IMEN 315 11 Biomechanics of Work IMEN 315

IMEN 315 인간공학 11. Biomechanics of Work 고려대학교 산업경영공학과

IMEN 315 인간공학 THE MUSCULOSKELETAL SYSTEM 고려대학교 산업경영공학과

IMEN 315 인간공학 q. Bones and Connective Tissues 고려대학교 산업경영공학과

IMEN 315 인간공학 Synovial Joint Fibrous Joint Cartilaginous Joint 고려대학교 산업경영공학과

IMEN 315 인간공학 BIOMECHANICAL MODELS 고려대학교 산업경영공학과

IMEN 315 인간공학 q. Single-Segment Planar Static Model (Q) Suppose a person is holding a load of 20 -kg mass with hands in front of his body and his forearm are horizontal. The load is equally balanced the two hands (A) W = mg = 20 kg*9. 8 m/sec 2 = 196 N Won-each-hand = 98 N ∑ Felbow = 0 - 16 N – 98 N + Relbow = 0 Relbow = 114 N ∑ Melbow = 0 - 16 N(0. 18 m) – 98 N(0. 36 m) + Melbow = 0 Melbow = 38. 16 N-m shoulder 30° HW: What are the reactive force and moment at shoulder? BW = 78 Kg Upper arm = 0. 028 BW distance from shoulder to elbow = 33 cm center of mass = 15 cm from the shoulder elbow 고려대학교 산업경영공학과

IMEN 315 인간공학 q. Low-Back Biomechanics of Lifting Clockwise rotational moment: Mload-to-torso (L 5/S 1) = Wload*h + Wtorso*b Counterclockwise rotational moment: Mback-muscle = Fback-muscle * 5(N-cm) ∑ ML 5/S 1 = 0 Fmuscle * 5 = Wload*h + Wtorso*b Fmuscle = Wload*h/5 + Wtorso*b/5 If h=40 cm and b=20 cm then Fmuscle = 8*Wload + 4*Wtorso Torso weight = 350 N, Load = 300 N (about 30 kg) then Fmuscle = 3, 800 N Normal range of strength capability of the erector spinal muscle is 2, 200 to 5, 500 N Compression force on L 5/S 1 (ignoring the abdominal force, f a) ∑ F L 5/S 1 = 0 Fcompression = Wload*cosα + Wtorso*cosα + Fmuscle Suppose α = 55°, torso weight 350 N, load = 450 N Fcompression = 450*cos 55° + 350*cos 55° + 5000 = 258 + 200 + 5000 = 5458 N 고려대학교 산업경영공학과