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Free Body Diagrams. Chapter 1 in Text. 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. - PowerPoint PPT Presentation
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04/22/23 Dr. Sasho MacKenzie - HK 376 1
Free Body Diagrams
Chapter 1 in Text
04/22/23 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.
04/22/23 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.
04/22/23 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…
04/22/23 Dr. Sasho MacKenzie - HK 376 5
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.
04/22/23 Dr. Sasho MacKenzie - HK 376 6
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
04/22/23 Dr. Sasho MacKenzie - HK 376 7
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.
04/22/23 Dr. Sasho MacKenzie - HK 376 8
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).
04/22/23 Dr. Sasho MacKenzie - HK 376 9
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
04/22/23 Dr. Sasho MacKenzie - HK 376 10
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
04/22/23 Dr. Sasho MacKenzie - HK 376 11
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
04/22/23 Dr. Sasho MacKenzie - HK 376 12
a) The Right Hand Ring
Fy = may
Givenax = 0ay = 0
Fx = max
04/22/23 Dr. Sasho MacKenzie - HK 376 13
b) The Gymnast
Fx = max
Fy = may
Givenax = 0ay = 0
04/22/23 Dr. Sasho MacKenzie - HK 376 14
c) The Right Hand Cable Givenax = 0ay = 0
Fx = max
Fy = may
04/22/23 Dr. Sasho MacKenzie - HK 376 15
d) Cable and RingGivenax = 0ay = 0
Fx = max
Fy = may
04/22/23 Dr. Sasho MacKenzie - HK 376 16
Down Hill SkierShow the complete free body diagram for the skier and skis system
Y
X
04/22/23 Dr. Sasho MacKenzie - HK 376 17
Y
X
New Reference FrameGivenax < 0 ay = 0
Fx = max
Fy = may
04/22/23 Dr. Sasho MacKenzie - HK 376 18
Pushing Book on TableDraw the complete FBD of a) The bookb) The table
X
Y
Givenax > 0ay = 0
04/22/23 Dr. Sasho MacKenzie - HK 376 19
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
04/22/23 Dr. Sasho MacKenzie - HK 376 20
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
04/22/23 Dr. Sasho MacKenzie - HK 376 21
Hammer Thrower
X
Y
Draw the complete FBD of a) The athlete ax < 0 ay = 0
b) The hammer ax > 0 ay < 0
04/22/23 Dr. Sasho MacKenzie - HK 376 22
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
04/22/23 Dr. Sasho MacKenzie - HK 376 23
b) The Hammer
mg
Fx1
Fx = max
Fx1 = max > 0Fx1 > 0
Fy = may
– mg = may < 0 0 > mg
Givenax >0ay < 0 X
Y