To solve part (b) of the problem, we calculate the forces involved in the scenario where the forearm is held in static equilibrium. Here’s how:

Problem Setup:

The system is in static equilibrium, meaning the sum of torques and forces acting on the forearm must be zero. The forces involved are:

1. $\mathbf{F}_g = 200 \, \text{N}$, acting vertically downward at a distance of $40 \, \text{cm}$ from the elbow joint (pivot).
2. $\mathbf{F}_{\text{Tendon}}$, at an angle of $80^\circ$ from the horizontal, located $4 \, \text{cm}$ from the elbow joint.
3. $\mathbf{F}_{\text{Forearm}}$, exerted by the upper arm at the elbow joint, acts to balance the forces horizontally and vertically.

Key assumptions:

• Ignore the weight of the forearm.
• Solve first for $\mathbf{F}_{\text{Tendon}}$ using the torque equilibrium equation.

Step 1: Torque Equilibrium

The torques are calculated about the elbow joint (pivot point). For equilibrium: $\sum \tau = 0$ $\mathbf{F}_g \cdot d_g = \mathbf{F}_{\text{Tendon}} \cdot d_{\text{Tendon}} \cdot \sin(80^\circ)$ Substitute known values: $200 \cdot 40 = \mathbf{F}_{\text{Tendon}} \cdot 4 \cdot \sin(80^\circ)$ $8000 = \mathbf{F}_{\text{Tendon}} \cdot 4 \cdot 0.9848$ $\mathbf{F}_{\text{Tendon}} = \frac{8000}{4 \cdot 0.9848} = 2030.2 \, \text{N}$

Step 2: Force Equilibrium

The forearm is in static equilibrium, so: $\sum F_x = 0 \quad \text{and} \quad \sum F_y = 0$

Horizontal force equilibrium:

$F_{\text{Forearm}, x} = F_{\text{Tendon}} \cdot \cos(80^\circ)$ $F_{\text{Forearm}, x} = 2030.2 \cdot \cos(80^\circ) = 2030.2 \cdot 0.1736 = 352.3 \, \text{N}$

Vertical force equilibrium:

$F_{\text{Forearm}, y} + F_{\text{Tendon}} \cdot \sin(80^\circ) = F_g$ $F_{\text{Forearm}, y} + 2030.2 \cdot 0.9848 = 200$ $F_{\text{Forearm}, y} = 200 - 1998.5 = -1798.5 \, \text{N}$

The total magnitude of $\mathbf{F}_{\text{Forearm}}$ is: $\mathbf{F}_{\text{Forearm}} = \sqrt{F_{\text{Forearm}, x}^2 + F_{\text{Forearm}, y}^2}$ $\mathbf{F}_{\text{Forearm}} = \sqrt{352.3^2 + (-1798.5)^2}$ $\mathbf{F}_{\text{Forearm}} = \sqrt{124113.3 + 3234572.25} = \sqrt{3358685.6}$ $\mathbf{F}_{\text{Forearm}} = 1833.8 \, \text{N}$

Final Answers:

1. $\mathbf{F}_{\text{Tendon}} = 2030.2 \, \text{N}$
2. $\mathbf{F}_{\text{Forearm}} = 1833.8 \, \text{N}$

Let me know if you'd like more details or clarification!

Related Questions:

1. How does changing the angle of $\mathbf{F}_{\text{Tendon}}$ affect the torque balance?
2. What would happen if the weight of the forearm is included in the calculations?
3. How is torque different from force in these kinds of problems?
4. What is the role of sine and cosine in torque calculations?
5. How does increasing the dumbbell weight affect the forces?

Tip:

Always pick the pivot point carefully when solving torque problems; it simplifies calculations significantly!