DA 125 N sign is supported by two ropes. One rope pulls up
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and to the right 01 = 35.5° above the horizontal with a tension T1, and the other rope pulls up and to the left 02 = 48.5° above the horizontal with a tension Tz, as shown in the figure. Find the tensions Ti and Tz.

To solve for the tensions \( T_1 \) and \( T_2 \) in the ropes supporting the sign, we need to apply the principles of equilibrium. Since the sign is in equilibrium, the sum of forces in both the horizontal and vertical directions must be zero.

### Step-by-Step Solution:

1. **Breakdown of Forces:**
   Each tension force can be split into horizontal and vertical components. Let’s denote the tension forces and their components as follows:

   - **Tension \( T_1 \):**
     - Horizontal component: \( T_1 \cos \theta_1 \)
     - Vertical component: \( T_1 \sin \theta_1 \)
     - Angle \( \theta_1 = 35.5^\circ \)

   - **Tension \( T_2 \):**
     - Horizontal component: \( T_2 \cos \theta_2 \)
     - Vertical component: \( T_2 \sin \theta_2 \)
     - Angle \( \theta_2 = 48.5^\circ \)

2. **Sum of Forces in the Horizontal Direction:**
   The horizontal components must cancel each other out:
   \[
   T_1 \cos \theta_1 = T_2 \cos \theta_2
   \]

3. **Sum of Forces in the Vertical Direction:**
   The vertical components must add up to the weight of the sign, which is 125 N (assuming the sign is in equilibrium and stationary):
   \[
   T_1 \sin \theta_1 + T_2 \sin \theta_2 = 125
   \]

4. **Solving the System of Equations:**
   From the horizontal force balance equation:
   \[
   T_1 = T_2 \frac{\cos \theta_2}{\cos \theta_1}
   \]

   Substitute this into the vertical force balance equation:
   \[
   \left(T_2 \frac{\cos \theta_2}{\cos \theta_1}\right) \sin \theta_1 + T_2 \sin \theta_2 = 125
   \]

   Simplifying to solve for \( T_2 \):
   \[
   T_2 \left( \frac{\cos \theta_2 \sin \theta_1}{\cos \theta_1} + \sin \theta_2 \right) = 125
   \]

   \[
   T_2 = \frac{125}{\frac{\cos \theta_2 \sin \theta_1}{\cos \theta_1} + \sin \theta_2}
   \]

5. **Calculating Tensions:**
   Now we can plug in the angles and compute \( T_2 \). Once we have \( T_2 \), we can find \( T_1 \) using the horizontal force equation.

Let’s compute these values.The tensions in the two ropes are:

- \( T_1 = 83.28 \, \text{N} \) (up and to the right at \( 35.5^\circ \))
- \( T_2 = 102.32 \, \text{N} \) (up and to the left at \( 48.5^\circ \))

These values ensure that the sign remains in equilibrium.

Would you like more details or have any further questions? Here are some related questions you might find useful:

1. How do you derive the formulas for horizontal and vertical force components?
2. What are the conditions for equilibrium in a system?
3. How would the tensions change if the angles were different?
4. What if the weight of the sign changed to 150 N?
5. How can you solve similar problems using a different approach, like vector diagrams?

**Tip:** Always ensure to break down forces into components when dealing with inclined forces and use trigonometric functions to analyze them accurately.

DA 125 N sign is supported by two ropes. One rope pulls up and to the right at θ1 = 35.5° above the horizontal with a tension T1, and the other rope pulls up and to the left at θ2 = 48.5° above the horizontal with a tension T2, as shown in the figure. Find the tensions T1 and T2.

Learn how to find the tensions T1 and T2 in ropes supporting a 125 N sign using principles of equilibrium and trigonometry. This solution breaks down the problem into horizontal and vertical force balances, applies trigonometric formulas, and solves for T1 and T2 step-by-step. Suitable for Grades 11-12.

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