Question 7.11: Standing with slightly elevated heel Suppose you stand on yo......
Standing with slightly elevated heel
Suppose you stand on your toes with your heel slightly off the ground. In order to do this, the larger of the two lower leg bones (the tibia) exerts a force on the ankle joint where it contacts the foot. The Achilles tendon simultaneously exerts a force on the heel, pulling up on it in order for the foot to be in static equilibrium. What is the magnitude of the force that the tibia exerts on the ankle joint? What is the magnitude of the force that the Achilles tendon exerts on the heel?
Sketch and translate First we sketch the foot with the Achilles tendon and the tibia. We choose the foot as the system of interest. Three forces are exerted on the foot: the tibia is pushing down on the foot at the ankle joint; the floor is pushing up on the ball of the foot and the toes; and the Achilles tendon is pulling up on the heel. We choose the axis of rotation as the place where the tibia presses against the foot.
Simplify and diagram Model the foot as a very light rigid body. The problem says that the foot is barely off the ground, so we will neglect the angle between the foot and the ground and consider the foot horizontal. The gravitational force exerted on the foot by Earth is quite small compared with the other forces that are being exerted on it, so we will ignore it. A force diagram for the foot is shown below. When you are standing on the ball and toes of both feet, the floor exerts an upward force on each foot equal to half the magnitude of the gravitational force that Earth exerts on your entire body: F_{\text {Floor on Foot }}=\frac{m_{\mathrm{Body}} \text{g}}{2} . The Achilles tendon pulls up on the heel of the foot, exerting a force T_{\text {Tendon on Foot }} .The tibia bone in the lower leg pushes down on the ankle joint exerting a force F_{\text {Tibia on Foot }} \cdot
Represent mathematically Let’s apply the conditions of equilibrium to this system. Note that the distance from the toes to the joint L_{\text {Toe }} is somewhat longer than the distance from the joint to the Achilles tendon attachment point L_{\text {Tendon }} . The torque condition of equilibrium becomes
+\left[T_{\text {Tendon on Foot }}\left(L_{\text {Tendon }}\right)\right]+F_{\text {Bone on Foot }}(0)-\frac{m_{\text {Body }} \text{g}}{2}\left(L_{\text {Toe }}\right)=0
\Rightarrow T_{\text {Tendon on Foot }}=\frac{m_{\text {Body }} \text{g}}{2}\left(\frac{L_{\text {Toe }}}{L_{\text {Tendon }}}\right)Now, apply the y-scalar component of the force condition of equilibrium:
\Sigma F_y=T_{\text {Tendon on Foot }}+\left(-F_{\text {Bone on Foot }}\right)+\frac{m_{\text {Body }} \text{g}}{2}=0\Rightarrow F_{\text {Bone on Foot }}=T_{\text {Tendon on Foot }}+\frac{m_{\text {Body } \text{g}}}{2}
Solve and evaluate The distance from the place where the bone contacts the foot to where the floor contacts the foot is about 5 times longer than the distance from the bone to where the tendon contacts the foot. Consequently, the force that the Achilles tendon exerts on the foot is about
T_{\text {Tendon on Foot }}=\frac{m_{\text {Body }} \text{g}}{2}\left(\frac{L_{\text {Toe }}}{L_{\text {Tendon }}}\right)=\frac{m_{\text {Body }} \text{g}}{2}(5)=\frac{5}{2} m_{\text {Body }} \text{g}
or two and a half times the gravitational force that Earth exerts on the body. Using g = 10 N/kg for a 70-kg person, this force will be about 1750 N. That’s a very large force for something as simple as standing with your heel slightly elevated! The force exerted on the joint by the leg bone would be
F_{\text {Bone on Foot }}=T_{\text {Tendon on Foot }}+\left(\frac{m_{\text {Body }} \text{g}}{2}\right)
= 1750 N + 350 N = 2100 N
This force is three times the weight of the person! The forces are much greater when moving. Thus, every time you lift your foot to walk, run, or jump, the tendon tension and joint compression are several times greater than the gravitational force that Earth exerts on your entire body.
Try it yourself: Estimate the increase in the magnitude of the force exerted by the Achilles tendon on the foot of the person in this example if his mass were 90 kg instead of 70 kg.
Answer: An increase of at least 500 N (110 lb).