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The arm in Figure P8.7 weighs 41.5 N. The force of gravity acting on the arm acts through point A. Determine the magnitudes of the tension force vector F t in the deltoid muscle and the force vector F s of the shoulder on the humerus (upper-arm bone) to hold the arm in the position shown.

To solve this problem, we need to apply the principles of static equilibrium. For the arm to be in equilibrium, both the sum of the forces and the sum of the moments (torques) must be zero. Let's break down the steps:

1. **Sum of the Forces**:
   \[
   \sum \vec{F} = 0
   \]
   This involves both the horizontal and vertical components.

2. **Sum of the Moments**:
   \[
   \sum \vec{M} = 0
   \]
   We will take moments about point A to eliminate the force \(\vec{F}_s\).

Given Data:
- Weight of the arm (\(W\)) = 41.5 N, acting downward through point A.
- \(d_1 = 0.080 \, \text{m}\)
- \(d_2 = 0.290 \, \text{m}\)
- The angle \(\theta = 12^\circ\)

### Sum of the Moments about Point A:
To find the tension in the deltoid muscle (\(\vec{F}_t\)), we calculate the moments about point A:

\[
\sum M_A = 0
\]

The torque due to the weight of the arm:
\[
\tau_W = W \cdot d_2
\]
The torque due to the tension in the deltoid muscle:
\[
\tau_{F_t} = F_t \cdot d_1 \cdot \cos(\theta)
\]

Since the arm is in equilibrium:
\[
W \cdot d_2 = F_t \cdot d_1 \cdot \cos(\theta)
\]

Plugging in the known values:
\[
41.5 \, \text{N} \cdot 0.290 \, \text{m} = F_t \cdot 0.080 \, \text{m} \cdot \cos(12^\circ)
\]

\[
F_t = \frac{41.5 \, \text{N} \cdot 0.290 \, \text{m}}{0.080 \, \text{m} \cdot \cos(12^\circ)}
\]

Now, let's calculate the exact value:

\[
F_t = \frac{41.5 \cdot 0.290}{0.080 \cdot \cos(12^\circ)}
\]

### Sum of the Forces:
Since the arm is in equilibrium, the sum of the forces in both the horizontal and vertical directions must be zero.

- In the horizontal direction:
  \[
  F_t \sin(12^\circ) = F_s \cos(\theta)
  \]

- In the vertical direction:
  \[
  F_t \cos(12^\circ) + F_s \sin(\theta) = W
  \]

Solving for \(\vec{F}_t\):

\[
F_t \approx \frac{41.5 \cdot 0.290}{0.080 \cdot 0.9781}
\]
\[
F_t \approx \frac{12.035}{0.078248}
\]
\[
F_t \approx 153.8 \, \text{N}
\]

### Solving for \( \vec{F}_s \):
Using the force equilibrium equations:

\[
F_s = \frac{F_t \sin(12^\circ)}{\cos(\theta)}
\]

Now substituting the values:

\[
F_s = \frac{153.8 \sin(12^\circ)}{\cos(90^\circ)}
\]
\[
F_s \approx \frac{153.8 \cdot 0.2079}{1}
\]
\[
F_s \approx 31.97 \, \text{N}
\]

So, the magnitudes of the forces are:
- \(\vec{F}_t \approx 153.8 \, \text{N}\)
- \(\vec{F}_s \approx 31.97 \, \text{N}\)

Would you like more details or have any questions?

Here are some additional related questions:
1. How would the force values change if the angle were increased to 20 degrees?
2. What would happen if the weight of the arm increased?
3. How does the moment arm distance affect the tension in the muscle?
4. Can you calculate the forces if the weight acted at a different point along the arm?
5. How would adding an external load at the wrist affect the tension and shoulder forces?

**Tip:** Always cross-check your final calculations and units to ensure accuracy in physics problems.

The arm in Figure P8.7 weighs 41.5 N. The force of gravity acting on the arm acts through point A. Determine the magnitudes of the tension force vector F_t in the deltoid muscle and the force vector F_s of the shoulder on the humerus (upper-arm bone) to hold the arm in the position shown.

Determine the tension force in the deltoid muscle and the shoulder force required to hold a 41.5 N arm in static equilibrium. This problem involves statics, force equilibrium, torque calculations, and trigonometry.

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