solve: About 50 years ago, the San Diego Zoo, in California, had the largest gorilla on Earth: its mass was about 3.10 × 102 kg. Suppose a gorilla with this
mass hangs from two vines, each of which makes an angle of 30.0° with
the vertical. Draw a free-body diagram showing the various forces, and
find the magnitude of the force of tension in each vine. What would happen to the tensions if the upper ends of the vines were farther apart?

### Free-Body Diagram
We start by considering the forces acting on the gorilla. Since the gorilla is hanging from two vines, the forces involved are:

1. **Weight of the gorilla (W)**: This force is directed vertically downward and is equal to the mass of the gorilla multiplied by the gravitational acceleration.
   
2. **Tension forces (T₁ and T₂)**: These forces are present in the two vines. They act along the direction of the vines and make an angle of 30.0° with the vertical. Since both vines are identical and the angles are the same, the tension in each vine will be the same (T₁ = T₂ = T).

#### Free-body diagram summary:
- Weight (W) acting vertically downwards.
- Tension forces (T₁ and T₂) acting along the vines, each at an angle of 30.0° with the vertical.

### Step-by-Step Solution:

1. **Mass of the gorilla (m)**:
   \[
   m = 3.10 \times 10^2 \, \text{kg} = 310 \, \text{kg}
   \]

2. **Gravitational force (W)**: 
   The weight of the gorilla is the force due to gravity:
   \[
   W = m \cdot g = 310 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 3038 \, \text{N}
   \]
   So the weight of the gorilla is 3038 N.

3. **Resolve forces vertically**:
   Since the gorilla is at rest and not accelerating, the net force in both the vertical and horizontal directions must be zero (equilibrium condition).
   In the vertical direction, the sum of the vertical components of the tension forces must balance the weight of the gorilla:
   \[
   2T \cdot \cos(30.0^\circ) = W
   \]
   Substituting the known values:
   \[
   2T \cdot \cos(30.0^\circ) = 3038 \, \text{N}
   \]
   \[
   2T \cdot \frac{\sqrt{3}}{2} = 3038 \, \text{N}
   \]
   \[
   T \cdot \sqrt{3} = 3038 \, \text{N}
   \]
   \[
   T = \frac{3038 \, \text{N}}{\sqrt{3}} \approx 1754.17 \, \text{N}
   \]

Thus, the tension in each vine is approximately **1754 N**.

### What would happen if the upper ends of the vines were farther apart?
If the upper ends of the vines were farther apart, the angle between the vine and the vertical would increase. As this angle increases, the vertical component of the tension force in each vine decreases, which means the total tension in each vine must increase to support the same weight. Therefore, **the tension in each vine would increase** if the upper ends of the vines were farther apart.

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Do you have any questions or would you like more details on any step? 

Here are five questions that extend from this problem:

1. What happens to the tension if the angle between the vine and the vertical is reduced?
2. How would the tension change if the gorilla weighed twice as much?
3. What if the gorilla was moving upwards at a constant speed—would the tension change?
4. How would the tension change if one vine was shorter than the other?
5. How can we apply these principles to a human hanging from two ropes?

Tip: Always resolve forces into components when dealing with angles to simplify the problem.

About 50 years ago, the San Diego Zoo, in California, had the largest gorilla on Earth: its mass was about 3.10 × 10^2 kg. Suppose a gorilla with this mass hangs from two vines, each of which makes an angle of 30.0° with the vertical. Draw a free-body diagram showing the various forces, and find the magnitude of the force of tension in each vine. What would happen to the tensions if the upper ends of the vines were farther apart?

Solve a physics problem involving a 310 kg gorilla hanging from two vines at a 30° angle. Learn how to calculate the tension in each vine using force resolution and trigonometry. Suitable for high school students in Grades 10-12.

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