1. Vectors and Free Body Diagrams

2. scalars (non-vectors)
A vector has a magnitude(amount) and a direction.
displacement
velocity
acceleration
momentum
force
scalars (non-vectors)
length, mass, time, speed
energy, density,
A vector representations looks like a "ray", but it is actually a line segment, with an arrow on one end to indicate direction.

3. There are two ways of doing vector addition.
The math way
(geometry)
The "drawing" way
(graphical)

4. Do vector addition the math way requires that each vector be broken up into it component X and Y parts.
You find all the X parts and add them.
You find all the Y parts and add them.
Up 4.3 (Y)
over 8 (X)

5. -4 -8 3 -4 6 4 10 10 -5 -2

6. To add these vectors the math way you add all the X - Y parts.-4 -8 3 -4 6 4

7. Gather all the X components.-4 -8 3 -4
(6)+(4)+(-4) = 6 6 4

8. Gather all the X components.-4 -8 3 -4
(6)+(4)+(-4) = 6 6
Gather all the Y components. 4
(3)+(-8)+(-4) = -9

9. horizontal +6 vertical -9
Gather all the X components. -4 -8 3 -4
(6)+(4)+(-4) = 6 6
Gather all the Y components. 4
(3)+(-8)+(-4) = -9
horizontal +6
vertical -9

10. To add these vectors the graphical way.
You move each vector (re-draw) so that one begins where the last ended. -4 -8 -4 4 3 6

11. -4 -4

12. Solve this the graphical way.

13. to the right 6 down 9 The resultant vector is: ( horizontal 6 )
( vertical -9 ) -9 6

14. Typically you are not given the X and Y parts of a vector.
80 Newton
45 degrees
You are given the direction and magnitude
(angle and size)

15. How to I find the X and Y components? 80 Newton 45 degrees
If it is drawn on a graph then I can just count.
11 up
11 right

16. Do vector addition to find the net push force on the box.
8.2 N and 3.3 N push is the same as
a single 11.5 N push

17. net pull is 11.5 N to the right.
Find the net pull.
8.2 N
3.3 N
8.2 N
3.3 N
11.5 N
net pull is 11.5 N to the right.

18. Is the resultant force still going to be 11.5 N (to the right) ?
The magnitude of these vectors have not changed, but their directions have.
8.2 N
3.3 N
Is the resultant force still going to be 11.5 N (to the right) ?

19. No it is not! How much is the resulting force (magnitude), and
which direction does it pull?

20. We can answer this graphically by adding the vectors.
How do we know the magnitude (how long) and the direction of the force?

21. We can answer this graphically by adding the vectors.
How do we know the magnitude (how long) and the direction of the force?

22. 10.3 N
If we have drawn the vectors to scale (using a ruler) then we can use the scale to measure the result.

23. 10.3 N
If we have drawn the vectors to scale (using a ruler) then we can use the scale to measure the result.

24. 10
10.3 N
And we can use a protractor to find the angle.

25. or sometimes we prefer to know the
X and Y components.

26. X = 10.2
or sometimes we prefer to know the
X and Y components.

27. Y = 1.7
X = 10.2
or sometimes we prefer to know the
X and Y components.

28. A person is pulling on a box with a force of 15 N at an 30 degree angle from the ground.
30

29. How much force is pulling the box forward?
What part of that 15 N force is actually being used to pull parallel to the surface?
15 N
30

30. IOW: how much force would be needed to hold the box in place?
30

31. Begin by making a "FREE BODY" diagram.
30

32. A person is pulling on a box with a force of 15 N at an 30 degrees angle from the ground.
Are we done drawing?
30

33. A person is pulling on a box with a force of 15 N at an 30 degrees angle from the ground.
30

34. This is the original vector.
15 N at an 30 degrees

35. This vector can be split up into its horizontal (X) and vertical (Y) components.

36. The horizontal (X) part is the side ways force acting on the object.

37. There is a 13 N force to the right.

38. If this box were sitting on a scale that read 10 N before he started pulling, what will it read now?
30
13 N
10 N

39. How much is the up-pull?
15 N
13 N
30
13 N
10 N

40. 7.5 N

41. net 2.5 N down force on the scale.
Since there is a 7.5 N up force and a gravitational 10 N down force, the result is a
net 2.5 N down force on the scale.
7.5 N
2.5 N
10 N

42. How hard does a zip line cable pull on the trees?
FYI: the pull force on both ends of the cable will always be the same magnitude.

43. We start by collecting some specific information (data)
The trees are 20 m apart.
The man has a weight of
180 pounds ( 800 N )
The cable sags 3 m
We assume the cable's weight is insignificant.

44. What do you think the force on the trees (cable) will be?
The trees are 20 m apart.
The man has a weight of
180 pounds ( 800 N )
The cable sags 3 m

45. Now let's make diagram of our situation.

46. The trees are 20 m apart.
scale
1cm = 1m
The cable sags 3 m

47. scale 1cm = 1m The trees are 20 m apart. The cable sags 3 m 10m 3m
90
17 3m
73
10.44m
17
Bring in the known measurements.
Calculate the hypotenuse.
Calculate the angles.

48. scale
1mm = 10 N
Let's make a FREE Body diagram showing all the force vectors.
First we will draw the down ward forced of 800 N caused by the hanging man.

49. scale
1mm = 10 N
Now lets add the other vector forces caused by the cables. We know that the angle from the horizontal is 17

50. So the vertical (X) component of each cable force vector must be 400 N
Each cable force has an upward component (X)
So the X from both cables together must equal 800 N
Each cable must lift 400 N

51. Now we can measure the cable's force vector.
So the vertical (X) component of each cable force vector must be 400 N
Each cable force has an upward component (X)
So the X from both cables together must equal 800 N
Each cable must lift 400 N

52. 1400 N
1400 N
If we have done everything correctly then this person should have no net force (balanced forces).
How can we check this?
800 N

53. 1400 N
1400 N
I suppose theoretically that if we add all the force vectors together we should end up with ZERO
800 N

54. 1400 N
1400 N
I suppose theoretically that if we add all the force vectors together we should end up with ZERO
800 N

55. True story: once a friend of mine wanted to barrow my zip line equipment. I reluctantly agreed, knowing that it can be dangerous if not set up correctly.

56. Well he did not set the zip line high enough, so he needed to pull the zip line tighter (less sag). Luckily the rider was only a few feet off the ground with the cable broke loose.
Why did it break?

57. This was his zip line setup: Distance between trees 40 m
Sag in zip line with rider (1 m)

58. The trees are 40 m apart.
The cable sags 1 m
20 m
20 m
1 m

59. The rider has a weight of 180 pounds ( 800 N )
Each cable must lift ½ the weight of 90 pounds. So the vertical vector of the cable force must be 90
90 pounds.
3500 pounds.
After doing some calculation I determined that the force needed (cable tension) to provide 90 pounds of lift on a 40 meter cable with so little sag was over 3500 pounds.
FYI: bouncing on the cable could easily produce double the force.

60. and if the cable only had 1 foot (
and if the cable only had 1 foot (.3 m) of sag then the force on the cable (tension) would become
90 pounds.
6000 pounds.

61. And if the cable only had 1cm (
And if the cable only had 1cm (.01 M) of sag then the force on the cable (tension) would become
90 pounds.
180,000 pounds.

62. how much force would it take to have zero sag?
90 pounds.
 pounds.

63. What is the tension on the rope?
A car is on the side of a hill being held in place by a rope tied to a tree.
What is the tension on the rope?
Mass of car = 1200 kg
angle of incline = 45

64. F = ma F = 1200 kg * 9.8 𝑚/ 𝑠 2 F = 11,760 N Mass of car = 1200 kg
angle of incline = 45
F = ma
F = 1200 kg * 9.8 𝑚/ 𝑠 2
F = 11,760 N ?
Tension ?
Normal
11,760
gravity

65. Mass of car = 1200 kg angle of incline = 45 ? Tension ? Normal 11,760
gravity

66. Mass of car = 1200 kg
angle of incline = 45 ?
Tension ?
Normal
I know that the two upward forces must add together to equal the gravity force.
11,760
gravity

67. and I know the amount of the gravity force
Mass of car = 1200 kg
angle of incline = 45 ?
Tension
I know the direction of the normal force and the direction of the tension force.
and I know the amount of the gravity force ?
Normal
11,760
gravity

68. Mass of car = 1200 kg
angle of incline = 45 ?
Tension ?
Normal
I also know that the two upward forces must add together to equal (balance) the gravity force.
8300
normal
11,760
gravity
8300
tension

69. I am going to check my answer using the Pythagorean's Theorem.
Mass of car = 1200 kg
angle of incline = 45 ?
Tension
I am going to check my answer using the Pythagorean's Theorem. ?
Normal
8300
normal
𝑎 2 + 𝑏 2 = 𝑐 2
11,760
gravity
2𝑎 2 = 𝑐 2
2𝑎 2 =138,297,600
𝑎 =8315
8300
tension

70. What is the tension on the rope?
A car is on the side of a hill being held in place by a rope tied to a tree.
What is the tension on the rope?
Weight of car = 9000 N
angle of incline = 20
20

71. Weight of car = 9000 kN
angle of incline = 20
9000
gravity

72. Weight of car = 9000 kN angle of incline = 20 ? Tension ? Normal 7780
gravity
4500
tension

73. lets see check our results the math way.
60
9000
gravity
4500
normal ?
Tension ?
Normal
90
30
7794
normal
7780
normal
Using a triangle solver on my phone
9000
gravity
4500
tension

74. If the car were sitting on a “truck” scale,
what weight would be indicated?
7780
same as the
normal force ?
Tension ?
Normal
7780
normal
9000
gravity
4500
tension
20

75. to do some practice problems!
Time for you
to do some practice problems!