The biomechanics of elbow extension

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SUMMARY

The discussion centers on the biomechanics of elbow extension, specifically analyzing the forces involved in the movement. The formula presented, 𝐹𝑃π‘₯ = 93.6650N sin (85 βˆ’ 90) + 28.11N sin(180 βˆ’ 85 βˆ’ 79), results in a calculated force of 𝐹𝑃π‘₯ = -0.4154 N. Participants emphasize the importance of considering torques in addition to forces and suggest a more detailed analysis of the shoulder joint's role in elbow extension mechanics.

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  • Familiarity with force and torque calculations
  • Knowledge of muscle force dynamics
  • Basic anatomy of the shoulder and elbow joints

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thegoose
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Homework Statement
I noticed that when I do a calculation of the form 85-90 it gives a result of -5, which affects my overall answer by giving a negative number. I think I didn't correctly determine the angle of my muscle strength. Would it be possible for this angle to be 0Β°? So, by doing the 85-0 calculation, we would get a positive number.
Relevant Equations
Σ𝐹⃗ = 0

X=

Σ𝐹π‘₯ = 0

βˆ’πΉπ‘ƒπ‘₯ + 𝐹𝑀 sin (𝛽 βˆ’ 90Β°) + 𝐹𝑒𝑙 sin (180 βˆ’ 𝛽 βˆ’ πœƒ) = 0

𝐹𝑃π‘₯ = 𝐹𝑀 sin ( 𝛽 βˆ’ 90Β°) +𝐹𝑒𝑙 sin(180 βˆ’ 𝛽 βˆ’ πœƒ)
𝐹𝑃π‘₯ = 93.6650N sin (85 βˆ’ 90) + 28.11N sin(180 βˆ’ 85 βˆ’ 79)

Fpx= -8.16344N +7.74816N

𝐹𝑃π‘₯ = -0.4154 N←
IMG_0116.jpg
 
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Please define all your variables.
I assume FM means the force exerted by a muscle, but it seems to be exerted directly down at the shoulder pivot, which is not going to achieve anything. You need to analyse torques, not just forces, and consider the shoulder joint in more detail.
 
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