Free Body Diagrams Chapter 1 in Text 5/25/2019 Dr. Sasho MacKenzie - HK 376

Free-body diagram An essential tool for evaluating every situation in biomechanics. The critical first step in analyzing any biomechanical event. Isolates the “body” (leg, arm, shoe, ball, block etc.) from all other objects. Only shows external forces acting on an object. 5/25/2019 Dr. Sasho MacKenzie - HK 376

Internal Forces Forces that act within the object or system whose motion is being investigated. Newton’s 3rd Law states that forces come in equal and opposite pairs. With internal forces, the forces act on different parts of the same system. These forces cancel each other out, and therefore don’t affect the motion of the system. Forces at the knee if you are looking at the motion of the entire leg. 5/25/2019 Dr. Sasho MacKenzie - HK 376

External Forces Only external forces can change the motion of an object or system. External forces are those forces that act on an object as a result of its interaction with the environment. These include friction, air resistance, gravity, pushing or pulling… 5/25/2019 Dr. Sasho MacKenzie - HK 376

Static Equilibrium The state of a system when all the external forces on that system sum to zero and the system is not moving. In other words, the system is at rest, and has no net force acting on it. 5/25/2019 Dr. Sasho MacKenzie - HK 376

Constructing a Free Body Diagram of a Man pushing a Book on a Desk Step 1 Decide which body or combination of bodies (system) to isolate as the free body diagram Step 2 The body (or system) is isolated by a diagram that represents its complete external boundary 5/25/2019 Dr. Sasho MacKenzie - HK 376

Gravity acts on the CM of an object. Step 3 Represent all external forces acting on the isolated body in their proper positions. Known external forces should be represented by vector arrows with their appropriate magnitude and direction indicated. Gravity acts on the CM of an object. 5/25/2019 Dr. Sasho MacKenzie - HK 376

Friction force will always oppose the direction of motion (velocity). Step 4 The coordinate axis should be shown on the diagram indicating positive and negative directions. Note The acceleration of the body will always be in the direction of the net force. Friction force will always oppose the direction of motion (velocity). 5/25/2019 Dr. Sasho MacKenzie - HK 376

The Man y x Fy1 and Fy2 = upward ground reaction forces Fx1 mg1 Fy1 and Fy2 = upward ground reaction forces mg1 = body weight Fx2 and Fx3 = friction force Fx1 = reaction force of book on subject Fx3 Fx2 Fy1 Fy2 5/25/2019 Dr. Sasho MacKenzie - HK 376

The Book y x Fy3 = upward table reaction force mg2 = book weight Fx1 mg2 Fy3 = upward table reaction force mg2 = book weight Fx4 = friction force Fx1 = push from instructor on book Fy3 Fx4 If book + accelerates Fx1 > Fx4 If book - accelerates Fx1 < Fx4 Zero acceleration Fx1 = Fx4 5/25/2019 Dr. Sasho MacKenzie - HK 376

Gymnast Performing Iron Cross Show the complete free body diagram for The right hand ring The gymnast The cable The cable and ring Static Equilibrium Y X The right hand ring Given ax = 0 ay = 0 5/25/2019 Dr. Sasho MacKenzie - HK 376

a) The Right Hand Ring Fy2 Fx2 Given ax = 0 ay = 0 mg Fx1 Fx = max Fy = may Fy2 – Fy1 – mg = may = 0 Fy2 = Fy1 + mg 5/25/2019 Dr. Sasho MacKenzie - HK 376

b) The Gymnast mg Given ax = 0 ay = 0 Fy1 Fy2 Fx1 Fx2 Fx = max Fy = may Fy2 + Fy1 – mg = may = 0 Fy2 + Fy1 = mg 5/25/2019 Dr. Sasho MacKenzie - HK 376

c) The Cable Fy1 Given ax = 0 ay = 0 Fx1 mg Fx2 Fx = max Fy = may Fy1 – Fy2 – mg = may = 0 Fy1 = Fy2 + mg 5/25/2019 Dr. Sasho MacKenzie - HK 376

d) Cable and Ring Fy1 Given ax = 0 ay = 0 Fx1 Fx = max mg Fx2 Fy = may Fy1 – Fy2 – mg = may = 0 Fy1 = Fy2 + mg Fy2 5/25/2019 Dr. Sasho MacKenzie - HK 376

Down Hill Skier Show the complete free body diagram for the skier and skis system Y X 5/25/2019 Dr. Sasho MacKenzie - HK 376

Y X New Reference Frame mgy mgx Fx2 Given ax < 0 ay = 0 Fy1 Fx1 Fx = max Fx1 + Fx2 – mgx = max < 0 Fx1 + Fx2 < mgx mgy mgx mg Fy = may Fy1 – mgy = may = 0 Fy1 = mgy 5/25/2019 Dr. Sasho MacKenzie - HK 376

Pushing Book on Table Draw the complete FBD of The book The table Given ax > 0 ay = 0 X Y 5/25/2019 Dr. Sasho MacKenzie - HK 376

a) The Book Y Given ax > 0 Fx2 ay = 0 X mg Fx1 Fy1 Fx = max Fx2 – Fx1 = max > 0 Fx2 > Fx1 Fy = may Fy1 – mg = may = 0 Fy1 = mg 5/25/2019 Dr. Sasho MacKenzie - HK 376

b) The Table Fy1 Y Given Fx1 ax = 0 ay = 0 X mg Fx = max Fx1 – Fx2 – Fx3 = max = 0 Fx1 = Fx2 + Fx3 Fx2 Fx3 Fy2 Fy3 Fy = may Fy2 + Fy3 – Fy1 – mg = may = 0 Fy2 + Fy3 = Fy1 + mg 5/25/2019 Dr. Sasho MacKenzie - HK 376

Hammer Thrower Draw the complete FBD of The athlete ax < 0 ay = 0 The hammer ax > 0 ay < 0 X Y 5/25/2019 Dr. Sasho MacKenzie - HK 376

a) The Athlete Y Given ax < 0 ay = 0 Fx1 X Fx = max Fx2 – Fx1 = max < 0 Fx2 < Fx1 mg Fy = may Fy1– mg = may = 0 Fy1 = mg Fx2 Fy1 5/25/2019 Dr. Sasho MacKenzie - HK 376

b) The Hammer Y Given ax >0 ay < 0 Fx1 X mg Fx = max Fy = may – mg = may < 0 0 > mg 5/25/2019 Dr. Sasho MacKenzie - HK 376