Joint contact reaction force - Part 1

What is the stifle joint contact reaction force?

From Newton’s third law, we know that for every action, there is an equal and opposite reaction. So, a joint reaction force must be the reaction of the joint to some kind of action. Let’s think abstractly and visualise what happens in the stifle joint surfaces at mid-stance.

An intuitive answer would be that the joint surfaces get compressed. And that is part of the answer. The other part would be that the two surfaces would tend to slip on each other. In other words, the tibial surface would tend to translate relative to the femur while both the tibial and femoral surfaces compress against each other. If you thought similarly, then your intuition served you well this time. However, intuition only provides momentum for our thinking process. We need to use this momentum to start working on the problem analytically and let our analysis lead us to the proof or disproof of our intuition. Only if we prove our intuition can we use it as a statement on which we can base conclusions and more questions.

Figure 1 illustrates a simplified idealised model of a pelvic limb contacting the ground. As the animal pushes the ground downwards and backwards, the ground reacts with an opposite force (upwards and forward). This Ground Reaction Force (GRF) stops the limb from moving into the ground while pushing the animal forward. The tibia is also pulled by the patellar tendon, the hamstring muscles and the gastrocnemius muscle in different directions. We will call the sum of the forces applied to the tibia the tibial force FT, which is the total force applied to the tibia at mid-stance.

Figure 1. GRF: Ground Reaction Force, FP: Patellar tendon force, FH: Hamstrings force, FG: Gastrocnemius force, FT: Tibial force. The tibial force is the sum of all forces applied to the tibia (blue vectors).

Before we get into the forces applied to the joint surfaces, let’s look at a simple problem. Let’s assume a horizontal rigid beam is fixed on a vertical wall (Figure 2). We will assume that the beam cannot be bent or compressed/stretched, and is fixed to the wall, so it cannot translate or rotate about the wall. The type of support the wall provides to this beam is called “fixed support”. We want to study the wall’s reaction if we try to push the beam downwards and towards the wall.

Figure 2

Let’s assume we push the beam at point A with a force FA. This force will create a moment MO, which is the tendency of the beam to rotate about point O. We have assumed that the wall will not allow the beam to rotate about it or translate relative to it. For the wall to achieve this, it must apply to the beam a force FO equal in magnitude and opposite in direction to the force FA and a moment M equal in magnitude and opposite in direction to the moment MO.

Let’s express all this mathematically.

It would be useful to analyse the forces into two perpendicular components, as in Figure 3. The reason why this is useful will become apparent in another post. Force FAx will tend to push the beam into the wall, and force FOx will react to this. Force FAy will tend to push the beam downwards and rotate it about point O. The wall will react with an upwards force FOy and a moment M.

FOx tends to compress the beam at its contact surface with the wall. FAy tends to translate the beam relative to the wall. 

If we know the magnitude of force FA and its direction (θ), we can calculate its components. Also, if we know the length (d) of the beam, we can calculate moment MO, which will be the product of force FAy and the length d (Figure 3, Equation 3). As a reminder, the moment a force generates about a point is the product of the force with the perpendicular distance between the line of the force and that point.

Figure 3 illustrates the mathematical analysis of what we have discussed so far. The symbol Σ means “Sum”. So, the first equation means that the sum of all forces in the x-axis is zero. This is from Newton’s first law, which states that the sum of the forces on a body must be zero for the body to be steady (in equilibrium). Our beam is steady in the x-direction. Similarly, the second equation means that the sum of all forces in the y-direction is zero, and the third equation means that the sum of all moments about point O is zero. This is because the beam does not move in the y-direction or rotate about point O.

Figure 3. The variables in red circles are the unknowns. We have a system of three equations in three unknowns, which is solvable (consistent).

So, if we know the forces and distance d, we can calculate the reaction forces and moment the wall applies to the beam. Now, let's imagine that the beam represents the tibia, the force we apply to the beam represents the tibial force FT, and the wall represents the femur.

The reaction force FO of the wall represents the stifle joint contact reaction force!

Like FO (FOx) tends to compress the beam at its contact surface with the wall, the stifle joint contact reaction force (JRF) tends to compress the joint surfaces and damage the cartilage. The action force from the tibia to the femur FT represents force FA in the beam-wall analogy. As FA (FAy) tends to translate the beam relative to the wall, FT tends to translate the tibia relative to the femur (Figure 4). Hopefully, everything has fallen into place by now.

Figure 4. A force FT is applied on the tibia. The femur reacts with a Joint Reaction Force (JRF).

Our mission was to investigate the wall’s reaction to our push. The wall’s reaction is described by forces FOx, FOy, and moment M. These are our unknowns. In Figure 3, we have three equations in these three unknowns. By solving the system, we can calculate the wall’s reaction to our push.

Now that we understand how to calculate the joint contact force in a simplified model, we have one more thing to consider. The femur is not joined to the tibia with a fixed-type support, as the former allows the latter to rotate about its contact point. A fixed support more closely represents the sacroiliac joint. A healthy stifle joint acts more like a hinge.

In the next post, we will get even closer to studying the stifle joint contact reaction force.