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T.E. #9: Simultaneous Ankle and STJ Equilibriums

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  #1  
Old 16th February 2008, 12:37 PM
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Default T.E. #9: Simultaneous Ankle and STJ Equilibriums

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Thought Experiment #9: Simultaneous Ankle Joint and Subtalar Joint Equilibrium Equations

When ground reaction force (GRF) acts on the plantar foot, the summation of GRF on all the locations of the plantar foot can be thought to be acting at a specific area of the plantar foot which is known as the center of pressure (CoP). In normal relaxed bipedal stance, the CoP is positioned anterior to the ankle joint axis and positioned lateral to the subtalar joint (STJ) axis. Also, in normal relaxed bipedal stance, the Achilles tendon will be exerting a tensile force on the posterior calcaneus due to the contractile activity of the gastrocnemius and soleus muscles. This central nervous system (CNS) controlled gastrocnemius-soleus contractile activity is one of the mechanisms which allows the individual to maintain upright balance by positioning the center of mass (CoM) of the individual over the plantar foot during relaxed bipedal stance.

The combination of these external forces (i.e. GRF) and internal forces (i.e. Achilles tendon tensile force) cause simultaneous ankle joint moments and STJ moments. Both the ankle joint and STJ moments are dependent on each other since an increase or decrease in GRF or repositioning of the CoP on the plantar foot will affect not only the ankle joint moments but also the STJ moments. If the individual wants to remain in upright balance during relaxed bipedal stance, then the contractile activity within the gastrocnemius and soleus muscles will need to be varied by the CNS in order to simultaneously maintain rotational equilibrium at both the ankle joints and STJ joints.

In the diagram below, the plantar aspect of the right foot of an individual that weighs 800 N is illustrated. Each foot is given to bear 400 N of GRF on its plantar aspect. It is given that the Achilles tendon tensile force is directed vertically upward, 2.0 cm medial to the STJ axis and 6.0 cm posterior to the ankle joint axis. No other ankle joint plantarflexors of dorsiflexors are active. The CoP is located plantar to the 3rd metatarsal shaft and is directed vertically upward at distance of 3.0 cm lateral to the STJ axis and an unknown distance of Z cm anterior to the ankle joint axis. Both the STJ and ankle joints are in rotational equilibrium.

Answer the following questions:

1. What is the Achilles tendon tensile force?

2. What is the distance from the CoP to the ankle joint axis?

3. Why must the CoP always be lateral to the STJ axis in the normal foot when the CoP is anterior to the ankle joint axis during relaxed bipedal stance?

4. If the CoP was directly plantar to the ankle joint axis, where would the CoP lie relative to the STJ axis? Why?

5. If the Achilles tendon was surgically moved to a point 4.0 cm medial to the STJ axis, where would the CoP need to be positioned on the plantar foot in order to maintain rotational equilibrum at the STJ axis during relaxed bipedal stance?

6. Assuming that the weight of the foot is 30 N, what would be the interosseous compression force at the ankle joint?

7. If the STJ axis was internally rotated relative to that illustrated below, so that the pronation moment arm to CoP was 4.5 cm (assuming the same magnitudes of CoP force and Achilles tendon tensile force as illustrated below), where would the posterior aspect of the STJ axis need to be positioned relative to the Achilles tendon in order to still maintain rotational equilibrium at both the ankle and STJ axes?

8. During the late midstance phase of gait, how might the kinematics of the foot illustrated below be different compared to the foot in question #7? Which of these two feet would have more tendency to undergo late midstance pronation during walking, all other factors being equal?
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Last edited by Kevin Kirby : 17th February 2008 at 10:33 AM.
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Old 16th February 2008, 03:25 PM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

Thanks Kevin for another TE. I don't mind being wrong as long as someone corrects me, so ...
1) 600 N

2) 9 cm

3) With CoP anterior to the ankle joint a tensile force must exist in the TA, which is posterior to ankle joint, to counterbalance. The resultant TA tensile force, with TA medial to STJ axis, must be counterbalanced by the CoP exerting a pronatory moment, only possible from a position lateral to the STJ axis.

4) If you want to maintain the normal relaxed bipedal stance; as CoP is exerting no moments about the ankle jt therefore no counterbalancing TA force is needed. If the CoP was still lateral to the STJ axis then a TA counterbalancing force would be necessary, however this would create moments about the ankle joint which would not be able to be counterbalanced by the "plantar to ankle jt" position of the CoP.

5) As long as this repositioning doesn't affect the perp. distance of TA insertion to the ankle jt, then 4.5 cm.

Thanks again, mark c
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Old 16th February 2008, 07:15 PM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

Quote:
Originally Posted by markjohconley View Post
Thanks Kevin for another TE. I don't mind being wrong as long as someone corrects me, so ...
1) 600 N

2) 9 cm

3) With CoP anterior to the ankle joint a tensile force must exist in the TA, which is posterior to ankle joint, to counterbalance. The resultant TA tensile force, with TA medial to STJ axis, must be counterbalanced by the CoP exerting a pronatory moment, only possible from a position lateral to the STJ axis.

4) If you want to maintain the normal relaxed bipedal stance; as CoP is exerting no moments about the ankle jt therefore no counterbalancing TA force is needed. If the CoP was still lateral to the STJ axis then a TA counterbalancing force would be necessary, however this would create moments about the ankle joint which would not be able to be counterbalanced by the "plantar to ankle jt" position of the CoP.

5) As long as this repositioning doesn't affect the perp. distance of TA insertion to the ankle jt, then 4.5 cm.

Thanks again, mark c

Mark:

Questions #1 and #2 are correct. Good job. Can you show everyone the equations you used to come to these answers?

Question #3 was: Why must the CoP always be lateral to the STJ axis in the normal foot when the CoP is anterior to the ankle joint axis during relaxed bipedal stance? Mark, your answer was basically correct. However, I would answer this question as follows:

The CoP must be lateral to the STJ in the normal foot when the CoP is anterior to the ankle joint axis during relaxed bipedal stance since when the CoP is anterior to the ankle joint, CoP creates an ankle joint dorsiflexion moment which must be counterbalanced by an ankle joint plantarflexion moment from Achilles tendon tension in order to maintain ankle joint rotational equilibrium and prevent the individual from falling forward or backward. The necessary Achilles tendon tensile force, due to the medial position of the Achilles tendon to the STJ axis will also create a STJ supination moment. In order for the foot to not supinate toward the maximally supinated position due to this Achilles tendon tensile force, the forefoot CoP must be positioned lateral to the STJ. A CoP which is positioned lateral to the STJ axis will produce a STJ pronation moment, counterbalance the STJ supination moment from Achilles tendon tensile force and create rotational equilibrium across the STJ so that the STJ can be in a stable rotational position during relaxed bipedal stance.

Question #4 was not answered correctly. Read the question again.

Question #5 was not answered correctly. Try again. Please show your work so everyone can learn.

Mark, you are almost there. Anyone else want to also give the questions a try??
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Old 16th February 2008, 08:31 PM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

Goodaye Kevin, thanks for the reply.

1) Let p be (TA tensile force)
((6 x 10 -2 m) x (p N)) Nm = ((9 x 10 -2 m) x (400 N)) Nm
p = 60 N

2) Let q be the perp. distance from the CoP to the ankle joint
((400 N) x (q m)) Nm = ((600 N) x (6 x 10-2 m))
q = 9 x 10 -2 m

3) exactly

4) & 5) I'm waiting for Asher's reply! too diffucult for me ... bugger, piddle, poop
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Old 16th February 2008, 09:42 PM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

Quote:
Originally Posted by Kevin Kirby View Post
Question #4 was not answered correctly. Read the question again.

Question #5 was not answered correctly. Try again. Please show your work so everyone can learn.

Mark, you are almost there. Anyone else want to also give the questions a try??
I will attempt to give it a go.

Question 4: If the COP was located directly to the ankle joint axis then no dorsiflexion plantarflexion moment would be occurring from the COP. Therefore there would be no need for the TA to exert a plantarflexion force, resulting in no supinatory force from this contraction around the STJ axis. (Therefore no COP would be located laterally to the STJ axis in the forefoot where position X is on the diagram). I would then assume the COP would be then located directly plantar to the ankle joint axis and STJ axis.

Question 5: The new TA force is 1200Nm after being relocated 4cm medially to the STJ axis.
The COP would need to be 9cm laterally from STJ axis (instead of 3cm when the TA is located 2cm medially from the STJ axis).

Not completely confident with these answers....
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Old 16th February 2008, 10:02 PM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

Quote:
Originally Posted by markjohconley View Post
Goodaye Kevin, thanks for the reply.

1) Let p be (TA tensile force)
((6 x 10 -2 m) x (p N)) Nm = ((9 x 10 -2 m) x (400 N)) Nm
p = 60 N

2) Let q be the perp. distance from the CoP to the ankle joint
((400 N) x (q m)) Nm = ((600 N) x (6 x 10-2 m))
q = 9 x 10 -2 m

3) exactly

4) & 5) I'm waiting for Asher's reply! too diffucult for me ... bugger, piddle, poop
Mark:

1. Should be 600 N, not 60 N, using your formula.
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Old 16th February 2008, 10:06 PM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

Quote:
Originally Posted by Dean Hartley View Post
I will attempt to give it a go.

Question 4: If the COP was located directly to the ankle joint axis then no dorsiflexion plantarflexion moment would be occurring from the COP. Therefore there would be no need for the TA to exert a plantarflexion force, resulting in no supinatory force from this contraction around the STJ axis. (Therefore no COP would be located laterally to the STJ axis in the forefoot where position X is on the diagram). I would then assume the COP would be then located directly plantar to the ankle joint axis and STJ axis.
Dean: You are correct. However, the correct name for the moment from CoP about the ankle is a dorsiflexion moment.

Quote:
Originally Posted by Dean
Question 5: The new TA force is 1200Nm after being relocated 4cm medially to the STJ axis.
The COP would need to be 9cm laterally from STJ axis (instead of 3cm when the TA is located 2cm medially from the STJ axis).

Not completely confident with these answers....
Dean, sorry wrong answer. I think that if you and Mark did the equations again carefully, you would both come up with the correct answer.
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Old 16th February 2008, 10:58 PM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

Quote:
Originally Posted by markjohconley View Post

4) If you want to maintain the normal relaxed bipedal stance; as CoP is exerting no moments about the ankle jt therefore no counterbalancing TA force is needed. If the CoP was still lateral to the STJ axis then a TA counterbalancing force would be necessary, however this would create moments about the ankle joint which would not be able to be counterbalanced by the "plantar to ankle jt" position of the CoP.
I left the concluding sentence off, which was “ Therefore the CoP would have to be directly plantar to the STJ axis”


The answer to part 1) should be

Let y be the TA tensile force magnitude, then to calculate y by considering moments about STJ axis

(2 x 10 -2 m) x (y N) Nm = (3 x 10 -2 m) x (400 N) Nm
y = 600
Therefore the TA tensile force is 600 N



Part 5)

The TA tensile force would not change as GRF and perp. distances to the ankle jt axis remain the same.

Thus to the STJ axis moments

Let m be the magnitude of the perp. distance of GRF from the STJ axis, then

((m cm) x (400 N)) Nm = ((4 cm) x (600 N)) Nm
m = 6
Therefore the GRF would have to be positioned 6 cm perpendicular distance lateral to the STJ axis.


mark c
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Old 16th February 2008, 11:21 PM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

Quote:
Originally Posted by markjohconley View Post
I left the concluding sentence off, which was “ Therefore the CoP would have to be directly plantar to the STJ axis”


The answer to part 1) should be

Let y be the TA tensile force magnitude, then to calculate y by considering moments about STJ axis

(2 x 10 -2 m) x (y N) Nm = (3 x 10 -2 m) x (400 N) Nm
y = 600
Therefore the TA tensile force is 600 N



Part 5)

The TA tensile force would not change as GRF and perp. distances to the ankle jt axis remain the same.

Thus to the STJ axis moments

Let m be the magnitude of the perp. distance of GRF from the STJ axis, then

((m cm) x (400 N)) Nm = ((4 cm) x (600 N)) Nm
m = 6
Therefore the GRF would have to be positioned 6 cm perpendicular distance lateral to the STJ axis.


mark c
Good job, Mark. The CoP would need to be positioned 6.0 cm lateral to the STJ axis in order to exert sufficient STJ pronation moment to counterbalance the increased STJ supination moment caused by the change in Achilles tendon supination moment arm from 2.0 cm to 4.0 cm.

Now on to the next question for Mark or any others following along:

6. Assuming that the weight of the foot is 30 N, what would be the interosseous compression force at the ankle joint in Thought Experiment #9? Please show the calculations.
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Old 16th February 2008, 11:29 PM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

Quote:
Originally Posted by markjohconley View Post
Part 5)

The TA tensile force would not change as GRF and perp. distances to the ankle jt axis remain the same.

Thus to the STJ axis moments

Let m be the magnitude of the perp. distance of GRF from the STJ axis, then

((m cm) x (400 N)) Nm = ((4 cm) x (600 N)) Nm
m = 6
Therefore the GRF would have to be positioned 6 cm perpendicular distance lateral to the STJ axis.


mark c
oh whoops. Nice job Mark!
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Old 17th February 2008, 10:31 AM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

Mark, Dean and Others Following Along:

I have added question #6 and two new questions, #7 and #8, to the original posting in this thread to tidy up things a bit. Please have a go at it. You guys are doing great.
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Old 18th February 2008, 04:54 AM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

Quote:
Originally Posted by Kevin Kirby View Post
Mark, Dean and Others Following Along:

I have added question #6 and two new questions, #7 and #8, to the original posting in this thread to tidy up things a bit. Please have a go at it. You guys are doing great.
Ok I will attempt.

Q6 Im a little confused on. Need a little guidance with this. Understand interosseous ankle compression, but stumped how to calculate this. Im gathering you need to incorporate the body weight. But a little stuck here.

Q7 The STJ would be need to be positioned 3cm from the TA in order for rotational equilibrium at the STJ and ankle joint.

My working:
CoP = 400 N x 4.5cm
= 1800Nm

Let x be moment arm (cms) of TA to STJ axis.
600x = 1800
x = 3cm

Q8 THe modified foot in Q7 with internally positioned STJ axis would tend to have increased pronation moments through late midstance of gait. This being due to the increase in GRF occuring as the centre of mass becomes positioned more anteriorly as body weight goes forward during late midstance. This increase in the GRF together with the increase in the moment arm of the CoP will increase the CoP lateral to the STJ axis (exacerbated in the foot with the internally positioned STJ axis in Q7 and resultant increased moment arm), being inclined to pronate more through midstance.
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Old 18th February 2008, 08:02 AM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

Quote:
Originally Posted by Dean Hartley View Post
Ok I will attempt.

Q6 Im a little confused on. Need a little guidance with this. Understand interosseous ankle compression, but stumped how to calculate this. Im gathering you need to incorporate the body weight. But a little stuck here.
Interosseous ankle joint compression force is simply that joint compresion force between the talus and tibia. You know the forces from GRF on the plantar foot, the gravitational force from the mass of the foot, and the vertical muscle contractile forces. You need no other information to solve this question.

Quote:
Q7 The STJ would be need to be positioned 3cm from the TA in order for rotational equilibrium at the STJ and ankle joint.

My working:
CoP = 400 N x 4.5cm
= 1800Nm

Let x be moment arm (cms) of TA to STJ axis.
600x = 1800
x = 3cm
Correct!

Quote:
Q8 THe modified foot in Q7 with internally positioned STJ axis would tend to have increased pronation moments through late midstance of gait. This being due to the increase in GRF occuring as the centre of mass becomes positioned more anteriorly as body weight goes forward during late midstance. This increase in the GRF together with the increase in the moment arm of the CoP will increase the CoP lateral to the STJ axis (exacerbated in the foot with the internally positioned STJ axis in Q7 and resultant increased moment arm), being inclined to pronate more through midstance.
Correct and good answer! The more internally rotated STJ axis, even though this foot may be in STJ rotational equilibrium in standing, will have greater magnitudes of external STJ pronation moments from GRF than a foot with a less internally rotated STJ axis during late midstance.

Could feet with more internally rotated STJ axes (laterally located at the heel and medially located at the forefoot) be feet that also tend to show more late midstance pronation?? Theoretically, yes, but we need research to show this is the case or not.

Good job Dean. Give #6 a try.
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Old 18th February 2008, 09:00 AM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

OK, everyone. Only one more question to go before they are all anwered. Good job so far to Mark and Dean.

Here it is:

6. Assuming that the weight of the foot is 30 N, what would be the interosseous compression force at the ankle joint (i.e. force between the superior surface of the talus and inferior surface of the tibia)?
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Old 18th February 2008, 12:03 PM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

Quote:
Originally Posted by Kevin Kirby View Post
OK, everyone. Only one more question to go before they are all anwered. Good job so far to Mark and Dean.

Here it is:

6. Assuming that the weight of the foot is 30 N, what would be the interosseous compression force at the ankle joint (i.e. force between the superior surface of the talus and inferior surface of the tibia)?
Hint: Newton's 3rd Law. Bigger hint, look at Kevin's avatar
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Old 18th February 2008, 06:00 PM
Dean Hartley Dean Hartley is offline
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

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OK, everyone. Only one more question to go before they are all anwered. Good job so far to Mark and Dean.

Here it is:

6. Assuming that the weight of the foot is 30 N, what would be the interosseous compression force at the ankle joint (i.e. force between the superior surface of the talus and inferior surface of the tibia)?
370N of interosseous compression force at the ankle.
Unsure sorry.

Anytime you want to jump in here Mark! My physics knowledge is rather limited although thought it would have something to do the "every action has an equal and opposite reaction etc"
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Old 18th February 2008, 10:14 PM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

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370N of interosseous compression force at the ankle.
Unsure sorry.

Anytime you want to jump in here Mark! My physics knowledge is rather limited although thought it would have something to do the "every action has an equal and opposite reaction etc"

Dean:

Your answer would be correct only if the Achilles tendon was not pulling vertically upward on the foot with a force of 600 N. The larger the contractile force of an extrinsic foot muscle, the greater the ankle joint interosseous compression force.

Here's another hint. When doing such a problem, make sure that all the forces acting on the foot add up to equal zero, or, in other words, make sure that all the forces acting in opposite directions to each other exactly equal each other. You can assume that all the opposing forces are equal since the foot is not moving and is in static equilibrium.
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Old 18th February 2008, 10:25 PM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

Dave Smith sent me in all the answers privately, and he got them all right. If no one else gets the answer to #6 within the next few days, I'll let Dave, Eric or Simon give you all the correct answer.
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Old 19th February 2008, 06:12 AM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

Hi everyone,

have been following with some interest.

Haven't had to do physics calculations for some time.

But I thought i'd just give it a go - probably way off the mark.

Anyway since 'compression' must have two forces acting in opposite directions and assuming to get an equilibrium/balance the forces must cancel each other out.

I've come up with 450N for the compression,as we know what the GRF, Gravity and AT forces are.


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Old 19th February 2008, 07:04 AM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

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Hi everyone,

have been following with some interest.

Haven't had to do physics calculations for some time.

But I thought i'd just give it a go - probably way off the mark.

Anyway since 'compression' must have two forces acting in opposite directions and assuming to get an equilibrium/balance the forces must cancel each other out.

I've come up with 450N for the compression,as we know what the GRF, Gravity and AT forces are.


AD
AD:

Sorry, your answer is incorrect. Force from Achilles tendon = 600 N upward. Weight from foot = 30 N downward. Force from GRF = 400 N upward. The problem can easily be solved using these three variables.

It helps if you know how to use free body diagrams. but the problem may also be solved intuitively using simple math.
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Old 19th February 2008, 04:00 PM
Dean Hartley Dean Hartley is offline
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

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Originally Posted by Kevin Kirby View Post
Dean:

Your answer would be correct only if the Achilles tendon was not pulling vertically upward on the foot with a force of 600 N. The larger the contractile force of an extrinsic foot muscle, the greater the ankle joint interosseous compression force.

Here's another hint. When doing such a problem, make sure that all the forces acting on the foot add up to equal zero, or, in other words, make sure that all the forces acting in opposite directions to each other exactly equal each other. You can assume that all the opposing forces are equal since the foot is not moving and is in static equilibrium.
so it may be 970N of ankle compression?
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Old 19th February 2008, 04:48 PM
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Default Re: T.E. #9: Simultaneous Ankle and STJ Equilibriums

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so it may be 970N of ankle compression?
Yes, Dean, very nicely done. The 400 N of GRF on the plantar foot pushes the foot upward. The Achilles tendon tensile force of 600 N pulls the foot upward. Therefore, there is a net upward force on the foot of 1,000 N. Since gravity accelerates the mass of the foot with a force of 30 N, then some other force must be pushing the foot downward with a force equal to: 1,000 N - 30 N = 970 N. That 970 N of downward force comes from the distal tibial pushing downward on top of the talus.

Therefore, in order to have static equilibrium, so that the foot is not accelerating, but at rest, upward forces acting on the foot must equal downward forces acting on the foot:

Quote:
Upward forces = downward forces

GRF + Achilles tendon force = (Acceleration of gravity)(mass of foot) + tibial compression force onto talus

400 N + 600 N = 30 N + 970 N
Question #6 emphasizes the point that the ankle joint is subjected to much greater compression forces than the body weight of the individual. This occurs due to the upward pulling forces from the extrinsic muscles of the foot which help to further compress the ankle joint, especially during the late midstance phase of gait.
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