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Answer but cable 1 is 30 while cable 2 is 45

Let's solve the problem, making sure we account for the correction: cable 1 has an angle of 30°, and cable 2 has an angle of 45°.

### Given:
- Weight of the mirrored sphere: \( W = 200.0 \, \text{N} \)
- Cable 1 angle: \( \theta_1 = 30^\circ \)
- Cable 2 angle: \( \theta_2 = 45^\circ \)

The two cables are holding the sphere in equilibrium, so the forces balance both horizontally and vertically. Let the tensions in the cables be \( T_1 \) (in cable 1) and \( T_2 \) (in cable 2).

### Step 1: Vertical force equilibrium
The vertical forces must add up to the weight of the sphere:
\[
T_1 \sin(\theta_1) + T_2 \sin(\theta_2) = W
\]
Substitute the given values:
\[
T_1 \sin(30^\circ) + T_2 \sin(45^\circ) = 200.0 \, \text{N}
\]
Since \( \sin(30^\circ) = 0.5 \) and \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \), the equation becomes:
\[
0.5 T_1 + \frac{\sqrt{2}}{2} T_2 = 200.0
\]
This simplifies to:
\[
0.5 T_1 + 0.707 T_2 = 200.0 \quad \text{(Equation 1)}
\]

### Step 2: Horizontal force equilibrium
The horizontal components of the tension forces must cancel each other out:
\[
T_1 \cos(\theta_1) = T_2 \cos(\theta_2)
\]
Substitute the given values:
\[
T_1 \cos(30^\circ) = T_2 \cos(45^\circ)
\]
Since \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \) and \( \cos(45^\circ) = \frac{\sqrt{2}}{2} \), the equation becomes:
\[
\frac{\sqrt{3}}{2} T_1 = \frac{\sqrt{2}}{2} T_2
\]
Simplifying this:
\[
\sqrt{3} T_1 = \sqrt{2} T_2
\]
\[
T_1 = \frac{\sqrt{2}}{\sqrt{3}} T_2
\]
\[
T_1 = 0.8165 T_2 \quad \text{(Equation 2)}
\]

### Step 3: Solve the system of equations
Substitute \( T_1 = 0.8165 T_2 \) from Equation 2 into Equation 1:
\[
0.5 (0.8165 T_2) + 0.707 T_2 = 200.0
\]
\[
0.40825 T_2 + 0.707 T_2 = 200.0
\]
\[
1.11525 T_2 = 200.0
\]
\[
T_2 = \frac{200.0}{1.11525} \approx 179.42 \, \text{N}
\]

Now, using Equation 2 to find \( T_1 \):
\[
T_1 = 0.8165 \times 179.42 \approx 146.51 \, \text{N}
\]

### Final Answer:
- \( T_1 \approx 146.51 \, \text{N} \) (tension in the cable at \( 30^\circ \))
- \( T_2 \approx 179.42 \, \text{N} \) (tension in the cable at \( 45^\circ \))

Would you like a detailed explanation of any steps? 

Here are 5 related questions to think about:
1. How would the tensions change if the angles were both increased?
2. What would happen if one cable were removed?
3. How is the vertical equilibrium related to the weight of the object?
4. How would different weights of the sphere affect the tension in the cables?
5. Can this system be solved graphically, and how would it differ from the algebraic approach?

**Tip:** Always remember that the sum of forces in both vertical and horizontal directions must be zero in equilibrium problems.

A mirrored sphere, weighing 200.0 N, is suspended from the ceiling of a ballroom by two cables at angles of 30° and 45°. What is the tension in each of these cables?

This page explains how to find the tension in cables holding a 200.0 N mirrored sphere at angles of 30° and 45°. The problem covers key concepts in statics and trigonometry, including vertical and horizontal force equilibrium.

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