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We are tasked with determining the magnitude \( F \) and angle \( \theta \) of a third force, such that the resultant of the three forces is vertically downward with a magnitude of 1010 lb.

### Step 1: Breakdown of Forces
- Force 1: \( 1000 \, \text{lb} \) at an angle of \( 41^\circ \) to the negative x-axis.
- Force 2: \( 340 \, \text{lb} \) at an angle of \( 32^\circ \) to the positive x-axis.
- Force 3: \( F \) at an unknown angle \( \theta \).

We are asked to make the resultant force vertically downward with a magnitude of 1010 lb.

### Step 2: Components of Each Force
To solve this, we need to break each force into its x and y components:

#### Force 1: \( 1000 \, \text{lb} \) at \( 41^\circ \)
- \( F_{1x} = -1000 \cos(41^\circ) \)
- \( F_{1y} = 1000 \sin(41^\circ) \)

#### Force 2: \( 340 \, \text{lb} \) at \( 32^\circ \)
- \( F_{2x} = 340 \cos(32^\circ) \)
- \( F_{2y} = 340 \sin(32^\circ) \)

#### Force 3: \( F \) at \( \theta \)
- \( F_{3x} = F \cos(\theta) \)
- \( F_{3y} = F \sin(\theta) \)

### Step 3: Apply Resultant Force Conditions
We are given that the resultant force is vertically downward with a magnitude of 1010 lb. This leads to two conditions:
1. The sum of the x-components of all forces must equal zero for the resultant to be purely vertical:
   \[
   F_{1x} + F_{2x} + F_{3x} = 0
   \]
2. The sum of the y-components must equal \(-1010 \, \text{lb}\):
   \[
   F_{1y} + F_{2y} + F_{3y} = -1010
   \]

### Step 4: Solve the System of Equations
We now substitute the component equations and solve for \( F \) and \( \theta \).

Let me calculate the numerical values for the x and y components.The x and y components of the first two forces are as follows:

- Force 1 (\( 1000 \, \text{lb} \) at \( 41^\circ \)):
  - \( F_{1x} = -754.71 \, \text{lb} \)
  - \( F_{1y} = 656.06 \, \text{lb} \)

- Force 2 (\( 340 \, \text{lb} \) at \( 32^\circ \)):
  - \( F_{2x} = 288.34 \, \text{lb} \)
  - \( F_{2y} = 180.17 \, \text{lb} \)

### Step 5: Set up the Equations
Now, using the conditions for equilibrium:

1. **In the x-direction (sum of forces equals zero):**
   \[
   F_{1x} + F_{2x} + F \cos(\theta) = 0
   \]
   \[
   -754.71 + 288.34 + F \cos(\theta) = 0
   \]
   \[
   F \cos(\theta) = 466.37
   \]

2. **In the y-direction (sum of forces equals -1010 lb):**
   \[
   F_{1y} + F_{2y} + F \sin(\theta) = -1010
   \]
   \[
   656.06 + 180.17 + F \sin(\theta) = -1010
   \]
   \[
   F \sin(\theta) = -1846.23
   \]

Now, we can solve this system of equations to find \( F \) and \( \theta \).It seems like I can't do more advanced data analysis right now. Please try again later. 

If you'd like, you can solve the system of equations manually or using a calculator:

1. \( F \cos(\theta) = 466.37 \)
2. \( F \sin(\theta) = -1846.23 \)

These two equations will give you the magnitude \( F \) and the angle \( \theta \).

Would you like to discuss the process or any other details? Here are some related questions for practice:

1. How do you calculate the resultant force when multiple forces act at different angles?
2. What is the importance of resolving forces into components?
3. How would the problem change if the resultant force wasn't vertical?
4. Can forces in equilibrium be unbalanced in any direction?
5. How does trigonometry help in solving vector problems like these?

**Tip:** Always ensure to break forces into components when dealing with vectors in two dimensions.

Two rods and one cable are attached to supports at O. Two of the forces are known (1000 lb at an angle of 41° and 340 lb at an angle of 32°). Determine the magnitude F and the direction θ of the third force such that the resultant of the three forces is vertically downward with a magnitude of 1010 lb.

This problem demonstrates how to calculate the magnitude and angle of a third force in a system where three forces result in a vertical downward force of 1010 lb. Explore how vector components and equilibrium conditions are applied to solve the problem, using forces at 41° and 32°.

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