You are able to hold out your arm in an outstretched horizontal position because of the action of the deltoid muscle. Assume the humerus bone has a mass M1=3.6kg , length L=0.66m and its center of mass is a distance L1=0.33m from the scapula. (For this problem ignore the rest of the arm.) The deltoid muscle attaches to the humerus a distance L2=0.15m from the scapula. The deltoid muscle makes an angle of θ=17∘ with the horizontal, as shown. (Figure 1) (Figure 2)Use g=9.8m/s2 throughout the problem.Find the tension T in the deltoid muscle. Express the tension in newtons, to the nearest integer.

To find the tension \( T \) in the deltoid muscle, we can use the principles of static equilibrium. The system is in equilibrium, meaning that the sum of torques around the scapula must equal zero.

### Given:
- Mass of the humerus, \( M_1 = 3.6 \, \text{kg} \)
- Length of the humerus, \( L = 0.66 \, \text{m} \)
- Distance of center of mass from the scapula, \( L_1 = 0.33 \, \text{m} \)
- Distance from the scapula to the deltoid muscle attachment, \( L_2 = 0.15 \, \text{m} \)
- Angle of the deltoid muscle with the horizontal, \( \theta = 17^\circ \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)

### Step 1: Calculate the weight of the humerus
The weight \( W \) of the humerus can be calculated as:
\[
W = M_1 \cdot g = 3.6 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 35.28 \, \text{N}
\]

### Step 2: Calculate the torque due to the weight of the humerus
The torque \( \tau_w \) exerted by the weight of the humerus about the scapula is:
\[
\tau_w = W \cdot L_1 = 35.28 \, \text{N} \cdot 0.33 \, \text{m} = 11.64 \, \text{Nm}
\]

### Step 3: Calculate the torque due to the tension in the deltoid muscle
The tension \( T \) in the deltoid muscle creates a torque \( \tau_t \) about the scapula:
\[
\tau_t = T \cdot L_2 \cdot \sin(\theta)
\]

### Step 4: Set the torques equal to each other (for equilibrium)
Since the system is in equilibrium:
\[
\tau_t = \tau_w
\]
\[
T \cdot L_2 \cdot \sin(\theta) = W \cdot L_1
\]
Substituting the known values:
\[
T \cdot 0.15 \cdot \sin(17^\circ) = 11.64
\]

### Step 5: Solve for \( T \)
First, calculate \( \sin(17^\circ) \):
\[
\sin(17^\circ) \approx 0.2924
\]

Now substituting this value into the equation:
\[
T \cdot 0.15 \cdot 0.2924 = 11.64
\]
\[
T \cdot 0.04386 = 11.64
\]
\[
T = \frac{11.64}{0.04386} \approx 265.1 \, \text{N}
\]

### Final Answer
The tension \( T \) in the deltoid muscle is approximately:
\[
\boxed{265} \, \text{N}
\]

If you have any questions or need further details, feel free to ask! Here are some related questions you might consider:

1. How would the tension change if the angle of the deltoid muscle increased?
2. What role does the length of the humerus play in this equilibrium?
3. How would the analysis change if we included the forearm in the model?
4. What assumptions were made in this calculation?
5. How would the torque change if the mass of the humerus was different?

**Tip:** Always visualize the forces and torques in a static equilibrium problem to ensure accurate setup of your equations!

You are able to hold out your arm in an outstretched horizontal position because of the action of the deltoid muscle. Assume the humerus bone has a mass M1=3.6kg, length L=0.66m and its center of mass is a distance L1=0.33m from the scapula. The deltoid muscle attaches to the humerus a distance L2=0.15m from the scapula. The deltoid muscle makes an angle of θ=17° with the horizontal. Use g=9.8m/s² throughout the problem. Find the tension T in the deltoid muscle. Express the tension in newtons, to the nearest integer.

Explore the physics problem of calculating the tension in the deltoid muscle given the mass and length of the humerus. This solution involves key concepts of torque and equilibrium, suitable for Grades 10-12 students.

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