Problem Set #3

Principles of Biomechanics 1 HK*2270 Problem Set #3 1. What is the perpendicular force produced at the end of a 1.650 m shovel if the moment of force acting on the shovel is 500 N.m? 2. What is the Moment about point A created by the forces depicted in the figure below? 3. What is the moment produced when you push perpendicularly 65.0 cm away from the hinges of a door with a force of 56.5 N? 4. a) Compute the horizontal distance to the center of mass from the feet of a 60.0 kg person who is lying on their back on a 2.00 m “reaction board”. Note the feet are at one end [B] and their head is at the other end [A]; the scale reading is 300 N. Assume that the weight of the board has been subtracted from the scale reading. b) When the person mentioned above raises one leg and both arms, where is the center of mass of the new scale reading is 350 N? Why? A tangent Radius= 25.0 cm B 300 N 200 N 200 N A B C D 30.0 cm 20.0 cm 20° 20° (a) (b) A tangent Radius= 25.0 cm B 300 N 200 N 200 N A B C D 30.0 cm 20.0 cm 20° 20° (a) (b) Hint for (b): Carefully consider the perpendicular distance for both the x and y components of the 2 forces depicted…

Principles of Biomechanics 2 6. a) Draw a FBD and calculate the tensile force in each still ring cable when an 80.0 kg athlete performs the gymnastics maneuver illustrated in Figure (a). Assume that the athlete is motionless and that the two cables are oriented vertically. (b) Assume this same gymnast moves into an “iron-cross” position and that the body mass is evenly distributed b/w the left and right hands. Draw a FBD and calculate the force moment at the gymnast’s right shoulder joint caused by the force at the hand, where the distance between the right shoulder and the hand is 75.0 cm. Use only the view from the frontal plane, as illustrated in Figure (b). (a) (b) 5. Make a free-body diagram of the lower leg and foot (i.e. as one segment) for the situation depicted in the figure to the right. Write out the equation of motions that would allow you to determine the forces and moment at the child’s knee. Assume that the child is balanced (no acceleration). Assume that the ground reaction force (F ground ) is known.

Principles of Biomechanics 4 READ the next 2 questions carefully… you are being asked to solve for the Net Joint Moment about a joint and are given information about the internal muscle forces around the joint. How would you set up this question differently that the earlier questions in this Problem Set? HINT: We did a similar examples in Lecture, hip joint and teeter-totter at the end of lecture ! 9. Assume an elbow is in a flexed position with the forearm at 90.0° to the upper arm. The biceps exerts a 200 N force, pulling at 100.0° at a point 5.00 cm from the joint axis. The triceps exerts 300N force in a direction 85° to the forearm at a point 3.00 cm from the joint axis. With the orientation/position of the arm as illustrated below, indicate if the moment being generated to maintain this static equilibrium about the elbow depicts flexion or extension. Assume there are no other tissues that can create a moment of force across the elbow. HINT: This question is focused on the moment produced by the muscle groups – consider this as you set up your static equilibrium equation: how is this moment generated by the muscles related to the Net Joint Moment? 10. a) Draw a space and free body diagram of the following scenario: A sagittal view of the right Refer to the static position of the foot as illustrated (below). The Tibialis Anterior muscle exerts a 325 N force, pulling at 124.0° at a point 4.25 cm from the ankle joint axis. The Achillies tendon exerts a 530 N force pulling at 79° from the horizontal at a point 3.50 cm from the ankle joint axis. Calculate the moment about the ankle. If the foot acts in the horizontal plane, is the moment being generated to maintain this static position of the ankle a plantarflexion or dorsifexion moment? Assume there are no other tissues that can create a moment of force across the ankle joint. HINT: the set up of Question 10 is similar to Question 8 (above). 124° 79° F Tibialis Anterior F Achillies 124° 79° Ankle

Principles of Biomechanics 5 ANSWERS 1. The perpendicular force produced at the end of the shovel is 303 N. 2. (a) Moment about Point A = -75.0 N.m (b) Moment about Point A = -37.6 N.m 3. Moment of force produced when you push the door is 36.7 N.m 4. (a) The tensile force in the sill ring cables is 785N or ~ 392N per cable. (b) The Moment at one shoulder is 294 N 5. F knee in x-direction = - F ground in x F knee in y-direction = +F COM - F ground in y Moment knee = – (F gy * d perp ) – (F gx * d perp ) - (F COM * d perp ) 6. (a) Height of center of mass of the individual is 1.019 m (b) The new height of the center of mass is 1.189 m. Why? The movement of the body segments shifts the COM towards the head! 7. F ankle in x-direction: -100.0 N; F ankle in y-direction: -675 N, Moment ankle : -113.8 N.m (c) As the moment calculated is negative, it indicates a plantarflexion moment generated at the ankle joint 8. The moment produced at the elbow joint is + 0.881 N.m. Given the axis system chosen (I chose +y to be along the humerus (upper arm) and +x to be along the forearm and to the right; +M to be in the counterclockwise direction) and as the moment calculated about the elbow joint is positive, it indicates elbow flexion. 9. F elbow in x-direction: +175.0 N; F elbow in y-direction: 418 N, Moment elbow : +143.0 N.m (c) As the moment calculated is positive it indicates a flexion moment generated at the elbow joint. 10. The moment at the ankle joint will be - 6.76 N.m. Given the axis system chosen (I chose +y to be vertical up, +x to be to the right; +M to be in the counterclockwise direction) and as the moment calculated about the ankle joint is negative, it indicates that the moment is generated by the ankle plantarflexors .