1. Inclined Planes 

2. Inclined Planes 
What natural tendency does this object have?

3. Inclined Planes To Slide DOWNHILL!!
What natural tendency does this object have?
To Slide DOWNHILL!!

4. Inclined Planes 
And what will naturally cause this tendency?

5. Inclined Planes GRAVITY!!
And what will naturally cause this tendency?
GRAVITY!!

6. Inclined Planes 
In your notebook, draw an arrow that represents the direction of the gravitational Force...

7. Inclined Planes 
In your notebook, draw an arrow that represents the direction of the gravitational Force… IS THIS CORRECT?

8. Inclined Planes
NO!! 
In your notebook, draw an arrow that represents the direction of the gravitational Force… IS THIS CORRECT?

9. Inclined Planes 
Try Again…Remember: in which direction does gravity ALWAYS pull an object?

10. Inclined Planes ?? 
Try Again…Remember: in which direction does gravity ALWAYS pull an object!

11. Inclined Planes
YES!! 
Try Again…Remember: in which direction does gravity ALWAYS pull an object!

12. Inclined Planes 
Let’s label this the object’s weight, or “mg” (representing mass x gravity)

13. Inclined Planes mg 
Let’s label this the object’s weight, or “mg” (representing mass x gravity)

14. So what force, then, pulls the object downhill?
Inclined Planes mg 
So what force, then, pulls the object downhill?

15. Inclined Planes mg 
Gravity is still responsible for pulling the object downhill…but only a portion of the gravitational force is directed that way…

16. …or we could call it a component of the object’s weight
Inclined Planes mg 
…or we could call it a component of the object’s weight

17. Inclined Planes mg 
…or we could call it a component of the object’s weight
(parallel to the inclined surface)

18. Inclined Planes mg mg 
…or we could call it a component of the object’s weight
(parallel to the inclined surface)

19. Inclined Planes Question: 20 N mg 50 N mgFf mg
50 N mg 
Question:
Suppose you were given that the parallel component of the object’s weight was 50 N, and Friction was 20 N. Describe what the object is doing. (first label the direction of friction)

20. Inclined Planes Answer: 20 N mg 50 N mgFf mg
50 N mg 
Answer:
Certainly, you could conclude that the object has a net force of 30 N downhill and would therefore be accelerating downhill.

21. Inclined Planes Question #2: 20 N mg mgFf mg mg 
Question #2:
Now suppose I ask for the coefficient of sliding friction on the surface of the incline...

22. How would you be able to answer this?
Inclined Planes
20 N Ff mg mg 
Question #2:
How would you be able to answer this?

23. Inclined Planes Question #2: 20 N mg mgFf mg mg 
Question #2:
How would you be able to answer this?
What equation would you need to remember?

24. Inclined Planes
20 N Ff mg mg 
Ff =  x Fnorm

25. …but what is the “normal force?”
Inclined Planes
20 N Ff mg mg 
Ff =  x Fnorm
…but what is the “normal force?”

26. Inclined Planes
20 N Ff mg mg 
Recall that the normal force is a “supporting force” by the surface that an object is pressed against…

27. PERPENDICULAR to the surface.”
Inclined Planes
20 N Ff mg mg 
Recall that the normal force is a “supporting force” by the surface that an object is pressed against…
PERPENDICULAR to the surface.”

28. Fnorm For each of the following examples, label the direction of:
the weight of the crate (Fg) and
The normal force (Fnorm)

29. Horizontal Surface Inclined Surface Examples Vertical Surface Fnorm Fg

30. Inclined Planes
20 N Ff mg mg 
Let’s label the Normal Force on the diagram. And let’s also define it as exactly opposite of the object’s Perpendicular Component of weight.

31. Inclined Planes 20 N mg mg mg
Fnorm
20 N Ff mg mg mg 
Let’s label the Normal Force on the diagram. And let’s also define it as exactly opposite of the object’s Perpendicular Component of weight.

32. Inclined Planes 40 N 20 N mg mg 40 N mg
Fnorm
20 N Ff mg mg
40 N mg 
If I told you that the perpendicular component of weight was 40 N, you could then figure out the coefficient of friction (since Ff was already given as 20 N)

33. Inclined Planes and  = 20 N/40 N  = 0.5 40 N 20 N mg mg mg
Fnorm
20 N Ff mg mg mg 
So if Ff =  x Fnorm
then, 20 N =  x 40 N
and  = 20 N/40 N
 = 0.5

34. Inclined Planes and  = 20 N/40 N  = 0.5 40 N 20 N mg mg mg
Fnorm
20 N Ff mg mg mg 
So if Ff =  x Fnorm
then, 20 N =  x 40 N
and  = 20 N/40 N
 = 0.5

35. Inclined Planes mg mg mg Let’s Focus in on:
Fnorm Ff mg mg mg 
Let’s Focus in on:
How to find the Parallel and Perpendicular Components of the object’s Weight...

36. mg mg mg 
Redraw the parallel component as if you were adding this vector to the perpendicular component (tip to tail)

37. mg mg  mg
You now have a right triangle that is similar to the triangle formed by the incline over the horizontal.

38. Which angle in the vector triangle is represented by angle ?mg mg  mg
Which angle in the vector triangle is represented by angle ?

39. Consider this right triangle for a moment… mg mg  mg
Consider this right triangle for a moment…
 +  = 90

40. And  and  form a right angle together somg mg  mg
And  and  form a right angle together so
 +  = 90

41. Therefore:  +  = 90 =  +  …somg mg  mg
Therefore:  +  = 90 =  +  …so
 +  =  +  …and
 = 

42. Angle  is given in the problem…  mg mg  mg
Angle  is given in the problem…
When finding the parallel and perpendicular components of an object’s weight, set it up this way...

43. mg  sin = mgll/mg … … mg sin = mgll
(Opp/hyp)
sin = mgll/mg … … mg sin = mgll
cos  = mg/mg … … mg cos  = mg 
(Adj/hyp)

44. mg  sin = mgll/mg … … mg sin = mgll
cos  = mg/mg … … mg cos  = mg 

45. mg  sin = mgll/mg … … mg sin = mgll
cos  = mg/mg … … mg cos  = mg 

46.  mg
433 N mg 
250 N mg
1. So Please find the parallel and perpendicular components of this block’s weight. Suppose it weighs 500 N and is resting on an incline that is 30 degrees above horizontal.

47.  mg
433 N mg 
250 N mg
1. So Please find the parallel and perpendicular components of this block’s weight. Suppose it weighs 500 N and is resting on an incline that is 30 degrees above horizontal.

48. mg  mg mg 
2. Do the same calculation now for the same 500 N object if it rests on an incline that is 50 degrees above horizontal.

49. Now Try a couple of book problems together: the first one is just finding those parallel and perpendicular components:
page 133 #34
The second is a friction/incline application:
Page 143 #100

50. #34
15° mg

51. #34
m = 0.44 kg
15° mg

52. 100. Ff mg
= Ff mg
mg sin = (mg cos) 
sin = 
cos
tan = 

53. 101. FT mg FT Ff mg 

54. 105. FA mg Ff So… FA = Ff + mg Fa = (mg cos) + mg sin
Constant Velocity mg Ff 
So… FA = Ff + mg
Fa = (mg cos) + mg sin