The biomechanics of elbow extension

Thread starter thegoose
Start date Nov 25, 2023
Tags

Biomechanics Physics Sin

Click For Summary

The discussion focuses on the biomechanics of elbow extension, specifically analyzing the forces involved. The equation presented calculates the force exerted in the x-direction, resulting in a negative value, indicating a need for further clarification of variables used. There is an emphasis on the importance of analyzing torques rather than just forces to understand the mechanics better. Additionally, the role of the shoulder joint in this context is highlighted as crucial for accurate biomechanical analysis. A comprehensive understanding of these factors is essential for effective evaluation of elbow extension mechanics.

Nov 25, 2023

thegoose

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 0°? So, by doing the 85-0 calculation, we would get a positive number.

Relevant Equations: Σ𝐹⃗ = 0

X=

Σ𝐹𝑥 = 0

−𝐹𝑃𝑥 + 𝐹𝑀 sin (𝛽 − 90°) + 𝐹𝑒𝑙 sin (180 − 𝛽 − 𝜃) = 0

𝐹𝑃𝑥 = 𝐹𝑀 sin ( 𝛽 − 90°) +𝐹𝑒𝑙 sin(180 − 𝛽 − 𝜃)

𝐹𝑃𝑥 = 93.6650N sin (85 − 90) + 28.11N sin(180 − 85 − 79)

Fpx= -8.16344N +7.74816N

𝐹𝑃𝑥 = -0.4154 N←

Physics news on Phys.org

Nov 25, 2023

haruspex

Science Advisor

Homework Helper

Insights Author

Gold Member

2024 Award

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.

Last edited: Nov 25, 2023

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

View full post »