91Ó°ÊÓ

First, draw a free-body diagram that includes the forces acting on the leg and their respective lever arms. This will help us visualize the relationship between the forces and distances involved in the problem. Include the force of the lower leg's weight, the force of the ankle weight, and the force exerted by the patella tendon.

Set up an equation for the net torque on the lower leg about the knee joint. The net torque should be equal to zero since the leg is not accelerating. The torque due to weight of the leg, torque due to the ankle weight, and torque due to the force from the patella tendon should be included in the equation. The equation looks like: \(\text{Torque due to leg's weight} + \text{Torque due to ankle weight} - \text{Torque due to patella tendon} = 0\)

To calculate the torque due to the force of the leg's weight, multiply the force of the weight by the perpendicular distance from the knee joint to the line of action of the weight. The force of the weight is equal to the mass of the leg times gravity: \(mg = 5.0 \:\text{kg} \cdot 9.81 \:\text{m/s}^2 = 49.05 \:\text{N}\). The perpendicular distance can be found using the right triangle formed by the angle between the leg and the vertical and the distance from the center of mass of the leg to the knee: \(d_\text{leg} = 0.22 \mathrm{m} \sin{(\theta)}\). So the torque due to leg's weight is: \(49.05 \:\text{N} \cdot d_\text{leg}\).

To calculate the torque due to the force of the ankle weight, multiply the force of the weight by the perpendicular distance from the knee joint to the line of action of the weight. The force of the weight is equal to the mass of the ankle weight times gravity: \(mg = 3.0 \:\text{kg} \cdot 9.81\:\text{m/s}^2 = 29.43 \:\text{N}\). The perpendicular distance can be found using the right triangle formed by the angle between the leg and the vertical and the distance from the ankle weight to the knee: \(d_\text{ankle} = 0.41 \mathrm{m} \sin{(\theta)}\). So the torque due to ankle weight is: \(29.43 \:\text{N} \cdot d_\text{ankle}\).

The torque due to the force from the patella tendon can be calculated by multiplying the force exerted by the patella tendon \(F_\text{tendon}\) by the perpendicular distance from the knee joint to the line of action of the force. The perpendicular distance can be found using the right triangle formed by the angle between the lower leg and the direction of the patella tendon force: \(d_\text{tendon} = 0.10 \mathrm{m} \sin{(20.0^\circ)}\). So the torque due to force from the patella tendon is: \(F_\text{tendon} \cdot d_\text{tendon}\).

Now plug the torque due to the leg's weight, torque due to the ankle weight, and torque due to force from the patella tendon into the torque equilibrium equation and solve for \(F_\text{tendon}\). For \(\theta = 30.0^\circ\): \(49.05 \:\text{N} \cdot 0.22 \mathrm{m} \sin{(30.0^\circ)} + 29.43 \:\text{N} \cdot 0.41 \mathrm{m} \sin{(30.0^\circ)} = F_\text{tendon} \cdot 0.10 \mathrm{m} \sin{(20.0^\circ)}\) Solve for \(F_\text{tendon}\), we get \(F_\text{tendon} \approx 316.57 \:\text{N}\) For \(\theta = 90.0^\circ\): \(49.05 \:\text{N} \cdot 0.22 \mathrm{m} \sin{(90.0^\circ)} + 29.43 \:\text{N} \cdot 0.41 \mathrm{m} \sin{(90.0^\circ)} = F_\text{tendon} \cdot 0.10 \mathrm{m} \sin{(20.0^\circ)}\) Solve for \(F_\text{tendon}\), we get \(F_\text{tendon} \approx 214.46 \:\text{N}\) So, the force exerted by the patella tendon to hold the leg at an angle of (a) \(30.0^{\circ}\) is approximately \(316.57 \:\text{N}\) and (b) \(90.0^{\circ}\) is approximately \(214.46 \:\text{N}\).