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### Problem Analysis
The arm is horizontal, and we need to calculate the torques about the elbow joint due to:
1. The weight of the forearm/hand.
2. The weight of the ball.
3. The force applied by the biceps.

We’ll use the following values provided:
- Weight of the ball, \( w_b = 1.80 \, \text{N} \)
- Mass of the forearm/hand, \( m_f = 1.30 \, \text{kg} \)
- Gravitational acceleration, \( g = 9.81 \, \text{m/s}^2 \)
- Force exerted by the biceps, \( F_b = 13.0 \, \text{N} \)
- Distance from elbow to ball, \( 2L = 25.0 \, \text{cm} \)
- Distance from elbow to center of mass of forearm, \( L = 12.5 \, \text{cm} \)
- Distance from elbow to point of biceps attachment, \( d = 3.60 \, \text{cm} \)

### Solution Steps
Let's calculate each torque one by one, using the formula \( \tau = r \cdot F \), where \( r \) is the distance from the axis of rotation (the elbow joint) and \( F \) is the force acting perpendicular to that distance.

#### 1. Torque Due to the Weight of the Forearm/Hand
The weight of the forearm/hand is \( F_f = m_f \cdot g = 1.30 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 12.753 \, \text{N} \).

- Distance from elbow to the center of mass of the forearm: \( L = 12.5 \, \text{cm} = 0.125 \, \text{m} \).
- Torque due to the forearm weight:
  \[
  \tau_f = L \cdot F_f = 0.125 \, \text{m} \times 12.753 \, \text{N} = 1.594125 \, \text{N} \cdot \text{m}
  \]

#### 2. Torque Due to the Weight of the Ball
The force due to the weight of the ball is \( w_b = 1.80 \, \text{N} \).

- Distance from elbow to the ball: \( 2L = 25.0 \, \text{cm} = 0.25 \, \text{m} \).
- Torque due to the weight of the ball:
  \[
  \tau_b = 2L \cdot w_b = 0.25 \, \text{m} \times 1.80 \, \text{N} = 0.45 \, \text{N} \cdot \text{m}
  \]

#### 3. Torque Due to the Biceps Force
The force applied by the biceps is \( F_b = 13.0 \, \text{N} \).

- Distance from elbow to point of biceps attachment: \( d = 3.60 \, \text{cm} = 0.036 \, \text{m} \).
- Torque due to the biceps force (acts in the opposite direction to the other torques, so it will be subtracted):
  \[
  \tau_biceps = d \cdot F_b = 0.036 \, \text{m} \times 13.0 \, \text{N} = 0.468 \, \text{N} \cdot \text{m}
  \]

#### 4. Net Torque Calculation
To find the net torque acting about the elbow, we set the sum of torques due to weights equal to the torque due to the biceps:
\[
\tau_{\text{net}} = \tau_biceps - (\tau_f + \tau_b) = 0.468 \, \text{N} \cdot \text{m} - (1.594125 \, \text{N} \cdot \text{m} + 0.45 \, \text{N} \cdot \text{m})
\]
\[
\tau_{\text{net}} = 0.468 - 2.044125 = -1.576125 \, \text{N} \cdot \text{m}
\]

### Answer
The magnitude of the net torque about the elbow joint is approximately:
\[
|\tau_{\text{net}}| = 1.576 \, \text{N} \cdot \text{m}
\]

Would you like a further breakdown, or do you have any other questions? 

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Here are five related questions to explore:
1. What would happen if the biceps force increased?
2. How does changing the distance \( d \) of the biceps attachment affect the torque?
3. How would the net torque change if the arm was not horizontal?
4. What is the effect of using a heavier object than the softball?
5. How would the torque change if the forearm was longer?

**Tip:** Remember that torque depends on both the force and its distance from the rotation point, meaning small changes in distance can significantly impact torque.

Using your elbow joint as the axis of rotation, determine the magnitude of the net torque acting about your elbow due to the weight of your forearm/hand, the weight of the ball, and the force applied by your bicep.

Learn how to calculate the net torque acting on an elbow joint in a horizontal position holding a 1.8 N ball, with forces exerted by the biceps and gravity. This physics problem demonstrates torque equilibrium concepts using distances and forces.

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