Consider holding a carton of milk in your hand. The force of the bicep muscle actsat an angle of 15 degrees to the vertical, while the weight of the arm and the milk both act downwards. The distance from the elbow to where the bicep muscle is attached via the distal bicep tendons to the radius and ulna bones is 5 cm. the distance from the elbow to the hand, holding the milk, is 35 cm. the forearm has a mass of 4 kg and the milk carton a mass of 2 kg. Assuming the forearm is kept perfectly horizontal, find the tension in the bicep muscle. As a function of the angle of the forearm with respect to the horizontal direction ( as the forearm is lowered) calculate the tension in the bicep muscle. Include a plot of tension in the bicep as a function of the angle of the forearm relative to the horizontal

To solve this problem, we need to break it down into individual components and apply the principles of equilibrium. Let's start by identifying the forces acting on the system:

1. Force from the bicep muscle: This force acts at an angle of 15 degrees to the vertical and is the force that we need to find (let's call it T).

2. Weight of the arm: This force acts vertically downwards and is equal to the product of the mass of the forearm and the acceleration due to gravity (4 kg * 9.8 m/s^2 = 39.2 N). Let's call it W1.

3. Weight of the milk carton: This force also acts vertically downwards and is equal to the product of the mass of the milk carton and the acceleration due to gravity (2 kg * 9.8 m/s^2 = 19.6 N). Let's call it W2.

Let's define the coordinates for our calculations:
- x-axis: Horizontal direction from the elbow towards the hand
- y-axis: Vertical direction pointing upwards

Now, let's draw a free-body diagram to visualize the forces acting on the system. The bicep force has horizontal and vertical components, while the weights act solely in the vertical direction.

```
|
| T sin(15°)
|
|---------
| | (35 cm)
| |
| |
| Forearm
| (4 kg)
---------|------------------
| |
| | W2
| |
| |
| |
| | W1
| |
```

From the diagram, we can see that in the vertical direction (y-axis), the forces balance each other: T sin(15°) = W1 + W2.

Next, let's consider the horizontal direction (x-axis). There is no horizontal acceleration, so the forces in this direction should also balance each other. The horizontal component of the bicep force can be calculated by T cos(15°).

Since the forearm is held horizontally, the horizontal component of the bicep force must equal the sum of the weights in the horizontal direction.

T cos(15°) = 0 (no horizontal acceleration) + 0 (no horizontal acceleration)

Now, we have two equations for our system:

Equation 1: T sin(15°) = W1 + W2
Equation 2: T cos(15°) = 0 + 0

Let's solve Equation 1 for T:

T = (W1 + W2) / sin(15°)

Substituting the known values: T = (39.2 N + 19.6 N) / sin(15°)

T ≈ 190.5 N

Now we can plot the tension (T) in the bicep muscle as a function of the angle of the forearm relative to the horizontal. To do this, we need to vary the angle and recalculate T using the formula we derived above.

The plot will show that as the forearm is lowered (angle decreases), the tension in the bicep muscle increases. The graph will have a positive slope, indicating an increasing trend.

I hope this explanation helps you understand how to solve the problem and create the plot.