incline planes for learners who are new.ppt

demonstration of inclines.

1. 
Inclined
Planes

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

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

4. 
And what will naturallycause this tendency?
Inclined
Planes

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

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

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

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

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

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

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

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

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

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

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

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

17. 
…or we could call it a componentof
the object’s weight
(parallelto the inclined surface)
Inclined
Planes
mg

18. 
…or we could call it a componentof
the object’s weight
(parallelto the inclined surface)
Inclined
Planes
mg
mg

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

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

21. 
Question #2:
Now suppose I ask for the coefficientof sliding
frictionon the surface of the incline...
Inclined
Planes
mg
F
f
mg
20 N

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

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

24. 
•F
f= 
xF
norm
Inclined
Planes
mg
F
f
mg
20 N

25. 
•F
f= xF
norm
…but what is the “normalforce?”
Inclined
Planes
mg
F
f
mg
20 N

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

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

28. F
norm
For each of the following
examples, label the direction of:
•the weightof the crate (F
g) and
•The normalforce (F
norm)

29. F
g
F
norm
Horizontal Surface
F
g
F
norm
Inclined Surface
F
g
F
norm
F
a
Vertical
Surface

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

31. 
Let’s label the Normal Forceon the diagram.
And let’s also define it as exactly opposite of
the object’s Perpendicular Componentof
weight.
Inclined
Planes
mg
mg
F
f
mg
F
norm
20 N

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

33. 
So if F
f= xF
norm
then, 20 N = x40 N
and = 20 N/40 N
= 0.5
Inclined
Planes
mg
mg
F
f
mg
F
norm
20 N
40 N

34. 
So if F
f= xF
norm
then, 20 N = x40 N
and = 20 N/40 N
= 0.5
Inclined
Planes
mg
mg
F
f
mg
F
norm
20 N
40 N

35. 
Let’s Focus in on:
How to find the Paralleland
PerpendicularComponentsof the
object’s Weight...
Inclined
Planes
mg
mg
F
f
mg
F
norm

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

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

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

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

42. 
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. sin= mg
ll/mg … … mg sin= mg
ll
cos = mg
/mg … … mg cos = mg


mg
mg
mg
(Opp/hyp)
(Adj/hyp)

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

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

46. 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.

mg
mg
mg
250 N
433 N

47. 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.

mg
mg
mg
250 N
433 N

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

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. 15°
mg
#34

51. 15°
mg
#34
m = 0.44 kg

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

53. 
F
f
mg
F
T
F
T
101.
mg

54. 
mg
F
f
F
A
105.
F
a= (mg cos)+ mg sin
So… F
A= F
f+ mg
Constant Velocity