Calculate the quadriceps muscle force

  • #1
Newby
4
0
Homework Statement
Calculate the quadriceps muscle force
Relevant Equations
πΉπ‘š=π‘ƒβˆ™π‘Ž+π‘Šβˆ™π‘/c
Hi guys,

I'm new here. So I have this biomechanics module which is completely out of my league (it's compulsory but it's not what I want to do). Could someone help with the following? It's about static and dynamic loads calculation.

Calculate the quadriceps muscle force applied through the patellar tendon. Lever arm of the muscle force is 4cm. The ground reaction force is 800N and its lever arm is 20cm.

This is the formula:
πΉπ‘š=π‘ƒβˆ™π‘Ž+π‘Šβˆ™π‘/c

On the slides P stands for weight of the barbell but we don't have it here so I don't know..
W stands for weight of the upper body
a - the lever arm of P
b- the lever arm of W
c- the lever arm of Fm
 
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  • #2
Welcome to the PF. :smile:

Can you use the "attach a file" feature to upload a diagram for this problem?

Are you familiar with free body diagrams (FBDs) and how to calculate the sum of moments about an axis of the body?
 
  • #3
Newby said:
This is the formula:
πΉπ‘š=π‘ƒβˆ™π‘Ž+π‘Šβˆ™π‘/c
I think you mean πΉπ‘š=(π‘ƒβˆ™π‘Ž+π‘Šβˆ™π‘)/c.
If there is no barbell in the question then you can drop that part, leaving
πΉπ‘š=π‘Šβˆ™π‘/c
You should be able to solve it from there.
 
  • #4
Thanks guys for the help, I'm good with the first one but now I'd really appreciate your feedback on this one.

Consider the following free body diagram. Using static analysis, solve for the muscle torque that will place this system in equilibrium, given mass of the leg and foot is 5.4kg; distance from the knee joint to the center of mass of the leg-foot system 0.232m; weight of the barbell 150N; and distance from the knee joint to the center of mass of the barbell 0.514m.

So we use the same formula once again. πΉπ‘š=(π‘ƒβˆ™π‘Ž+π‘Šβˆ™π‘)/c.
I believe P is 150N, a=51.4cm, c=23.2cm, W=5.4kg (52.7N). But how do I find b?

And then I have a formula for torque which is T=Fβ‹…r; r=distance from line of gravity. So I believe in my case r will be distance from the knee joint to the center of mass of the barbell (0.514m)?
 
  • #5
Newby said:
Consider the following free body diagram.
I will if you post it.
 
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  • #6
haruspex said:
I will if you post it.
Thanks for your help.
 

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  • #7
Newby said:
c=23.2cm
No.
What force acts at 23.2cm from the joint?
 
  • #8
haruspex said:
No.
What force acts at 23.2cm from the joint?
Could you put me on the right track? I need to fully understand at least one of these tasks to be able to carry on by myself.
 
  • #9
Newby said:
Could you put me on the right track? I need to fully understand at least one of these tasks to be able to carry on by myself.
It's a simple enough question: what, according to the information given, is 23.2cm from the joint?
 
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