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

2. 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

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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

13. 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

14. 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

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

16. 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

17. 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

18. 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

19. 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

20. 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

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

22. 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

23. 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