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Free Body Diagrams

Chapter 1 in Text

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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.

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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.

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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…

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Static Equilibrium

• The state of an 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.

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Constructing a Free Body Diagram of a Man pushing a Book on a

DeskStep 1Decide which body or combination of bodies (system) to isolate as the free body diagram

Step 2The body (or system) is isolated by a diagram that represents its complete external boundary

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Step 3Represent 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.

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Step 4The coordinate axis should be shown on the diagram indicating positive and negative directions.

NoteThe acceleration of the body will always be in the direction of the net force.

Friction force will always oppose the direction of motion (velocity).

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The Man

Fx4

Fx2

Fy3Fy2

Fy1

Fx3

Fy2 and Fy3 = upward ground reaction forces

Fy1 = body weight

Fx2 and Fx3 = friction force

Fx4 = reaction force of book on subject

x

y

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The Book

Fx4 Fy1

Fy2 Fx2

Fy2 = upward table reaction force

Fy1 = book weight

Fx2 = friction force

Fx4 = push from instructor on book

If book + accelerates Fx4 > Fx2

If book - accelerates Fx4 < Fx2

Zero acceleration Fx4 = Fx2

x

y

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Gymnast Performing Iron CrossShow the complete free body diagram for a) The right hand ringb) The gymnastc) The cabled) The cable and ring

Y

X Givenax = 0ay = 0

Static Equilibrium

The right hand ring

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a) The Right Hand Ring

Fy = may

Givenax = 0ay = 0

Fx = max

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b) The Gymnast

Fx = max

Fy = may

Givenax = 0ay = 0

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c) The Right Hand Cable Givenax = 0ay = 0

Fx = max

Fy = may

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d) Cable and RingGivenax = 0ay = 0

Fx = max

Fy = may

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Down Hill SkierShow the complete free body diagram for the skier and skis system

Y

X

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Y

X

New Reference FrameGivenax < 0 ay = 0

Fx = max

Fy = may

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Pushing Book on TableDraw the complete FBD of a) The bookb) The table

X

Y

Givenax > 0ay = 0

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a) The Book

mgFy1 Fx1

Fx2

Fx = max

Fx2 – Fx1 = max > 0Fx2 > Fx1

Fy = may

Fy1 – mg = may = 0Fy1 = mg

Givenax > 0ay = 0 X

Y

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b) The TableFy1

Fx1

mg

Fy2 Fy3

Fx2 Fx3

Fx = max

Fx1 – Fx2 – Fx3 = max = 0Fx1 = Fx2 + Fx3

Fy = may

Fy2 + Fy3 – Fy1 – mg = may = 0Fy2 + Fy3 = Fy1 + mg

Givenax = 0ay = 0 X

Y

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Hammer Thrower

X

Y

Draw the complete FBD of a) The athlete ax < 0 ay = 0

b) The hammer ax > 0 ay < 0

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a) The Athlete

mg

Fx1

Fx2

Fy1

Fx = max

Fx2 – Fx1 = max < 0Fx2 < Fx1

Fy = may

Fy1– mg = may = 0Fy1 = mg

Givenax < 0ay = 0 X

Y

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b) The Hammer

mg

Fx1

Fx = max

Fx1 = max > 0Fx1 > 0

Fy = may

– mg = may < 0 0 > mg

Givenax >0ay < 0 X

Y