Calculating joint reaction forces

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Homework Help Overview

The discussion revolves around calculating joint reaction forces in a biomechanics context, specifically focusing on the forces acting at the ankle. Participants are examining the implications of force direction and the correct application of summation equations in static equilibrium.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the definitions of positive and negative directions for forces, questioning how these definitions affect the summation of forces. There is a debate on whether the calculations presented in class are mathematically sound and how to correctly interpret the given values.

Discussion Status

There is an ongoing exploration of the definitions of force directions and their implications on the calculations. Some participants suggest that the confusion arises from how the positive direction is defined, while others emphasize the need to clarify the static nature of the forces involved. Multiple interpretations of the calculations are being discussed without a clear consensus.

Contextual Notes

Participants note that the problem involves static forces, leading to the equation of motion being zero (ma = 0). There is also mention of the need for clarity regarding the direction of the forces involved, particularly for Fkx, which remains unspecified.

Haku
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Homework Statement
If the joint reaction force at the ankle in the x direction is -125N, calculate the joint reaction force at the knee in the x direction (assuming those are the only two horizontal forces acting on it)
Relevant Equations
Summation
I have this question in my biomechanics class, and the way the teacher has solved it raised some questions with me.
This is the snippet of work from the lecture slides:
Screen Shot 2021-10-12 at 8.20.09 AM.png

But, if we see the red variables acting as 'placeholders' for the value of respective forces, and the value of the force at the ankle in x direction is -125N.
When summing all ankle forces in x direction it looks like the direction is accounted for twice, because what the summation is if you leave the variables as placeholders for the actual values is:

Fax + Fkx = 0.

=>

Fkx = -Fax = -(-125N) = 125N.

What I get from the calculation done in the picture is the following:

-Fax + Fkx = 0

<=>

-(-125N) + Fkx = 0

=>

Fkx = -125N

But in this case, you have applied the direction of the joint reaction force at the ankle twice right?
Which calculation is correct? Imo the former calculation is more mathematically sound, but I am not sure since it is contradictory to calculation given in class.
 
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I would not describe it as having applied a force twice. Rather, it is a confusion over how the positive direction has been defined.
Where it is clear that two forces will act in opposite directions, it is not unusual to define each to be positive in the direction it will act. That would give the summation equation in the snippet, but in that model the value should have been inserted as +125N.

Since the given force is specified as negative, it is more natural to use the same direction as positive for all forces. That leads to Fax + Fkx = 0, so, again, an answer of +125N.
 
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haruspex said:
I would not describe it as having applied a force twice. Rather, it is a confusion over how the positive direction has been defined.
Where it is clear that two forces will act in opposite directions, it is not unusual to define each to be positive in the direction it will act. That would give the summation equation in the snippet, but in that model the value should have been inserted as +125N.

Since the given force is specified as negative, it is more natural to use the same direction as positive for all forces. That leads to Fax + Fkx = 0, so, again, an answer of +125N.
Right, but we don't know the direction of force for Fkx. We just know that they are static, so ma = 0. Hence the sum of the forces is 0.
So should it just be Fax + Fkx = 0, then since we know Fax = -125N we can solve for Fkx right?
 
Haku said:
Right, but either way, with Fax defined as -125N, the resulting Fkx should be +125N right?
Yes.
 
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