To solve this problem, let's break it down into three parts as requested.

Part (a): Free-Body Diagrams (FBDs)

1. Free-Body Diagram of the Leg

   • Forces: There is a horizontal traction force of 50 N acting on the leg. The direction is horizontal because the doctor specified that the traction force should be horizontal.
   • Action/Reaction Pairs: The tension $T$ in the rope exerts an equal and opposite force on the leg, creating this traction force.

2. Free-Body Diagram of the Pulley

   • Forces: The pulley is in static equilibrium, so the forces must balance. There are two components of tension in the rope on either side of the pulley at an angle $\theta$ with respect to the horizontal.
   • Action/Reaction Pairs: Each tension force from the rope acts on the pulley, balanced by the reaction forces on the rope by the pulley.

3. Free-Body Diagram of the Mass

   • Forces: The mass hanging from the rope experiences a downward gravitational force $F_g = mg$, where $m = 4.2 \, \text{kg}$ and $g = 9.8 \, \text{m/s}^2$.
   • Action/Reaction Pairs: The weight provides tension in the rope, which is transmitted to the leg through the pulley.

Part (b): Newton's Second Law Equations for Each Component

1. For the Leg (horizontal equilibrium):

   • Let $T$ be the tension in each side of the rope.
   • Since the traction force $F_{\text{traction}} = 50 \, \text{N}$ is horizontal: $F_{\text{traction}} = 2T \cos(\theta) = 50 \, \text{N}$
   • In this case, the vertical components of the tension on each side of the rope will cancel out, ensuring that the net force on the leg remains horizontal.

2. For the Pulley (static equilibrium):

   • Horizontal: The horizontal components of tension on either side of the rope must be equal and opposite to balance each other. $T \cos(\theta) - T \cos(\theta) = 0$
   • Vertical: The vertical components of tension support the weight of the hanging mass. $2T \sin(\theta) = mg$

3. For the Mass (vertical equilibrium):

   • Since the mass is in equilibrium, the tension in the rope supports the gravitational force. $T = mg$

Part (c): Solving for the Traction Angle $\theta$

To find $\theta$, we can use the two equations from the components:

1. From the leg's equilibrium: $2T \cos(\theta) = 50 \, \text{N}$

2. From the pulley's vertical equilibrium: $2T \sin(\theta) = mg$

Substituting $T = mg$ into the equations:

1. Using $T = mg$: $2mg \cos(\theta) = 50$

2. Rearranging to solve for $\cos(\theta)$: $\cos(\theta) = \frac{50}{2 \cdot mg} = \frac{50}{2 \cdot 4.2 \cdot 9.8}$

Calculating: $\cos(\theta) \approx \frac{50}{82.32} \approx 0.607$ 3. Then, $\theta = \cos^{-1}(0.607) \approx 52.5^\circ$

Thus, the proper angle $\theta$ is approximately 52.5 degrees.

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Do you have any questions or would you like additional details on any step? Here are five related questions:

1. How does the angle $\theta$ affect the traction force in such a setup?
2. Why is it necessary for the rope to make the same angle on both sides of the pulley?
3. How would the required angle change if the doctor prescribed a different traction force?
4. What would happen if we used a different mass for