Introduction to Biomechanical Analysis

Transcription

1 Introduction to Biomechanical Analysis LEARNING OBJECTIVES: At the end of this laboratory exercise the student will be able to: Identify forces used during activities Identify moments used during activities Describe how changes in direction or magnitude in forces and moments influence the difficulty of an activity Use principles of static equilibrium to compute muscle forces and joint reaction forces generated during activity NOTE: Answers to the Thought Questions from the chapter may be found at the end of this document. Identifying the Forces 1. Place a coin with a string taped to its center on a large piece of paper with an x y coordinate system drawn on the paper. Complete the following activities: Draw a graphic representation of pulling on the string to move the penny along the x axis. (Label the magnitude of the unknown force, P. ) Pull the coin at some angle to the x axis. Draw a graphic representation and give polar coordinates. Add a second string to the center of the coin. You and your partner pull the strings to make the penny move at a 45º angle to the x axis. Draw the graphic representation of adding these vectors. Now pull the strings separately to get the coin to the same location. Observe and explain how changes in the magnitude and direction of pull affect motion of the coin. Does the coin rotate or translate? Why? 2. Observe your partner pushing open a door. Draw a representation of this activity labeling orientation, direction, and point of application. (Use unknown D for magnitude.) Draw an example of a type of door that uses a force couple to open it. How do you give yourself a mechanical advantage opening a heavy door? Oatis, Kinesiology: The Mechanics and Pathomechanics of Human Movement, 3rd ed. Page 1 of 7

2 3. Find the x and y components of each of the following vectors: R: 173 mph, 30º R: 141 ft/sec, 45º R: 200 lb, 80º 4. The three heads of the triceps are pulling on the olecranon in the following way: Lateral head: 5 lb, 10º to the left of vertical Medial head: 10 lb, 15º to the right of vertical Long head: 2 lb, vertical What is the total extension force in the triceps? Find the total force and direction. 5. Draw a first-class lever system. If 150 lb is 1 foot away from the fulcrum, what force, applied 1 foot from the fulcrum, is required to lift the weight? Draw a diagram of the activity including all forces. What force is required to lift the same load if the lifting force is applied 2 feet from the fulcrum? Draw a diagram of the activity including all forces. 6. Balance yourself and your partner as you both stand on a rocker board. Measure the approximate distance from the center of each of your bases of support to the balance point. Draw a free-body diagram. Using your weight, calculate your partner s approximate body weight. 7. Have your subject sit on a balance ball. Have enough help to ensure safety of the subject. Observe and use mechanical terms to explain your subject s reaction to rolling: Backward Forward 8. Assess the difficulty of standing on one leg Without assistance With a cane (or crutch) on the stance side With a cane (or crutch) on the unloaded side Which is hardest? Easiest? Briefly explain your answer. 9. Place a heavy book bag in a wheelchair. Compare the following: Tilt the wheelchair to lift the front wheels just off the ground. Hold the wheelchair tilted as though you were going to clear an 8-inch curb. Hold the wheelchair tilted far back as though you were picking the wheelchair up after tipping over. Of the above, which is hardest? Easiest? Briefly explain your answer, mechanically. This activity can be repeated with someone sitting in the wheelchair. Be sure you have enough help to ensure safety. Observe and explain the subject s reactions. Reconsider the above activity if a person with bilateral transfemoral amputations were seated in the wheelchair. What differences would there be? Oatis, Kinesiology: The Mechanics and Pathomechanics of Human Movement, 3rd ed. Page 2 of 7

3 Reconsider this activity if your patient had bilateral long leg casts. What differences would there be? 10. With your partner sitting on a plinth, resist knee extension with your hand at the distal tibia. In the same position, resist knee extension with your hand at the tibial tubercle. Which is easier for your subject? Which is easier for you? Why? 11. With your partner seated, resist shoulder flexion with your hand at the distal arm. Now resist at the distal forearm with forearm pronated. Which is easier for your subject? Which is easier for you? Why? 12. Determine your subject s ease of holding a heavy load in the hand with the elbow flexed to 90º and again with the elbow flexed to 30º. Draw a free-body diagram for the above situations. Calculate the muscle force in each. If there is a difference, explain why. (Assume a 2-inch anatomical moment arm for the elbow flexors.) Use Table 1.2 to determine the weight of the forearm. Measure the distance from the lateral epicondyle to the center of the weight for the moment arm of the weight. Measure the distance from the lateral epicondyle to your best estimate of the forearm and hand s center of mass for that moment arm. Assume an 80º angle of application for the elbow flexors. 13. Hold a heavy weight in your hand with your shoulder abducted to 90º and the elbow: Extended Flexed to 90º Flexed maximally Draw a free-body diagram of each situation above and calculate the muscle forces used in each instance. Use Table 1.2 to determine the weight of the upper extremity. Assume a 3-inch anatomical moment arm for the shoulder abductors and an angle of application of 15º. 14. Observe your partner walking. At heel strike the floor reaction force on the heel is 180 lb (120% of body weight of a 150-lb subject). This reaction force is acting at a right angle to the foot. What is the force required of the dorsiflexors to prevent the foot from slapping on the floor? Anatomical moment arm of the dorsiflexors: 1.5 inches Angle of application of the dorsiflexors: 60º Distance from floor contact to ankle: 2 inches Sin 60º = 0.87 Cos 60º = Given the following two forces, solve for the knee joint reaction force (sum of these two forces): Oatis, Kinesiology: The Mechanics and Pathomechanics of Human Movement, 3rd ed. Page 3 of 7

4 Hamstrings: Force = 1000 N, angle from horizontal = 45 degrees Quadriceps: Force = 700 N, angle from horizontal = 30 degrees 16. What is the force a physical therapist needs to apply to the back of a patient s head given the following conditions: Force of neck extensors = 200 N Moment arm of neck extensors = 3 cm Perpendicular distance from neck COR to PT s hand = 15 cm Use the following figure to solve problems 17 and With the elbow flexed 20 degrees, the moment arm of the biceps is equal to 10 mm. Solve for the force exerted by the biceps given the following assumptions: the biceps is the only muscle active the center of mass of the forearm is located 200 mm from the elbow center of rotation the mass of the forearm is 2 kg 18. With the elbow flexed 90 degrees, the moment arm of the biceps is equal to 30 mm. Solve for the force exerted by the biceps given the same assumptions as the previous problem. Oatis, Kinesiology: The Mechanics and Pathomechanics of Human Movement, 3rd ed. Page 4 of 7

5 Use the following figure to solve problems What is the torque about the ankle COR caused by the force of the Achilles tendon (F)? 20. Which of the designated forces (a, b, c, or d) could possibly counterbalance the torque caused by the Achilles tendon force (F) and keep the system in static equilibrium? Why? 21. Assuming that the force of the Achilles tendon (F) and the counterbalance force you selected in the previous problem are the only two forces acting on the foot, solve for the counterbalance force. Use the following figure to solve problems A cyclist applies the same force, F = 250 N, to a crank that is 15 cm long. If the crank is horizontal (see A in the figure above), determine the torque applied to the center of rotation of the crankset. Will the crank turn? Why or why not? 23. With the same force and crank, if the crank is vertical (see C in the figure above), determine the torque applied to the center of rotation of the crankset. Will the crank turn? Why or why not? Oatis, Kinesiology: The Mechanics and Pathomechanics of Human Movement, 3rd ed. Page 5 of 7

6 24. With the same force and crank, if the crank is at a 45 degree angle (see B in the figure above), determine the torque applied to the center of rotation of the crankset. Will the crank turn? Why or why not? Thought Problems with Answers 1. Your patient is performing knee extension strengthening exercises in the seated position by lifting a 5 lb load wrapped around the ankle. List three ways to decrease the external moment applied to the knee. What is likely to happen to the knee joint reaction force when the external moment is decreased? ANSWER: The external moment can be reduced by decreasing the weight wrapped around the ankle, by moving the ankle weight proximally or by placing the subject in a side-lying position so that the weight of the leg and ankle (and ankle weight) are no longer producing a flexion moment on the knee. This is known as the gravity minimized position. 2. A person is sitting in a chair, holding a dumbbell in one hand. ANSWER: You are given the following information: The humerus is aligned vertically (although it might not look like it in the figure) The elbow flexion angle ( ) is 110 degrees The length of the forearm (L) is 35 cm (from COR of elbow to dumbbell) The dumbbell has a mass (MD) of 5.0 kg The mass of the forearm is negligible The moment arm of the triceps (MAT) is 5.0 cm The moment arm of the biceps (MAB) is 7.0 cm a) Draw a free body diagram for this problem. Be sure to label your forces. b) Solve for the triceps muscle force with no co-contraction (biceps force = 0) c) Solve for the triceps muscle force with co-contraction (biceps force = 10% of triceps force) d) How does co-contraction of the biceps affect the load on the triceps? Does this make sense? Oatis, Kinesiology: The Mechanics and Pathomechanics of Human Movement, 3rd ed. Page 6 of 7

7 a) Free-body diagram of elbow (red dot): F D =M D * g θ F T F B F D b) M = 0 F D * (L) * sin θ + (F B ) * (MA B ) (F T ) (MA T ) = 0 Note: F B = 0 because there is no co-contraction F T = (F D * (L) * sin θ) / MA T F T = (5kg * 9.8 m/s 2 * 35 cm * sin (110ᴼ)) / 5.0 cm = 320 N c) F D * (L) * sin θ + (F B ) * (MA B ) (F T ) (MA T ) = 0 Note: F B = (0.1) F T F T = (F D * (L) * sin θ) / (MA T 0.1 MA B ) F T = (5kg * 9.8 m/s 2 * 35 cm * sin (110ᴼ)) / 4.3 cm = 380 N d) Co-contraction increased the triceps force by approximately 15%. This makes sense because these muscles are antagonists and generate torques in opposite directions. Oatis, Kinesiology: The Mechanics and Pathomechanics of Human Movement, 3rd ed. Page 7 of 7