Calculating joint reaction forces

AI Thread Summary
The discussion revolves around the calculation of joint reaction forces at the ankle in a biomechanics context. Participants analyze the summation of forces, particularly focusing on the directionality of the forces involved. The calculations presented lead to two interpretations: one suggesting Fkx should be +125N and the other indicating it as -125N, highlighting a confusion over the definition of positive direction. The consensus is that the forces should be treated consistently, with Fax defined as -125N leading to Fkx being +125N. Ultimately, the key takeaway is that clarity in defining force directions is crucial for accurate calculations.
Haku
Messages
30
Reaction score
1
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.
 
Physics news on Phys.org
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.
 
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.
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top