# How to calculate the force exerted by the "tendon" on the "tibia" bone?

• Iwawa48
Iwawa48
Thread moved from the technical forums to the schoolwork forums
TL;DR Summary: how to calculate the force exerted by the "tendon" and the resistance occurring on the "tibia" if a 70kg man leans on his tip of the foot.

Hello,
This is the question:
"A 70 kg person stands on the tip of the foot. Assuming a configuration as in Figure 5.6 calculate a) the force P exerted by the tendon; b) the resistance R acting on the tibia; c) the percentage deformation of the tibia, assuming that the section of the tibia is 3.8x10-4m^2 and the Young's modulus is Y=1.6x10^10N/M^2"
This is the image 5.6

This was my reasoning:
I first calculate the weight:
2 W = m* g = 70kg *9.8m/ s = 686N
Remembering that a rigid body is in equilibrium when the results of forces and moments are zero, I write the following conditions:

P is congruent with the result of the exercise, instead R is different; the exercise results in 1372N which by my reasoning is impossible.
What am I doing wrong?
To calculate point c) I would have to apply Hooke's law which depends on the value of R, and of course this result is also wrong.
Thank you and best regards
Gaetano

Iwawa48 said:
"A 70 kg person stands on the tip of the foot. ...

2 W = m* g = 70kg *9.8m/ s = 686N
Ther problem statement sounds like standing on one foot only to me so you would drop that factor 2

Iwawa48 said:
Remembering that a rigid body is in equilibrium when the results of forces and moments are zero, I write the following conditions:
Both equilibrium equations seem wrong. Think again and use the force convention from the diagram, where all 3 forces are the ones acting on the foot.

For the moment equilibrium on the foot:
What pivot did you choose?
What are the lever arms of each force around that pivot?
Which forces produce clockwise and which counter-clockwise moments?

For the force equilibrium the foot:
Which forces point up, which down?

Assume standing on the tip of one foot, not two.
70 kg supported by the toe, W = m⋅g = 70 * 9.8 = 686 N.
The bottom of the tibia is a fulcrum.
Torque about fulcrum is zero; P * 5 cm = W * 15 cm ;
P = 686 * (15 / 5) = 2058 N.
R = W + P = 2058 + 686 N = 2744 N.

MatinSAR, Lnewqban and A.T.
Baluncore said:
Assume standing on the tip of one foot, not two.
70 kg supported by the toe, W = m⋅g = 70 * 9.8 = 686 N.
The bottom of the tibia is a fulcrum.
Torque about fulcrum is zero; P * 5 cm = W * 15 cm ;
P = 686 * (15 / 5) = 2058 N.
R = W + P = 2058 + 686 N = 2744 N.
In other words, your tibia is compressed by 4 times your body weight here. And that is just the optimal static 2D case which doesn't include the muscle forces for moments in the other planes, antagonistic muscle co-contraction, and additional ground reaction for accelerating your body up, like in jumping off.

Last edited:
Lnewqban
Baluncore said:
The bottom of the tibia is a fulcrum.
Torque about fulcrum is zero; P * 5 cm = W * 15 cm ;
Yes, though taking the toes as the axis is also valid. But then the equation is R*15cm = P*(5+15)cm. The OP seems to have been inconsistent about the choice.

MatinSAR

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