PBSet#12_Solutions_2023

Principles of Biomechanics 1 HK*2270 Problem Set #12 P ART A. D YNAMIC E QUILIBRIUM Question 1. Given the FBD of a foot in swing phase and the following information, solve for the reaction forces and net joint moment at the ankle for a 1.76 m tall female with a mass of 57.2 kg. Her foot length is 25% of her overall height . ax = 9.07 m/s 2 ay = -6.62 m/s 2 α = 21.69 rad/s 2 α ay ax 9.85 cm 8.60 cm 1.95 cm 0.8 cm +y +x M + +y +x M + Ankle Toe

3 Question 3. You are given the following DYNAMIC data about a long jumper. For the Lower Leg & Foot combined segments You are given the following DYNAMIC data about a long jumper. For the Lower Leg & Foot combined segments Accelerations from COM: a x = - 7.20 m/s 2 a y = - 2.43 m/s 2 α = - 3.20 rad/s 2 Moment of inertia about combined Lower Leg & Foot COM = 0.876 kg*m 2 Vertical Ground reaction force under the ball of foot = 162.5 N, oriented 48° below the horizontal Mass of person = 64.3 kg In absolute space: Knee location to Lower Leg & Foot COM = 14.5 cm posterior; 11.5 cm above combined COM Heel of foot location to Lower Limb & Foot COM = 16.70 cm anterior and 8.70 cm below combined COM A. Complete the free body diagram of the situation described above in the space provided. B. Solve for the joint reaction forces at the knee. Free Body Diagram Person’s mass: 64.3kg. Mass proportion for the foot from anthropometric tables: 0.061 Mass of lower leg & foot: (0.061) * 64.3 kg = 3.9223 kg Weight of lower leg & foot: 3.9223 kg * 9.81 = 38.4778 N Since F Gnd = 162.5 N, F gnd-x = 162.5 cos 48 = 108.734 N F gnd-y = 162.5 sin 48 = 120.761 N Σ F x = ma x : -F gnd-x + R Kx = m LL&oot * a x - 108.734 + R Kx = (3.9223) * (- 7.20) -108.734 + R Kx = - 28.241 F Kx = + 80.493 N Σ F y = ma y : F gnd-y + R Ky – Fcom = m LL&foot * a y 120.761 + R Ky – 38.4778 = ( 3.9223 ) * (- 2.43) 82.283 + R Ky = - 9.531 F Ky = - 91.814 N

4 C. Solve for the net joint moment at the knee. D. Based on the moment calculated for this dynamic situation, is the knee tending to FLEX or EXTEND (circle one) E. Thus the _______________________________ muscles must be the major contributors to the production of the moment in this dynamic situation KNEE FLEXORS or KNEE EXTENSORS (circle one) Σ Mabout COM = ICOM α M K - M Gx + M Gy – MR Kx + MR Ky = I com α M K - (F Gx )(0.0870) + (F Gy )(0.1670) - (R Kx )(0.1150) + (R Ky )(0.1450) = I com α M K - ( 108.734 )(0.0870) + ( 120.761 )(0.1670) - ( 80.493 )(0.1150) + ( 91.814 )(0.1450) = ( 0.876 ) ( -3.20 ) M K - ( 9.4599 ) + ( 20.1671 ) - (9.2567) + (13.3130) = - 2.8032 M K + 14.7635= - 2.8032 M K = - 17.5667 N.m

7 Question 4. You are given the following DYNAMIC data about a lifting task. For the combined Forearm & Hand Accelerations from the COM: a x = 2.10 m/s 2 a y = 3.54 m/s 2 α = 5.67 rad/s 2 Moment of inertia about the Forearm & Hand COM = 0.078 kg*m 2 Force of the basket on the hand is 147 N, oriented 35° below the horizontal Mass of person = 59.2 kg In absolute space: Elbow location to Forearm & Hand COM = 10.30 cm posterior and 12.40 cm above Forearm & Hand COM Basket location to Forearm & Hand COM = 9.34 cm anterior and 7.4 cm below Forearm & Hand COM A. Complete the free body diagram of the situation described above in the space provided. B. Solve for the joint reaction forces at the elbow. Free Body Diagram Person’s mass: 59.2 kg. Mass proportion for the forearm & hand from anthropometric tables: 0.022 Mass of forearm & hand: (0.022) * 59.2 kg = 1.3024 kg Weight of forearm & hand: 1.3024 kg * 9.81 = 12.777 N Since F basket = 147 N, F basket-x = 147 cos 35 = 120.415 N F basket-y = 147 sin 35 = 84.316 N Σ F x = ma x : - F basket-x + R Ex = m f+h * a x - 120.415 + R Ex = (1.3024) * (2.10) - 120.415 + R Ex = (2.735) R Ex = 123.150 N Σ F y = ma y : F basket-y + R Ey - W f+h = m f+h * a y 84.316 + R Ey – 12.777 = (1.3024) * (3.54) 71.539 + R Ey = (4.6105) R Ey = - 66.929 N