We can solve this problem using the principles of static equilibrium.
First, let's label the given quantities:
Mass of person (m) = 65 kg
Gravitational acceleration (g) = 9.81 m/s²
Ratio of D to d = 2
When the person is in equilibrium, the upward force from the Achilles tendon (FT) and the downward force from the lower leg bone (FB) should balance the downward gravitational force (Fg) acting on the person. Also, the torques about any point should sum to zero.
1. Balancing vertical forces:
FT + FB = Fg
FT + FB = mg
2. Balancing torques:
For this, we can consider torques about point B, where the FB force acts. The distances of the Achilles tendon force (FT) and gravitational force (mg) are d and D, respectively, from point B.
Torque due to FT = d * FT
Torque due to mg = D * mg
Since D is twice as long as d, D = 2d. Equating the torques in equilibrium, we get:
d * FT = 2d * mg
Dividing both sides by d, we get:
FT = 2mg
Now we can plug in the values to find the tension FT in the Achilles tendon:
FT = 2 * (65 kg) * (9.81 m/s²)
FT ≈ 1273 N
This is the tension in the Achilles tendon, pulling upwards.
Now, we can use the vertical force equation to find FB:
FB = mg - FT
FB = (65 kg * 9.81 m/s²) - 1273 N
FB ≈ 390 N
This is the downward force exerted by the lower leg bone on the foot.
So, the tension in the Achilles tendon is 1273 N (upwards), and the force exerted by the lower leg bone on the foot is 390 N (downwards).
The Achilles tendon is attached to the rear of the foot as shown in (Figure 1) . A person elevates himself just barely off the floor on the "ball of one foot." Assume the person has a mass of 65kg and D is twice as long as d.
Find the tension FT in the Achilles tendon (pulling upward).
Find the (downward) force FB exerted by the lower leg bone on the foot.
1 answer