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We are given two forces acting at point O, with the following information:

- Force 1: \(1000 \, \text{lb}\) at an angle of \(41^\circ\) from the horizontal (along the left rod).
- Force 2: \(340 \, \text{lb}\) at an angle of \(32^\circ\) from the horizontal (along the right rod).

We are tasked with finding:

1. The magnitude \(F\) of the third force.
2. The angle \(\theta\) of the third force, such that the resultant force is vertically downward with a magnitude of \(1010 \, \text{lb}\).

### Solution Approach

We will resolve each of the forces into their horizontal (\(x\)) and vertical (\(y\)) components, and apply the conditions:

- The sum of the horizontal components must be zero for the resultant to be vertical.
- The sum of the vertical components must equal the given vertical force of \(1010 \, \text{lb}\), directed downward.

#### Step 1: Resolve each force into components

1. **Force 1 (\(1000 \, \text{lb}\)) at \(41^\circ\):**
   - Horizontal component: \(F_{1x} = 1000 \cos(41^\circ)\)
   - Vertical component: \(F_{1y} = 1000 \sin(41^\circ)\)

2. **Force 2 (\(340 \, \text{lb}\)) at \(32^\circ\):**
   - Horizontal component: \(F_{2x} = 340 \cos(32^\circ)\)
   - Vertical component: \(F_{2y} = 340 \sin(32^\circ)\)

3. **Force 3 (\(F\) at angle \(\theta\)):**
   - Horizontal component: \(F_{3x} = F \cos(\theta)\)
   - Vertical component: \(F_{3y} = F \sin(\theta)\)

#### Step 2: Apply equilibrium conditions

- **Horizontal equilibrium:**
  \[
  F_{1x} + F_{2x} + F_{3x} = 0
  \]
  \[
  1000 \cos(41^\circ) + 340 \cos(32^\circ) + F \cos(\theta) = 0
  \]

- **Vertical equilibrium:**
  \[
  F_{1y} + F_{2y} + F_{3y} = -1010
  \]
  \[
  1000 \sin(41^\circ) + 340 \sin(32^\circ) + F \sin(\theta) = -1010
  \]

Let's first calculate the horizontal and vertical components of the given forces and solve for \(F\) and \(\theta\).We have the following values for the sum of the known force components:

- The sum of the horizontal components: \( F_{1x} + F_{2x} = 1043.05 \, \text{lb} \)
- The sum of the vertical components: \( F_{1y} + F_{2y} = 836.23 \, \text{lb} \)

### Equilibrium conditions:

1. **Horizontal equilibrium:**
   \[
   1043.05 + F \cos(\theta) = 0
   \]
   Solving for \(F \cos(\theta)\):
   \[
   F \cos(\theta) = -1043.05
   \]

2. **Vertical equilibrium:**
   \[
   836.23 + F \sin(\theta) = -1010
   \]
   Solving for \(F \sin(\theta)\):
   \[
   F \sin(\theta) = -1010 - 836.23 = -1846.23
   \]

Now, we can use these two equations to solve for the magnitude \(F\) and angle \(\theta\).It seems like I can't do more advanced data analysis right now. Please try again later.

If you'd like, I can guide you through the process of solving the equations manually. Would you like that or any further clarification on the steps?

Here are five related questions to explore this topic further:
1. How do we determine the components of a force vector in physics?
2. What is the significance of resolving forces into horizontal and vertical components?
3. How does the concept of equilibrium apply in this problem?
4. What is the role of trigonometry in solving static equilibrium problems?
5. How can we check the correctness of force magnitudes and directions in mechanical systems?

**Tip:** In statics problems, always verify equilibrium by checking both force components and moments.

Two rods and one cable are attached to the supports at O. If two of the forces are as shown, determine the magnitude F and direction θ of the third force so that the resultant of the three forces is vertically downward with a magnitude of 1010 lb.

This problem involves determining the magnitude and direction of a third force so that the resultant of three forces is vertically downward with a magnitude of 1010 lb. Key concepts include vector decomposition and static equilibrium.

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