Physics Help Forum Magnitude and direction of arm

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 May 11th 2017, 06:54 PM #1 Junior Member   Join Date: May 2017 Posts: 2 Magnitude and direction of arm Would like to know if this working is correct or not and help on part b? I have for my answer of A) (after finding theta etc): 0 = ((12*9.81)*0.0365)+((1.5*9.81)*0.155)-(Fy*0.05) Fy = (42.9678+2.2808)/0.05 Fy = 904.972N F being brachialis group muscles: F=Fy/cos(theta) F = 904.972/cos(13.39) F = 930.26N direction from positive x axis: 90+13.39 = 103.39degrees No idea where to go from here for part b Attached Thumbnails
 May 11th 2017, 07:11 PM #2 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 552 I assume the arm is in equilibrium (not moving)? Looks like a problem involving equilibrium and torque.
 May 11th 2017, 07:14 PM #3 Junior Member   Join Date: May 2017 Posts: 2 I'm under the same assumption - it does say "slow arm curls" but without a statement of speed I've just assumed equilibrium
 May 11th 2017, 07:38 PM #4 Senior Member   Join Date: Nov 2013 Location: New Zealand Posts: 552 It appears as though the briachilis muscle is attached a small distance from the elbow joint and I assume they are suggesting that the briachilis muscle is the primary and dominating support for the forearm. If I understand the question correctly the attachment of the briachlis muscle to the forearm is the pivot point. Therefore the force and torque on the elbow joint E exactly counterbalances the sum of the torque (force * distance * cosine angle between them) on the other side. Therefore the force on joint E will be roughly upward (opposite gravity) if my interpretation of the question is correct. Make sense?

 Tags arm, direction, magnitude

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