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

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The discussion focuses on calculating the force exerted by forearm flexors to maintain a 45° position of a 35N hand and forearm. The key parameters include the distance from the elbow joint to the center of gravity (15cm) and the distance from the joint to the muscle attachment (3cm). The initial calculations led to an incorrect force of 239.04N, while the correct force required is 192.5N. The error stemmed from misapplying the sine and cosine functions in the torque equations.

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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.