1. FORCES AND FREE BODY DIAGRAMS
AIM:
How can we consolidate concepts learned about FBD’s and apply this knowledge to solve problems in statics (link to physics)? /

2. FOCUS: Draw/create force vector diagrams
Draw/create free body diagrams
Calculate direction and magnitude
Identify sense of a vector
Identify connection to physics
Applications?
RECALL
Newton’s Laws
Types of Forces
Convention – Direction of forces

3. Types of forces Push or pull Tension Reaction or Normal - support
Gravity
Friction

4. Statics – Directions Chart - Forces+Y -X -Y +X

5. Free Body Diagrams
Free Body Diagram
Principles of EngineeringTM
Unit 4 – Lesson Statics
Visual representation of force and object interactions.
Individual objects or members are isolated from their environment or system, illustrating all external forces acting upon it
Every object exists surrounded by other objects, perhaps resting against something, perhaps moving through its surroundings. While an object is in a given environment, it interacts with other objects around it, exerting forces on those objects, and having forces exerted on it.
A free body diagram isolates an object from its environment, or system, and symbolically examines all of the forces acting on the object.
This allows engineers to focus on the forces acting upon the object. From the free body diagram, engineers develop equations which can describe equilibrium, motion, momentum, strength, and many other physical properties.

6. Free Body Diagram Components
Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson Statics
Free Body Diagram Components
Force
A straight line push or pull acting upon an object
Vector quantity has direction and magnitude
Direction is illustrated by arrowhead
Magnitude is illustrated by length of line segment

7. Free Body Diagram Components
Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson Statics
Free Body Diagram Components
Moment
The twisting effort about a point or axis when a force is applied at a distance
Arc with an arrowhead acting about a point indicating direction of CW or CCW

8. Moment Review Moment (M) = Force (F) x distance (d)
Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson Statics
Moment Review
Moment (M) =
Force (F) x distance (d)
Distance (d) is called the moment arm. It must be measured perpendicular to the line of action of the force.
Point of Rotation F d
Line of Action

9. Free Body Diagram Procedure
Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson Statics
Free Body Diagram Procedure
A stack of three books, each weighing 5 lb, is sitting on top of a table. Draw the Free Body Diagram (FBD) of the top book.

10. Free Body Diagram Procedure
Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson Statics
Free Body Diagram Procedure
1. Sketch the isolated object.
What is the isolated object?
Top Book
Free body diagrams should be drawn completely isolated from all other objects around it. We “free” the body from its system. Instead of taking time to draw the object in detail, we substitute a simple geometric shape, such as a square or a circle. For these exercises the shape we draw is considered dimensionless.

11. Free Body Diagram Procedure
Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson Statics
Free Body Diagram Procedure
2. Sketch the applied and norm forces.
When an object is in contact with and is supported by a second object, the second object can be replaced with a normal force which is perpendicular to the surface of the second object.
Free body diagrams should be drawn completely isolated from all other objects around it. We “free” the body from its system. Instead of taking time to draw the object in detail, we substitute a simple geometric shape, such as a square or a circle. For these exercises the shape we draw is considered dimensionless.

12. Free Body Diagram Procedure
Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson Statics
Free Body Diagram Procedure
2. Sketch the applied force and norm forces.
Applied Force – Weight of top book
Free body diagrams should be drawn completely isolated from all other objects around it. We “free” the body from its system. Instead of taking time to draw the object in detail, we substitute a simple geometric shape, such as a square or a circle. For these exercises the shape we draw is considered dimensionless.
Norm Force – Reaction force pushing up on the book, causing it not to fall

13. Free Body Diagram Procedure
Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson Statics
Free Body Diagram Procedure
3. Label objects and forces.
W=5 lbf
PLTW – DE book
Free body diagrams should be drawn completely isolated from all other objects around it. We “free” the body from its system. Instead of taking time to draw the object in detail, we substitute a simple geometric shape, such as a square or a circle. For these exercises the shape we draw is considered dimensionless.
N=5 lbf

14. Free Body Diagram Procedure
Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson Statics
Free Body Diagram Procedure
4. Label dimensions.
For more complex free body diagrams, proper dimensioning is required, including length, height, and angles.
W=5 lbf
45°
8 ft
10 ft
38.6°
PLTW – DE book
Free body diagrams should be drawn completely isolated from all other objects around it. We “free” the body from its system. Instead of taking time to draw the object in detail, we substitute a simple geometric shape, such as a square or a circle. For these exercises the shape we draw is considered dimensionless.
N=5 lbf

15. Free Body Diagram Practice
Free Body Diagrams
Free Body Diagram Practice
Principles of EngineeringTM
Unit 4 – Lesson Statics
Create a FBD for the children's sled pictured below.
Fapp
Fapp θ θ Ff W Ff W FN FN

16. Free Body Diagram Practice
Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson Statics
Create a FBD for the refrigerator pictured below.
Fapp
Fapp W W FN Ff θ FN Ff θ

17. Free Body Diagram Practice
Free Body Diagrams
Free Body Diagram Practice
Principles of EngineeringTM
Unit 4 – Lesson Statics
Create a FBD for the pulley system pictured below.
Ftens
FBD of M1:
WM1
Ftens
Ftens
FBD of the movable pulley:
Free body diagrams are frequently used while solving pulley system problems. It is useful to isolate the parts of a pulley system to discover relationships between the parts. This pulley system consists of an upper fixed pulley, which is supporting block M1, and a lower movable pulley. The movable pulley is supporting block M2. A single rope or cord connects these pulleys. Both pulleys are considered mass-less and frictionless.
A free body diagram of M1 would start with a small shape. Two forces are acting on M1: The weight of M1 and the tension force of the rope.
We can also isolate the pulleys. Looking at the lower movable pulley, we draw the free body. The weight of the block M2 is pulling down on the pulley. Two strands of the rope are pulling up on the pulley, and each of the strands is exerting a tension force.
The tension force in the rope is equal in each free body diagram within the system since there is a single tension force in the entire rope.
WM2 + W pulley M2 M1
Tension Forces (Ftens ) are equal throughout the system.

18. Free Body Diagram Reactions
Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson Statics
Free Body Diagram Reactions
Different types of support reactions:
Cable, rope, or chain
Pin
Roller
Built-in end – Cantilever
To aid in completing free body diagrams, connections are often identified with letters.

19. Cable Support Cable, rope, chain – Replace with a tension force only.
Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson Statics
Cable Support
Cable, rope, chain – Replace with a tension force only.

20. Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson Statics
Cable Support
A sign with weight W is hung by two cables as shown. Draw the FBD of the sign and cables.

21. Cable Support FBD of sign and cables Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson Statics
Cable Support
FBD of sign and cables

22. Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson Statics
Pin Support
Pin – Replaced with TWO reaction forces, one vertical (y) and one horizontal (x).
Force
Joint / Pin A
Reaction
X Direction A A
RFAX
Joint / Pin A
Y Direction
RFAY
Force
Reaction

23. Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson Statics
Roller Support
Roller – Replaced with ONE reaction force, perpendicular to surface A A
Joint / Roller A
Y Direction
RFAY
Force
Reaction

24. Common Support Reactions
Free Body Diagrams
Common Support Reactions
Principles of EngineeringTM
Unit 4 – Lesson Statics
Beams and truss bridges are usually supported with one pin support and one roller support. This is called a simply supported object.
Create a FBD for the simply supported beam A B
RFAX
RFAY
RFBY

25. Free Body Diagrams
Built-in End Support
Principles of EngineeringTM
Unit 4 – Lesson Statics
Built-in-end (cantilever) – Replaced with TWO forces: one horizontal and one vertical, and ONE moment
Create a FBD for the built-in-end cantilever. A
MAccw
RFAX
RFAY

26. Summary Support Reactions
Free Body Diagrams
Summary Support Reactions
Principles of EngineeringTM
Unit 4 – Lesson Statics
Contact – Replace with a normal force
Cable, rope, chain – Replace with tension force
Pin – Replace with two reaction forces; one vertical and one horizontal
Roller – Replace with one reaction force perpendicular to surface
Built-in end (cantilever) – Replace with one horizontal force, one vertical force, and one moment

27. Free Body Diagrams
Truss Bridge FBD
Principles of EngineeringTM
Unit 4 – Lesson Statics
Supported with a pin at one end and a roller at the other
Draw the FBD of the entire truss bridge. A B C D E
500 lb

28. Truss Bridge FBD FBD of the entire truss bridge B C D E 500 lbf RFAX
Free Body Diagrams
Truss Bridge FBD
Principles of EngineeringTM
Unit 4 – Lesson Statics
FBD of the entire truss bridge B C D E
500 lbf
RFAX
RFAY
RFCY

29. STOP
STOP
Next - Moments

30. DO NOW WELCOME Create Free Body Diagrams for the three systems shown 12
Ftens
PHYSICS θ
Fapp
ARCHITECTURE
WSIGN
This is an opportunity for you to create a free body diagram. This system #2 consists of a bookshelf and two books. The blue physics book is leaning against the side of the shelf. The pink architecture book is leaning against the physics book. Your job is to create a free body diagram of the physics ** book.
WELCOME

31. Pulley Systems 3. FBD of M1: FBD of the movable pulley:
Tension Forces (Ftens ) are equal throughout the system.

32. STOP
STOP
Next - Moments

33. FBD - TRUSSES

34. Statics – Forces - Math C A B Let us look at the above example
100lbs 5’ C A
10’ B
50lbs
Let us look at the above example
What is not known?
The force that A is exerting.
What do you do first?
Free Body Diagram

35. Statics – Math Forces Free Body Diagram+X -X -Y
Statics – Math Forces
100 lbs 5’
RAY
RAX
10’
50lbs
Free Body Diagram
Please note that distances are included
Next: math

36. +Y +X -X -Y
Statics – Math Forces
100 lbs 5’
RAY
RAX
10’
50lbs
Remember that Newton’s Law in a static object is F=0
0=RAY+RBY+FCY
0=RAY+50Lbs+-100Lbs
RAY=50Lbs

37. +Y +X -X -Y
Statics – Math Forces
100 lbs 5’
RAY
RAX
10’
50lbs
RAX=0 (There are no X forces acting against the pin)
This is an unusual problem because usually we do not know any of the force values acting against the load.

38. Statics – Forces - Math C A B This is a more realistic problem.8’
100lbs C A
10’ B
This is a more realistic problem.
What is not known?
The force that A and B is exerting.
What do you do first?
Free Body Diagram

39. Statics – Math Forces Free Body Diagram
100 lbs 8’
RAY
RAX
10’
RBY
Free Body Diagram
Please note that distances are included
Next: math

40. FBD - TRUSSES
Figuring out the reactions
Lets do some math!

41. Statics - math Remember that Newton’s Law in a static object is F=0
0=RAY+RBY+FCY
0=RAY+RBY+-100Lbs
UNSOLVABLE!!!!!
RAX=0 There are no X forces acting against the pin

42. Statics - math 0=RAY+RBY+-100Lbs
We need to solve for either RAY or RBY
To solve for one of these we need to find the moments at one of these points.

43. STOP
STOP
Next - Moments

44. MOMENTS
AIM:
How can we determine the property which cause a body to rotate (MOMENT of A FORCE) , how it is identified and how its magnitude is calculated.
RECALL
Types of forces
Application on FBD’s
Newton’s Law in a static object F=0

45. Statics - Moments Definition: Moments are just like forces. M=0
The object’s tendency to rotate around a point.
Moments are just like forces.
Since the object is not moving there are moments that are acting against each other to keep the object from not moving
You need to add all of them up since the object is not moving they are equal to 0
M=0
The moments are used to find a force at a specific point
How do we find the moments? Lets check it out.

46. Equilibrium Rotational equilibrium:
Free Body Diagrams
Equilibrium
Principles of EngineeringTM
Unit 4 – Lesson Statics
Rotational equilibrium:
The state in which the sum of all the clockwise moments equals the sum of all the counterclockwise moments about a pivot point
Remember
Moment = F x D
©iStockphoto.com

47. Moments
A moment of a force is a measure of its tendency to cause a body to rotate about a point or axis.
It is the same as torque.
A moment (M) is calculated using the formula:
Moment = Force * Distance
M = F * D
Always use the
perpendicular
distance
between the force
and the point!

48. What distance would you use here?
Moments
What distance would you use here?
Moment = Force * Distance
M = F * D
F = 20 lb
D = 2 ft
Find the moment about point P: P F
2 ft
M = F * D
M = (20 lb) * (2 ft)
M = 40 ft-lb

49. Moments F P Moment = Force * Distance M = F * D
Can you visualize the result of
the force acting on the beam? P F
The beam has a tendency to rotate clockwise about point P.

50. Moments P F Moment = Force * Distance M = F * D F = 20 lb D = 0.5 ft
Find the moment about point P: P F
2 ft
0.5 ft
M = F * D
M = (20 lb) * (0.5 ft)
M = 10 ft-lb

51. Moments P F Moment = Force * Distance M = F * D
Can you visualize the result of
the force acting on the beam?
Moment = Force * Distance
M = F * D P F
The beam has a tendency to rotate clockwise about point P.

52. Moments
You have seen how a see-saw will balance on its fulcrum if equal weight is applied at both ends.
If two people of unequal weight are on the see-saw, it will rotate about its fulcrum to the side with the heavier person.
Fulcrum

53. Moments
A see-saw can also be balanced by moving the smaller person further from the fulcrum.
If a 45-pound child sat on one end 3 ft from the fulcrum, then how far from the fulcrum would a 30-pound child have to sit in order for the see-saw to be balanced?
3 ft
45 lbs
30 lbs x

54. Moments M1 = M2 45 lbs * 3 ft = 30 lbs * x 135 ft-lbs = 30 lbs * x
4.5 ft = x
3 ft
45 lbs
30 lbs x

55. Moments Typically it is assumed:
A moment with a tendency to rotate counter clockwise (CCW) is considered to be a positive moment.
A moment with a tendency to rotate clockwise (CW) is considered to be a negative moment.

56. Statics – Directions Chart -Moments_ +
If the moment is going in this direction (Clockwise) it is considered to be negative
If the moment is going in this direction (Counterclockwise) it is considered to be positive
These are given to indicate direction and do not actually mean that value is negative. This convention is used because of the right hand rule.

57. Statics – Directions Chart –Moments and Forces+Y + -X +X _ -Y

58. Statics - Moments CONCLUSION Moments are just like forces. M=0
The object’s tendency to rotate around a point.
Moments are just like forces.
Since the object is not moving there are moments that are acting against each other to keep the object from not moving
You need to add all of them up since the object is not moving they are equal to 0
M=0
The moments are used to find a force at a specific point

59. STOP
STOP

60. Perform Activity 2.1.5 –Calculating Moments.
CLASS ASSIGNMENT
Perform Activity –Calculating Moments.

61. Now we can SOLVE THE PROBLEM! Figuring out the reactions
Lets do some math!

62. Statics – Forces - Math C A B This is a more realistic problem.8’
100lbs C A
10’ B
This is a more realistic problem.
What is not known?
The force that A and B is exerting.
What do you do first?
Free Body Diagram

63. Statics – Math Forces Free Body Diagram
100 lbs 8’
RAY
RAX
10’
RBY
Free Body Diagram
Please note that distances are included
Next: math

64. Statics - math Remember that Newton’s Law in a static object is F=0
0=RAY+RBY+FCY
0=RAY+RBY+-100Lbs
UNSOLVABLE!!!!!
RAX=0 There are no X forces acting against the pin

65. Statics - math 0=RAY+RBY+-100Lbs
We need to solve for either RAY or RBY
To solve for one of these we need to find the moments at one of these points.

66. Statics – Math Forces Now the formula should read+Y +X -X -Y + _
100 lbs 8’
RAY
RAX
10’
RBY
Now the formula should read
MA=(DAB*RBY)-(DAC*FC)
MA=(10’*RBY)-(8’*100Lbs)
Remember that the sum of the moments = 0
0=(10’*RBY)-(8’*100Lbs)
0=(10’*RBY)-800FtLbs
-800FtLbs=-(10’*RBY)
-10’ ’
RBY=80Lbs

67. Statics - math Let us go back to the original problem+Y +X -X -Y + _
Statics - math
Let us go back to the original problem
Remember that Newton’s Law in a static object is F=0
0= RAY+RBY+FCY
Now we have RBY and can solve for the problem
0= RAY+80Lbs-100Lbs
RAY= Lbs
RAX=0 There are no X forces acting against the pin

68. STOP
STOP
Next -TRUSSES