1. What is the weight of a 15 kg rock?
147 N
A skateboard (mass 12 kg ) accelerates from rest to 3.8 m/s over a distance of 9.4m. What is the skateboard ’s rate of acceleration? 
0.768 m/s2
What force was applied to the skateboard in to achieve this rate of acceleration?
9.22 N
An unbalanced force of 25 N is applied to a 12.5-kg mass. What should be the acceleration experienced by the mass?
2 m/s2

2. 11/13 Forces Assignment Parts 2 & 3 due Friday
NOW: Pick up Force Notes II. ON Force ntoes one, answer example J.
Last Friday: A quiz over Fun with Forces & Fun with Weight.
Monday: finished lab.
Tuesday: reviewed lab, turned it in and took a quiz.
If you were absent for any of these you need to make them up today after school or tomorrow morning.
PM Test corrections & retakes Mon-Thur this week. You must show completed PM notes to be eligible for corrections & retakes. Retakes pm only. I have pm duty until 2:55
There is an issue with the website. I will fix it today if possible

3. = - 3 m/s2 = 10 s d = vit+ .5at2 = 30(10)+ .5 (-3)(10)2 = 150 m
Ex J The brakes of a 1000-kg car exert 3000 N.
a. How long will it take the car to come to a stop from a velocity of 30 m/s?
m = 1000 kg
F = N
vi = 30 m/s
vf = 0 m/s
= - 3 m/s2
= 10 s
b. How far will the car travel during this time?
d = vit+ .5at2
= 30(10)+ .5 (-3)(10)2
= 150 m

4. Forces Part II Friction

5. When surfaces are pressed together, we can identify four forces

6. Friction Force: FF
Force opposing motion
Measured in Newtons (N)

7. Applied Force: FA
The push or pull applied to the object
Measured in Newtons (N)

8. Fw Force of Weight or Gravity
(Mass in kg) (Acceleration due to Gravity)
Kg x 9.8 m/s2
Measured in Newtons (N)

9. Normal Force: FN
Usually a 3rd Law reaction to gravity, that is equal and opposite of Force of Weight (Fw)
Perpendicular to the surface.
Measured in Newtons (N)
FN is NOT FNET
Don’t confuse them just because they begin with N!

10. A 50kg object is moving horizontally at a constant velocity
A 50kg object is moving horizontally at a constant velocity. Is there acceleration? Is there a net force? FN FF FA FW

11. Ex L No Friction Fnet = ma
A 50 kg object experiences an applied force of 400 N, what is the Fnet? What is the acceleration?:
Fnet = FA
Fnet = Net Force, results in acceleration
a = Fnet /m = (400N )/ (50 kg) = 8 m/s2

12. Example L FN FA FW

13. Friction
Is a force that opposes motion.

14. Ex M Friction and Constant velocity. Is there a net force?
Fnet = 0
FA+ -FF = ma or FA + - FF = Fnet
A 50 kg object moves at a constant velocity when acted upon by an applied force of 400 N. What is the FF? What is the Fnet? What is the acceleration?:
FF = FA
Fnet = FA + - FF = 0 N
If Fnet = 0 then acceleration = ?

15. Example M Friction and constant velocityFN FF FA FW

16. How does friction affect net force?
Fnet = ma
ma = FA+ FF
Where:
Fnet = Net Force
FF = Friction Force
FA = Applied Force
You are actually subtracting FF from FA, since Ff is in the opposite direction
Friction will reduce the net force

17. Do all surfaces provide the same amount of friction
Do all surfaces provide the same amount of friction? How is this described? FN FF FA FW

18. Friction What does it depend on?
Depends on the surface of the materials.
Depends on how tightly the surfaces are pressed together.
FF = Force of Friction

19. Coefficient of Friction
The coefficient of friction is a measure of how difficult it is to slide a material of one kind over another; the coefficient of friction applies to a pair of materials, and not simply to one object by itself
Coefficient of Friction Reference Table - Engineer's Handbook

20. When Surfaces are Pressed Together
Coefficent of Friction µ can be calculated
It is a ratio of FF and FN

21. Coefficient of friction
What is it equal to?
What is the unit for Coefficient of friction?
If Coefficient of friction is small, what does that mean about the FF?
If Coefficient of friction is large, what does that mean about the FF?
The higher the coefficient of friction, the more difficult to slide

22. I call these FAWN problems

23. Example N
A 400 N force is applied to a 50kg object. Calculate the acceleration of the object if  = 0.3. FN FF FA Fg

24. Example Cont’d. FA = 400 N Fg = m·g (50kg)(9.8m/s2) Fg = 490 N
FN = 490 N also
FF = (0.3)(490N) = 147 N

25. Example cont’d.
FN = 490N
FF = 147N
FA = 400N
Fg = 490N

26. To solve for Acceleration must calculate Net Force
FNET = ma= 400N – 147N
ma = 253N
FNET = ma = FA+ -FF

27. Now use Net Force and mass of object in F=ma formula
a = F/m
a = 253N/50kg
a =5.06 m/s2

28. Ex O Putting it all together
A 50 kg object accelerates horizontally at 0.2 m/s2 from rest for 5 seconds. If the coefficient of friction is 0.01, what is the Fnet? What is the FF ? What is the FA ?How far does it travel in 5 seconds?:
Fnet = (50 kg) (0.2 m/s2) = 10 N
FF = μ FN
FN = (50kg)(9.8 m/s2) = 490 N
FF = (0.01) (490 N) =4.9N
FA = Fnet – -FF
FA = 10 N N = 14.9 N

29. How does friction affect net force? Putting it all together
A 50 kg object accelerates horizontally at 0.2 m/s2 from rest for 5 seconds. If the coefficient of friction is 0.01, what is the Fnet? What is the FF ? What is the FA ?How far does it travel in 5 seconds?:
Fnet = (50 kg) (0.2 m/s2) = 10 N
Set up DVVAT
d = ?
vi = 0 m/s
vf = ?
a =0.2 m/s2
d = (.5)(.2)(5)2
d = 2.5m

30. As the coefficient of friction decreases and the object and the applied force remain the same, What happens to: FA FW FN
FNET

32. TENSION aka FT
is the magnitude of the pulling force exerted by a string, cable, chain, or similar object on another object.
It is the opposite of compression. It is a “response force”
That is to say, if one pulls on the rope, the rope fights back by resisting being stretched
Ropes, strings, and cables can only pull. They cannot push because they bend.
is measured in newtons
is always measured parallel to the string on which it applies.

33. What does the rope provide?
A lift (vertical force) and a pull (horizontal force)
If there was no angle, would there be any vertical force? No
If the angle was at 90°, how would that affect the force components?
Force would only be in the vertical plane
How would you calculate the horizontal and vertical force components if the angle of the rope with the floor was 57° and the Force of tension (FA) in the rope was 400 N?

34. Ex P. This crate is be pulled with a rope across a friction based horizontal surface at a constant velocity. The rope exerts a tension of 400 N at an angle of57°. What is the coefficient of friction?
50 kg
57°

35. 50 kg B. Determine FAY or FVert A. Determine FAX or FHoriz
A box is pulled into motion with a rope across a horizontal surface. The rope makes an angle of 57° to the floor. The Force of tension (FA) in the rope is 400N
FAY
400 N
50 kg
57º
FAX
B. Determine FAY or FVert
= sin (57) (400N)
= N
A. Determine FAX or FHoriz
= cos (57) (400N)
= N

36. I work in a circle Determine FW Determine FAX Determine FAY
Determine FN
The FAY supplies part of the upward force. The total upward force is FAY + FN and together these are equal but opposite the FW. FN = FW – FAY
Determine FF
Determine coefficient of friction

37. 50 kg 57° Determine FN Determine Fw = FW- FAY = (50 kg)(9.8 m/s2)
FAY N
400 N
57°
50 kg
FAX N
Determine FN
= FW- FAY
= 490 N – N
= N
Determine Fw
= (50 kg)(9.8 m/s2)
= 490 N

38. 50 kg 57° Determine µ= FF/ FFN Determine FF = 217.86N/ 154.53 N
FAY N
400 N
57°
50 kg
FAX N
Determine µ= FF/ FFN
= N/ N
= 1.41
Determine FF
FF= Fax (Constant Velocity)
FF= N

39. The End