N Angular Kinetics: Calculating Force Needed to Maintain 45° Position

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The discussion centers on calculating the force required by forearm flexors to maintain a 45-degree position with a 35N hand and forearm. The key parameters include the center of gravity located 15cm from the elbow joint and the muscle attachment point at 3cm. Initial calculations using torque equations led to an incorrect force estimate, prompting a reevaluation of the approach. Participants noted that the calculation of resultant force (R) was unnecessary for the problem at hand. The correct force exerted by the forearm flexors should be 192.5N, highlighting the importance of accurate torque analysis in biomechanics.
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A 35N hand and forearm are held at a 45 degree angle to the vertically oriented humerus. The centre of gravity of the forearm and hand is located at a distance of 15cm from the joint centre at the elbow, and the elbow flexor muscle attach at an average distance of 3cm from the joint centre. How much force must be exerted by the forearm flexors to maintain this position?

Okay this is my attempt at the question:

Wt= 35N
dwt=0.15m
df=0.03m

fm(sin45) x (0.03m) = (35N) X (0.15m)
fm (0.02) = 5.25
fm= 262.5

Rv= 262.5 sin 45 - 35N
Rv=185.62 - 35
Rv=150.62

Rh= 262.5 Cos 45
Rh= 185.62

R= SqRoot (185.62)sqr + (150.62)sqr
R=239.04 N

But my answer is wrong can anyone see where I went wrong? My answer should be 192.5
 
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bionut said:
How much force must be exerted by the forearm flexors to maintain this position?

Okay this is my attempt at the question:

Wt= 35N
dwt=0.15m
df=0.03m

fm(sin45) x (0.03m) = (35N) X (0.15m)
I'd have said:
fm(cos45) x (0.03m) = (35N) X (0.15m) x cos45

I haven't figured out why you calculate R. The question doesn't seem to ask for it.
 

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