M = 6.5 kg in the palm of his hand. Lowerarm has mass m = 2.2 kg. As Figure 12-3
ID: 1733689 • Letter: M
Question
M = 6.5 kg in the palm of his hand. Lowerarm has mass m = 2.2 kg. As Figure 12-34shows, his upper arm is vertical and his lower arm ishorizontal.What is the magnitude of (a) the forceof the biceps muscle on the lower arm and (b) theforce between the bony structures at the elbow contactpoint?
Here is my attempt:
a) Net torque should be equal to zero so: (T = torque here)
Sum of T = M (0.12m)-(21.56N)(0.15m)-(63.7N)(0.34m) = 0
M=207.43N
b) sum of T = -(207.43N)(0.19m)+F(0.15m)-(63.7N)(0.18m) = 0
F=339.18N
I followed the steps found here:/physics-topic-5-412836-cpi0.aspx but both answers I computed areincorrect. Also, if the pic isn't showing up properly please let meknow! Thanks in advance.
Explanation / Answer
I found the book, so now I know what the pic looks like.The numbersthat you have are different than the ones I found, so simply changethose numbers and do the same thing. So there is a force actingupward on the bicep, downward on the elbow contact point, anddownward for both weights. Then the Sum of the moments about theelbow contact point is 0=(4cm)(F)-(15cm)(1.8)(9.8)-(33)(7.2)(9.8)and then we solve to find that F= 648.27 N. Now the other is simply0=-F+648.27-(9.8)(1.8)-(7.2)(9.8) and then F= 560.07
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