agonist and antagonist relationship l.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Agonist and Antagonist Relationship PowerPoint Presentation
Download Presentation
Agonist and Antagonist Relationship

Loading in 2 Seconds...

play fullscreen
1 / 28

Agonist and Antagonist Relationship - PowerPoint PPT Presentation


  • 385 Views
  • Uploaded on

Agonist and Antagonist Relationship. Agonist – is a muscle described as being primarily responsible for a specific joint movement while contracting Antagonist – is a muscle that counteracts or opposes the contraction of another muscle Simply, these are relative terms describing “opposites” .

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Agonist and Antagonist Relationship' - omer


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
agonist and antagonist relationship
Agonist and Antagonist Relationship
  • Agonist – is a muscle described as being primarily responsible for a specific joint movement while contracting
  • Antagonist – is a muscle that counteracts or opposes the contraction of another muscle
  • Simply, these are relative terms describing “opposites”
slide2
If an agonist muscle is considered a concentric contractor for a movement then the antagonist muscle is the eccentric contractor for the same movement.
  • Generally, concentric and eccentric contractions do not occur at the same time for a given movement.
slide3
What determines which one is working is the purpose of movement, acceleration (speeding-up) or deceleration (slowing-down).
  • Examples
steps to determine contraction type
Steps to determine contraction type
  • Identify the joint movement
  • Identify the agonist (concentric contractor) and antagonist (eccentric contractor) for the joint movement
  • Determine if the movement is speeding-up (accelerating) or slowing-down (decelerating)
    • If speeding-up then agonist working concentrically
    • If slowing-down then antagonist working eccentrically
elbow and radioulnar joint movements
Elbow and Radioulnar Joint Movements
  • elbow - flexion and extension
  • Radioulnar (forearm) - pronation and supination
slide6

Biceps Brachii*O: Long Head – supraglenoid tubercle above the superior lip of glenoid fossaShort Head – coracoid process and upper lip of glenoid fossaI: Tuberosity of radius and bicipital aponeurosisA: Flexion of elbow, supination of forearm (radioulnar), weak flexion of shoulder, and weak abduction of shoulder

slide7
BrachialisO: Distal ½ anterior shaft of humerusI: Coronoid process of ulnaA: True flexion of the elbow
slide8

BrachioradialisO: Distal 2/3 of lateral condyloid (supracondyloid) ridge of humerusI: Lateral surface distal end of radius at the styloid processA: Flexion of elbow, pronation from supinated position to neutral (thumb up), supination from pronated position to neutral

slide9

Triceps brachiiO: Long head – infraglenoid tubercle below inferior lip of glenoid fossa of scapulaLateral head – upper ½ posterior surface of humerusMedial head – distal 2/3 of posterior surface of humerusI: Olecranon process of ulnaA: All heads: extension of elbow Long head: extension, adduction, and horizontal abduction of shoulder

slide10

AnconeusO: Posterior surface of lateral condyle of humerusI: Posterior surface of olecranon process and proximal ¼ of ulnaA: extension of elbow

slide11

Pronator teresO: Distal part of medial condyloid ridge of humerus and medial side of proximal ulnaI: Middle third of lateral surface of radiusA: Pronation of forearm (radioulnar) and weak flexion of elbow

slide12
Pronator quadratusO: Distal fourth anterior side of ulnaI: Distal fourth anterior side of radiusA: Pronation of forearm
slide13

SupinatorO: Lateral epicondyle of humerus and neighboring posterior part of ulnaI: Lateral surface of proximal radius just below the headA: Supination of forearm

ligaments of the elbow
Ligaments of the Elbow
  • Radial collateral ligament – provides lateral stability of the elbow; resists lateral displacement of elbow
  • Ulnar collateral ligament – provides medial stability of the elbow; resists medial displacement of elbow
  • Annular ligament – stabilizes the head of radius to the ulna and allows smooth articulation with the ulna
introduction to linear kinetics
Introduction to Linear Kinetics
  • Linear Kinetics – the study of linear forces associated with motion (ex. force, momentum, inertia).
  • Linear Kinematics – the study of linear motion.
  • Force = mass x acceleration

Force → Acceleration → Velocity → Displacement

slide20
Vector – is a quantity that has both magnitude (how much) and direction.
  • Used as a measuring tool for linear variables which have both magnitude and direction.
  • Illustrated by an arrow where the tip represents direction and the length representing magnitude.
slide21
Muscle Force – can be measured with vectors, since muscle force pulls on bone in a linear fashion.
  • Vector composition – is a process of determining a single vector (usually called resultant) from two or more vectors.
slide22
Vectors can typically be analyzed as having horizontal (x) and vertical (y) components.
  • In this case, these perpendicular component vectors can be used to form a right triangle.
  • A common trigonometric principle used is the Pythagorean theorem, where A2 + B2 = C2.

C

θ

A

θ

B

slide23
Furthermore, the following equations are derived from the Pythagorean theorem:

sin θ = opposite / hypotenuse

cos θ = adjacent / hypotenuse

tan θ = opposite / adjacent

C

θ

A

θ

B

sample problem
Sample Problem

If the muscle force generated by the biceps brachii is 20 lbs, how much rotary (y) force is generated by the muscle? How much dislocating (x) force?

slide25
Known: angle of pull = 45 degrees

Muscle force (resultant) = 20 lbs

Unknown: rotary (y) force

dislocating (x) force

slide26
Rotary force (y) calculation:

sin θ = opposite / hypotenuse

sin 45 = rotary (y) / 20 lbs

sin 45 x 20 lbs = rotary (y)

rotary (y) = 14.14 lbs

slide27
Dislocating force calculation:

cos θ = adjacent / hypotenuse

cos 45 = dislocating (x) / 20 lbs

cos 45 x 20 lbs = dislocating (x)

dislocating = 14.14 lbs

pythagorean check
Pythagorean Check

A2 + B2 = C2

(rotary force)2 + (dislocating force)2 = (muscle force)2

(14.14 lbs)2 + (14.14 lbs)2 = (muscle force)2

399.88 lbs = (muscle force)2

√ 399.88 = muscle force

20 lbs = muscle force