1. A child has picked up a bucket of muddy water and is carrying it with an outstretched arm so that he doesn’t splash his clothes. Given the following information, calculate the resultant joint forces and moments at the elbow and shoulder. What general effect would motion, such as swinging the arm further upwards, have on these values? Why? (10)

Assumptions:

Mass of hand negligibile

Center of mass of arm segments located at mid-length

• Elbow

Vertical:

F_bucket + F_forearm + R_ev= 0

-10 kg * g - 1 kg * g + R_ev = 0

R_ev = 107.8 N

Horizontal: no external forces, R_eh = 0

Moments:

M_bucket + M_forearm + M_elbow = 0

-10 kg * g * 0.25 m(sin 60) - 1 kg * g * 0.125 m(sin 60) + M_elbow = 0

M_{elb ow} = 22.3 Nm (counterclockwise -- elbow flexors acting to prevent hyperextension)

• Shoulder

Vertical:

F_arm + R_ev + R_sv = 0

-2 kg * g - 107.8 N + R_sv = 0

R_sv = 127.4 N

Horizontal: no external forces, R_sh = 0

Moments:

R_ev*d + M_arm + M_elbow + M_shoulder = 0

-107.8 N * 0.33 m (sin 60) - 2 kg*g*0.33 m (sin 60) - 22.3 Nm + M_shoulder = 0

55.9 Nm (counterclockwise -- shoulder flexors active)

• With motion, there will be linear and angular acceleration. If the accelerations are positive, the resultant moments and forces will be increased (made more positive) .

1. A man lifts a 50 kg mass from the ground (Figure A). What is the required force in the erector spinae muscles of the back? What would the forces be if he used correct lifting procedures (bent his knees ? Figure B)? What would be the effect of increasing abdominal pressure during the lifting exercise? (15)

Figure A:

Figure B:

• Assume lever arm length of arm is the same as that for mass

Moments: M_w + M_torso + M_arm + M_muscle = 0

-50 kg * g * 25 cm - 40 kg * g * 18 cm - 5 kg * g * 25 cm + F_muscle* 3 cm = 0

M_muscle = 205.3 Nm (counterclockwise)

F_muscle = - 6843.7 N

Vertical: F_w + F_torso +F_arm +F^v_muscle + R^v_back = 0

-50 kg * g - 40 kg * g - 5 kg * g - 6843.7 N*sin 45 + R_back= 0

R^v_back = 5768.3 N

R_back = R^v_back /sin(45) = 8154.4 N

• Moments: M_w + M_torso + M_arm + M_muscle = 0

  -50 kg * g * 15 cm - 40 kg * g * 5 cm - 5 kg * g * 10 cm + F_muscle * 3 cm = 0

  M_muscle = -98 Nm (counterclockwise)

  F_muscle = -3266.7 N

  Vertical: F_w + F_torso + F_arm + F_muscle + R_back = 0

• If abdominal pressure is increased, the pressure acting on the cross-sectional area of the abdomen exerts an upward, resisting pressure (F = P*A). This acts to reduce the muscle force by generating a positive moment and also reduces the joint reaction force at the spine through both the reduced muscle force and a resistive force.

3. Describe how trabecular bone obtains its optimized strength to weight ratio. How does osteoporosis affect the trabecular bone of the vertebrae? What is the effect of these changes on the structural properties of the vertebral body? Explain the theory of how a compression fracture of a vertebral body progresses. (8)

• Forms a 3D structure based on the direction of primary applied loading with beams and connecting struts. The beams lie along the primary loading direction, the struts act to shorten the beams to minimize bending. In complex loading situations, the trabeculae can be organized along both principal compressive and tensile loading directions.
• Osteoporosis results in the progressive removal of trabeculae, typically with the struts being the first to resorb.
• This reduces the connectivity of the structure, resulting in reduced strength and modulus.
• Compression fractures in the vertebrae occur due to sequential bending failure of the vertical, beam trabeculae. As the horizontal trabeculae resorb, the effective length of the vertical beams increases and they are more likely to buckle. As individual trabeculae fail, the stress on the remaining structure increases. Trabecular failure progresses until the cortical shell also buckles, and compression failure occurs.

4. Describe how and why a muscle’s active and passive behavior shows contradictory relationships with strain rate. (4)

• In active muscle, the generated force is reduced as the contraction rate (strain rate) increases. This is due to the fact that the cross-bridges form less efficiently as the reaction rate increases. In passive muscle, the viscoelastic behavior results in a positive relationship between stress and strain. A larger force is required to cause a defined deformation at higher strain rates.

5. Three main muscles act to flex the arm: biceps, brachioradialis, and brachialis. Together they exert a force of 300 N in order to maintain the elbow at a 90 degree angle while the hand supports a 20 N weight. What are the two methods that can be used to estimate the forces in each muscle? Give a brief description of each. Pick one method and describe the steps in its application. (7)

• Optimization: an output criteria is selected for optimization, for instance total muscle force or energy, and the equilibrium equations are solved iteratively until the output is optimized given assumed or known constraints
• Reduction: the number of unknowns is reduced through physiologic or anatomic assumptions until the equation is made determinant.
• Example: Identify cross-sectional area of each of the three muscles involved in flexion. Assume constant maximum stress and that force exerted will be proportional to cross-sectional area. Determine the fraction of the force that each muscle will exert. Reduce 5 equations to 3 by equating total force to the sum of the forces in each of the muscles.

6. An individual is standing in the anatomic, neutral position. She moves her right arm in the following sequence of movements:
   1. Abduction of arm by 45 degrees
   2. Flexion of forearm by 90 degrees
   3. Pronation of hand by 90 degrees
   4. Internal rotation of shoulder by 90 degrees

• The arm is at a 90 degree angle to the torso, the forearm at a 90 degree angle to the arm, with the palm of the hand pointed downward.

7. A soccer player on the U.S national team prepares to kick a penalty shot. Just before he contacts the ball, the following conditions are measured:

   Ground reaction force:	0 N
   Mass of foot:	0.9 kg
   Angle of foot:	30 degrees plantarflexion
   Resultant lever arm for foot center of mass:	13 cm
   Foot acceleration:	2 m/s2 at an angle of 0 degrees
   Moment of inertia of foot:	0.0033 s2/Nm
   Angular acceleration:	9 rad/s2 counterclockwise

   Calculate the force and moment at the ankle. (6)

   SF = ma

SM = Ia

Vertical forces: (a^v = 0)

mg + R^v_a = 0 R^v_a = 8.82 N

Horizontal forces:

R_a^h = ma R^h_a = 1.8 N

Moments:

M_a - mg*0.13*cos30 = Ia

M_a = 0.993 Nm (counterclockwise)

 

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1. Define Wolff’s law. Give two pieces of experimental evidence for the theory. (4)

   Wolff’s law states that bone remodels in response to loads experienced.

   1. Astronauts lose bone in hypogravity.

   2. Rats in hypergravity add bone mass.

   3. Tibial graft remodels like metacarpal

   4. Pig radius increases in diameter after removal of ulna

    

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2. You have been assigned by your HMO to determine the one test that will be used to assess fracture risk in vivo. Discuss the factors that influence fracture risk. What test would you recommend and why? (10)

 

• Fracture risk:

  Bone material properties and geometry

  Soft tissue thickness (energy absorption)

  Probability of falling

  Age

  Balance

  Cognition

  Direction of fall

  Height of patient

• Test selection: any test would be admissable if it accurately discussed the benefits and limitations.

  Ideally, you should discuss the limitations of density in estimating bone strength (ie. scalar vs. anisotropic properties).