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)? 12-21-11 / 12-23-11

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

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

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

Free Body Diagrams Free Body Diagram Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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.

Free Body Diagram Components Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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

Free Body Diagram Components Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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

Moment Review Moment (M) = Force (F) x distance (d) Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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

Free Body Diagram Procedure Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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.

Free Body Diagram Procedure Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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.

Free Body Diagram Procedure Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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.

Free Body Diagram Procedure Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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

Free Body Diagram Procedure Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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

Free Body Diagram Procedure Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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

Free Body Diagram Practice Free Body Diagrams Free Body Diagram Practice Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Create a FBD for the children's sled pictured below. Fapp Fapp θ θ Ff W Ff W FN FN

Free Body Diagram Practice Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Create a FBD for the refrigerator pictured below. Fapp Fapp W W FN Ff θ FN Ff θ

Free Body Diagram Practice Free Body Diagrams Free Body Diagram Practice Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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.

Free Body Diagram Reactions Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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.

Cable Support Cable, rope, chain – Replace with a tension force only. Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Cable Support Cable, rope, chain – Replace with a tension force only.

Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Cable Support A sign with weight W is hung by two cables as shown. Draw the FBD of the sign and cables.

Cable Support FBD of sign and cables Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Cable Support FBD of sign and cables

Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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

Free Body Diagrams Principles of EngineeringTM Unit 4 – Lesson 4.1 - Statics Roller Support Roller – Replaced with ONE reaction force, perpendicular to surface A A Joint / Roller A Y Direction RFAY Force Reaction

Common Support Reactions Free Body Diagrams Common Support Reactions Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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

Free Body Diagrams Built-in End Support Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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

Summary Support Reactions Free Body Diagrams Summary Support Reactions Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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

Free Body Diagrams Truss Bridge FBD Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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

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 4.1 - Statics FBD of the entire truss bridge B C D E 500 lbf RFAX RFAY RFCY

STOP STOP Next - Moments

DO NOW WELCOME Create Free Body Diagrams for the three systems shown 1 2 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

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

STOP STOP Next - Moments

FBD - TRUSSES

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

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

+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

+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.

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

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

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

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

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.

STOP STOP Next - Moments

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

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.

Equilibrium Rotational equilibrium: Free Body Diagrams Equilibrium Principles of EngineeringTM Unit 4 – Lesson 4.1 - 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

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!

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

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.

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

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.

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

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

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

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.

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.

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

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

STOP STOP

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

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

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

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

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

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.

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’ -10’ RBY=80Lbs

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= 20Lbs RAX=0 There are no X forces acting against the pin

STOP STOP Next -TRUSSES