During isometric exercise tor strengthening the biceps muscle, an external force

Question

During isometric exercise tor strengthening the biceps muscle, an external force of 200 N is applied perpendicular to the forearm 0.3 m from the center of rotation of the elbow joint. Assuming that the elbow is flexed at 90 degree , and biceps muscle is the only active muscle with a moment arm of 0.04 m about the elbow joint. Discard the weight of the arm. How large a force (B) must the biceps muscle generate to counteract the torque due the external force? How large is the joint reaction force (R)?

Explanation / Answer

a) so we do sum of toques about elbow
remember torque=rF sin , where is the angle between r and F
-200*0.3sin90+B*0.04*sin(90)=0

so B=200*0.3/0.04=1500 N

b) now we use that sum of forces in x direction is zero

200-B+R=0

R=B-200=1500-200=1300N

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