1. Research & Revise Activity 2
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Activity 2: Muscles and muscle groups.
Which muscle group must be active?
Multiple muscle problem
Maximal muscle torque
Muscle torque vs. arm position
Muscle force-length effect on torque
These activities are designed to examine more completely the results from activity one.
ELIMINATE STABILIZING VS DISLOCATING MUSCLE FORCES (Pandy, 2002)
***********done (Tennant, 2002)

2. 1. Shoulder Abductors 2. Shoulder Adductors
Research & Revise PRS Q1
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Activity 2 –muscles and muscle groups: Which muscle group must be active at the shoulder when performing the iron cross?
1. Shoulder
Abductors
This question is used as a pretest to probe the students’ preconceptions. To answer this question they need to know two things: 1) that muscles can only develop tension, and 2) that the sum of moments about the shoulder joint must equal zero to maintain equilibrium. If they can mentally rotate the right arm to the horizontal iron cross position, they will realize that the external force on the ring tends to rotate the arm clockwise about the right shoulder joint. Therefore the muscle torque must be directed in a counterclockwise direction. Since muscles can only generate tension, the shoulder abductors, if active, would produce a clockwise torque while the adductors would produce the needed counterclockwise torque.
SEE IF WE CAN SUPPLEMENT WITH LITERATURE READINGS (ESPECIALLY ANATOMY BOOKS) SO THAT STUDENTS CAN GO AND LOOK
UP THESE MUSCLES AND LEARN ABOUT THEIR ANATOMY BY THEMSELVES (Pandy, 2002)
2. Shoulder
Adductors

3. Research & Revise Activity 2 – PRS Q2
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Activity 2 – Isolate the total arm+ring (arm-forearm-hand-ring combination) and draw a free body diagram. Neglect the weight of the total arm, lump the muscles into a single resultant force, and assume the shoulder joint is a frictionless hinge joint.
Compare your FBD with these:
The students should use the method of sections to isolate the total arm-ring combination by making an imaginary slice through the shoulder joint and all of the shoulder muscles, and making another slice through the cable attached to the ring. Ask the students to draw a single resultant muscle force for all of the muscles shown. The purpose of drawing the FBD is to emphasize which muscle group must be active if equilibrium is to be achieved.
LET’S NOT DO THE PRS….SHOW OF HANDS OR HAVE STUDENTS WRITE DOWN THEIR ANSWERS (Pandy, 2002)
****************done (Tennant. 2002)

4. Research & Revise PRS Q2
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Which of the following is a correct FBD of a gymnast’s arm + jig in holding the iron cross position? Assume the shoulder joint is frictionless and neglect the weight of the arm and the jig. T=Tension of the rope, M= Muscle force, J= Joint Reaction Force. T
(A)
(B) T M X M X J J
Instructors note:
The choices include several common solutions students have generated in the past. Polling students’ responses will help diagnose the difficulties your students may currently posses.
Answer 1 has no net moment about the joint, but the muscle force is in the wrong direction, since muscles can only develop active tension.
Answer 2 has a net clockwise moment, so cannot be in equilibrium.
Answer 4 would provide a net clockwise moment and implies that the muscle can generate a compressive force – both incorrect concepts.
Answer 3 is correct, indicating that the adductor muscle group must be active if the gymnast is in equilibrium. T T
(C)
(D) J J X X M M

5. Research & Revise PRS Q2
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Which of the following is a correct FBD of a gymnast’s arm + jig in holding the iron cross position? Assume the shoulder joint is frictionless and neglect the weight of the arm and the jig. T=Tension of the rope, M= Muscle force, J= Joint Reaction Force.
Since the muscles act at an angle and pull away from the shoulder joint, the correct answer is: T
(C) J X M

6. 1. Shoulder Abductors 2. Shoulder Adductors Research & Revise PRS Q3
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Activity 2 –muscles and muscle groups: Repeat - Which muscle group must be active at the shoulder when performing the iron cross?
1. Shoulder
Abductors
This is simply a repeat of PRS Q1 to see if students now understand that the adductor group must be active at the shoulder. Answer 2 is correct.
2. Shoulder
Adductors

7. Research & Revise Muscle Force 1
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Activity 2 – muscles and muscle groups
Assume that the two muscles marked below are the only active muscles at the shoulder. Can you find the force developed by each muscle to maintain equilibrium? (Consider the shoulder to be a frictionless hinge joint.)
1 Lat Dor
2 Pec Maj
Instructors Note:
The correct approach to this problem would be for the students to draw a FBD of the arm-forearm-hand-ring combination, write out the three equilibrium equations and identify the unknowns in the problem. This is the precursor to exploring statically indeterminate problems in the next part of this activity.

8. Research & Revise Muscle Force 2
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Activity 2 – muscles and muscle groups
How many unknown scalar quantities are in your free body diagram?
How many independent scalar equilibrium equations are applicable for this 2D problem?
1 Lat Dor
2 Pec Maj
They should come up with four unknown scalar quantities: the magnitude of each muscle force (directions are known) and the two components (or magnitude and direction) of the joint reaction force.
There are only three equilibrium equations, so the problem cannot be solved using statics alone. It is “statically indeterminate.”

9. Research & Revise Muscle Force 3
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Activity 2 – muscles and muscle groups
How many unknown scalar quantities are in your free body diagram?
A) B) C) 3. D) 4.
1 Lat Dor
2 Pec Maj
They should come up with four unknown scalar quantities: the magnitude of each muscle force (directions are known) and the two components (or magnitude and direction) of the joint reaction force.
There are only three equilibrium equations, so the problem cannot be solved using statics alone. It is “statically indeterminate.”

10. Research & Revise Muscle Force 3
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Activity 2 – muscles and muscle groups
How many unknown scalar quantities are in your free body diagram?
The unknown scalar quantities are the two muscle forces, so the correct answer is:
B) 2.
1 Lat Dor
2 Pec Maj

11. Research & Revise Muscle Force 4
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Activity 2 – muscles and muscle groups
How many independent scalar equilibrium equations are applicable for this 2D problem?
A) 1 B) C) D) 4
1 Lat Dor
2 Pec Maj
They should come up with four unknown scalar quantities: the magnitude of each muscle force (directions are known) and the two components (or magnitude and direction) of the joint reaction force.
There are only three equilibrium equations, so the problem cannot be solved using statics alone. It is “statically indeterminate.”

12. Research & Revise Muscle Force 4
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Activity 2 – muscles and muscle groups
How many independent scalar equilibrium equations are applicable for this 2D problem?
The only applicable equation in this problem is derived from summing the moments about the shoulder joint, so the correct answer is:
A) 1
1 Lat Dor
2 Pec Maj

13. Research & Revise Muscle Force 5
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Activity 2 – muscles and muscle groups
How would you define a statically indeterminate problem?
Can you think of ways the body might distribute forces between individual muscles to make this a statically determinate problem?
1 Lat Dor
2 Pec Maj
A statically indeterminate problem is one in which the number of independent equilibrium equations is less than the number of scalar unknowns in an equilibrium problem.
Reduction methods attempt to relate the force in one muscle to the force in the other. For instance:
forces are equal: F2 = F1
stresses are equal: F2=F1(A2/A1)
force proportional to number of red muscle fibers
force in one muscle is maximal (e.g., F2 = F2max)
Optimization methods are used to adjust muscle forces so that an objective function phi is minimized. For example:
phi = F1 + F2
phi = F1n + F2n
phi = joint reaction force
phi = F1 + F2 + J
phi = muscle fatigue

14. Research & Revise Maximum Torque
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Activity 2 – muscles and muscle groups
Assume that each of the three adductor muscles shown on the next slide generates its maximum possible force, Fmax,i. Derive expressions for the resultant muscle force, resultant direction, and its moment arm relative to the shoulder joint. Is the resultant muscle torque sufficient to hold a 5’ 10”, 160 lb gymnast in the iron cross position? Neglect weight of the arm.
This is actually an over-determined problem, since the only unknowns are the two components of the joint reaction force, and we have three independent equilibrium equations. The objective here is to see if enough torque can be generated by the shoulder muscles to overcome the torque induced by the cable. Anthropometric data in the next two slides are needed before this question can be answered.
This question is approached by first using anthropometric data to find the moment about the shoulder joint provided by the external force in the cable. Only the vertical component of the cable force (W/2) contributes to the external moment, so the cable angle does not need to be specified. Next, the maximum possible muscle torques are computed for each of the three muscles. The best way to compute these is to first find unit vectors for the lines of action of each muscle, multiply by the maximum muscle force, then take the cross product M= rxF, where r can be the vector from the shoulder (0,0) to the origin or the insertion or any other point along the line of action of each muscle. The sum of muscle torques is then compared with the external torque. One can show that the athlete has enough muscle strength to hold this position.

15. Research & Revise Muscle Resource
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In class:
What is maximum torque provided by Teres Major?
Activity 2 – muscles and muscle groups, resources
Origin & Insertion coordinates for arm in "iron cross" position
Shoulder joint is at (0,0)
Muscle
Insertion (x,y)
(cm)
Origin(x,y)
Maximum Force (N)
Teres Major
(+2.78, 0)
(-11, -14)
1800
Latissimus Dorsi
(+1.5, 0)
(-8, -28)
2700
Pectoralis Major
(+2.06, 0)
(-9, -19)
4000 y
insertion
These data are needed to find muscle forces and directions.
origin

16. Research & Revise Segment Lengths
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Activity 2 – muscles and muscle groups, resources
Segment Lengths
(Expressed as a Percentage of Total Body Height)
Segment Males Females
Head and Neck
Whole Trunk
Thorax
Abdomen
Pelvis
Total Arm
Upper Arm
Forearm
Hand
Forearm and Hand
Total Leg
Thigh
Leg
Foot
These data are needed to find the moment applied by the ring cable.

17. Research & Revise Arm Position 1
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Activity 2 – muscles and muscle groups
What factors influence maximal muscle torque?
Would you expect maximal muscle torque at the shoulder to vary with arm position  ?
Our conclusion from activity one was that the young man could not develop sufficient torque with his shoulder muscles to hold the iron cross position, but he could develop enough torque to hold a modified iron cross position (theta = 50 degrees). We now wish to see if the young man can provide the same maximum torque when his arms are horizontal (90 degrees) as he can when they are at 50 degrees. If so, we could easily design a jig that could reduce the externally applied torque just enough so he would have strength to hold the full iron cross position.

18. Research & Revise Arm Position 1
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Activity 2 – muscles and muscle groups
If maximum muscle force F is independent of , would maximal muscle torque vary with arm position  ?
Why or why not?
However, we will find that even if the young man can exert the same maximum shoulder adductor muscle force when his arms are at 50 degrees as he can when they are at 90 degrees, the muscle torque will be dependent on the angle he holds his arms. This is because the muscle moment arm depends on arm position.

19. How does muscle moment arm d vary with position  ?
Research & Revise
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Activity 2 – muscles and muscle groups
How does muscle moment arm d vary with position  ? s i
Muscle moment arm, d i d i
This shows graphically what happens to moment arm d as arm angle is changed for a single muscle, the latissimus dorsi. Clearly there is a point where the moment arm is a maximum. How can we find this? We will use the latissimus dorsi as a representative shoulder adductor in the following examples.
Muscle Length  o

20. Compute muscle length:
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Research & Revise
Activity 2 – muscles and muscle groups
How does Muscle Length vary with position  ?
The first step in deriving an expression for the moment arm is to find muscle length (lm, in red) as a function of arm angle. The law of cosines is used for this. The x and y coordinates of the insertion point of the latissimus dorsi muscle can then be computed from muscle length and arm angle.
PRESENT THIS FORMULATION DIFFERENTLY; USE DL/DTHETA METHOD SHOWN IN BME 342 (Pandy, 2002) ****DONE (Tennant, 2002)
Compute muscle length:
Note: Shoulder Joint
is at (0,0)

21. How does Muscle Force and Muscle Torque vary with position  ?
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Research & Revise
Activity 2 – muscles and muscle groups
How does Muscle Force and Muscle Torque vary with position  ?
A unit vector along the line of action of the latissimus dorsi muscle is found from the coordinates of its insertion and origin (note: this will not be true for all muscles, since some wrap around anatomical structures). The maximum force vector is found by multiplying the magnitude of the maximum muscle force (from table on a previous slide) by the unit vector in direction of the muscle force. The muscle torque for maximum muscle force can be computed by taking the cross product of the position of the insertion (relative to the shoulder) and the maximal muscle force. Thus the moment arm d is equal to xiuy – yiux. This is plotted on the next slide.
PRESENT THIS FORMULATION DIFFERENTLY (Pandy, 2002) ****DONE (Tennant, 2002)
Moment Arm:
Muscle Torque:

22. Sarcomere Force vs. Length
Research & Revise
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Activity 2 – muscles and muscle groups
Experimental Observations:
Sarcomere Force vs. Length
Muscle Force vs. Length
Experiments clearly show that the active component of muscle force depends on muscle length, and the force is maximal when there is maximal overlap between actin binding sites and myosin heads. Maximal overlap often (though not always) occurs near the resting length of the muscle.

23. Research & Revise
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Activity 2 – muscles and muscle groups
Muscle Force vs. Muscle Length (Quantitative Approach)
Does the force-length relationship for muscle influence the actual maximum torque that can be developed by a muscle?
How does this known physiological force-length relation influence muscle torque as arm position is changed?

24. Muscle Force vs. Muscle Length (Quantitative Approach)
Research & Revise
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Activity 2 – muscles and muscle groups
Muscle Force vs. Muscle Length (Quantitative Approach)
This provides a quantitative relationship between active muscle force and muscle length. It was determined empirically. Maximum muscle strength is assumed to be directly proportional to muscle volume. PCSA = Physiological Cross Sectional Area of Muscle. One could use this relation to predict the muscle mass necessary to provide the strength required to perform a particular task.
Add to this from FBD, assuming there is only one muscle.
Add calculations for statics problem
Plot force needed at different angles then plot force the muscle can produce at different angles using force-length properties (available force vs. needed force)

25. Muscle Force vs. Muscle Length
Research & Revise
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Activity 2 – muscles and muscle groups
Muscle Force vs. Muscle Length
This is a graph of relative muscle force vs relative muscle length using the empirical relationship in the previous slide. The muscle model predicts that the muscle is only capable of producing active tension when the muscle length is between 40% and 140% of the optimal muscle length. The equation in the previous slide can only be used to predict muscle force in that range.

26. Activity 2 – muscles and muscle groups
Research & Revise
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Activity 2 – muscles and muscle groups
Through statics find the force necessary to hold the position assuming that the relative magnitudes of the forces remain the same

27. Activity 2 – muscles and muscle groups
Research & Revise
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Activity 2 – muscles and muscle groups
Through statics find the maximum mass of the gymnast for the muscle forces given to be sufficient to hold the iron cross position.