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Psyc 552 Ergonomics & Biomechanics Lecture 10
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The Roger Problem – Weights Body mass: 50 kg Body weight: (50kg)(9.8m/s 2 )=490N Load mass:10kg Load weight: (10kg)(9.8m/s 2 ) = 98N Weight of forearm – hand segment (50kg)(.022)(9.8m/s 2 ) = 10.78N Weight of upper arm segment: (50kg)(.028)(9.8m/s 2 ) = 13.72N
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The Roger Problem – Lengths Shoulder (0, 13) Elbow (2, 2) COM Hand (16, 4)
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The Roger Problem – Lengths
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1 Square =.75” SL FH = [14 2 +2 2 ] -1/2 = 14.14 units – (14.14)(.75)(2.54cm)=26.9cm SL UA = [2 2 +11 2 ] -1/2 = 11.18 units – (11.18)(.75)(2.54cm)=21.3cm COM FH =(26.9cm)(.682)=18.34cm COM UA =(21.3cm)(.436)=9.29cm
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The Roger Problem – Forces R e = (98N x.5) + 10.78N = 59.78N M e = cos(8)[(26.9cm x 49N) + (18.34cm x 10.78N) = 1501 Ncm or 15.01 Nm R s = 59.78N + 13.72N = 73.5N M s = cos(280.3)[(21.3cm x 59.78N) + (9.29cm x 13.72N)] + 1501Ncm = 1751Ncm or 17.51Nm
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Internal Forces
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Low back models
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Low back models (cont)
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Lifting methods
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