A 68 -kg woman stands straight with both feet flat on the floor. Her center of gravity is a horizontal distance of \(3.0 \mathrm{cm}\) in front of a line that connects her two ankle joints. The Achilles tendon attaches the calf muscle to the foot a distance of \(4.4 \mathrm{cm}\) behind the ankle joint. If the Achilles tendon is inclined at an angle of \(81^{\circ}\) with respect to the horizontal, find the force that each calf muscle needs to exert while she is standing. [Hint: Consider the equilibrium of the part of the body above the ankle joint.]

Let F be the force exerted by each calf muscle, and let T be the tension in the Achilles tendon. We will consider the equilibrium of the part of the body above the ankle joint (the woman's body). The weight of the woman (W) acts at the center of gravity and is equal to the mass of the woman times the acceleration due to gravity (W = 68kg * 9.8m/s^2). Draw a diagram with the woman's feet on the floor and the ankle joint as the pivot point. Label the vertical and horizontal distances of center of gravity and the Achilles tendon attachment.

The torque due to the weight will be equal to the weight multiplied by the horizontal distance from the ankle joint to the center of gravity. The horizontal distance is given as 3.0cm (0.03m). Therefore, the torque due to the weight, \(\tau_W\), will be: $$\tau_W = W \times d_W = (68 kg \times 9.8 m/s^2) \times 0.03 m$$ $$\tau_W = 20.0016 Nm$$

For this step, we will use the concept of torque to find the torque due to the force exerted by the calf muscle (\(\tau_F\)). The tendon attaches to the foot at a distance of 4.4cm (0.044m) behind the ankle joint. The angle of the tendon with the horizontal is given as 81 degrees. Since the torque is the cross product of the force and the distance, we have to resolve the force into horizontal and vertical components. The vertical component of the force will produce a torque, while the horizontal component will not. The vertical component of the force, \(F_v\), can be found using the sine of the angle: $$F_v = F \times \sin{81^{\circ}}$$ The torque due to the force exerted by the calf muscle is given by the product of the vertical component of the force and the distance of the tendon attachment from the ankle joint: $$\tau_F = F_v \times d_F = (F \times \sin{81^{\circ}}) \times 0.044m$$

Since the body is in equilibrium, the net torque must be zero. The torque due to the weight of the woman and the torque due to the force exerted by the calf muscle must balance each other: $$\tau_W = \tau_F$$ $$20.0016Nm = (F \times \sin{81^{\circ}}) \times 0.044m$$ Now we can solve for the force exerted by the calf muscle: $$F = \frac{20.0016 Nm}{(0.044m \times \sin{81^{\circ}})}$$ $$F = 555.4 N$$ Since the woman is standing on both feet, each calf muscle needs to exert half of the force calculated: $$F_\text{each} = \frac{555.4 N}{2}$$ $$F_\text{each} = 277.7 N$$ Thus, each calf muscle needs to exert a force of 277.7 N while the woman is standing.