Question

A load of 10 kg is held in one hand, with the lower arm horizontal. The lower arm has mass 1.5 kg The distance from the elbow to the hand is 30 cm and the COM can be assumed to be 40% of this distance (from E to H). a) What is the external static elbow moment? (31.2 Nm) b) If the biceps muscle acts with a perpendicular moment arm of 5cm, what force must it generate in order to maintain static equilibrium? (624.0 N) c) What joint force acts on the elbow? (511 N) d) If the load is accelerated down with a = 2 m/s-, and the lower arm also accelerates down with no rotation about the elbow, what is the new external elbow moment? (24.8 Nm) e) If the load is accelerated to the right with a = 2 m/s, what is the new external elbow moment? (31.2 Nm)

Answer 1

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F = Bicep muscle force > Sun long a) External stakie elbow moment = (1.5 × 12 + 10x 30) kgf im =(19+ 30) x 9-81 XL N-m fibow. 1.4x30 I um 15 lng = 318 x 9.91 30cm [g = 9 81 mist taken] = 131-196 N-m Fw moment es will brinon about elbow E f.* 5 um = 31196 Nm. = F = 31-196 Nm 623.92 N s the fasse geresto 05m by musch. 99 Net vertical force at elbow F(115410) *9-91 + 623.923 N = 51 OSN If the luw is a & lower arm both accelepho dron a 2 m/s apparent fine cutting at Pfit will be due to apparent g valce of (9.81-2) = 7.9, mis - New external moment = farlier external moment for = 31-196 x 71 Nm = 24 84 Nm In this can the charpe in aerderation of 2 m/s to the right as aparast zero vale earlier makes no unpact in for at vertical downward fins at P AH aw frile gerecht in homentar druction will maite zen moment contribution about elbow E as momach am lryths an zero for bor. - New external moment will remeni sume as earlic ca 186 Nm