determine the coefficient of friction μ?
VOTE ON:
How do you determine the coefficient of friction μ?
A) Since friction = μ*FN, use FN = mg since it’s the opposite of weight.
B) Since friction = μ*FN, use FN = Y20 because they are equal from 3rd law.
C) Since friction = μ*FN, use the earlier setups to get the FN

18. SO the coefficient is

19. Look back at the paper from two classes ago

20. Look back at the paper from two classes ago

25. Lets Try Some HW