The problem involves two forces applied to the end of a screw eye, with the goal of removing the post by creating a resultant force that acts directly vertically upward. Let's break this down:

Part A:

You are asked to find the angle $\theta$ such that the resultant force is directly upward and has a magnitude of 690 N. From the image, one force is 500 N at a 30° angle, and the other force $F$ is applied at angle $\theta$.

To solve for $\theta$, we use vector components to ensure the horizontal components cancel each other and the vertical components sum up to 690 N.

1. Resolve the forces into components:

   • The horizontal component of the 500 N force is: $F_{x1} = 500 \cos(30^\circ)$
   • The vertical component of the 500 N force is: $F_{y1} = 500 \sin(30^\circ)$
   • For the second force $F$ at angle $\theta$: $F_{x2} = F \cos(\theta)$ $F_{y2} = F \sin(\theta)$

2. Condition for horizontal balance: $F_{x1} = F_{x2}$ $500 \cos(30^\circ) = F \cos(\theta)$

3. Condition for vertical force summing to 690 N: $F_{y1} + F_{y2} = 690$ $500 \sin(30^\circ) + F \sin(\theta) = 690$

By solving these two equations simultaneously, you will get the angle $\theta$.

Part B:

Once $\theta$ is known, you can determine the magnitude of the force $F$ by using the horizontal or vertical component equations. Specifically, use: $F = \frac{500 \cos(30^\circ)}{\cos(\theta)}$

Let me calculate the exact values for $\theta$ and $F$.The components of the first force $F_1 = 500 \, \text{N}$ are:

• Horizontal component: $F_{x1} = 433.01 \, \text{N}$
• Vertical component: $F_{y1} = 250.00 \, \text{N}$

Next, we can use these values to solve for $\theta$ and $F$ by setting up the system of equations for horizontal and vertical balances. Let's solve them.It seems like I can’t do more advanced data analysis right now. Please try again later.

Would you like to proceed with the calculations manually? I can guide you through the steps to solve for $\theta$ and $F$, or I can answer any questions you might have about the process.

Here are five related questions to expand on this topic:

1. How do horizontal and vertical components affect force equilibrium?
2. What is the method to resolve forces into components?
3. How does knowing the resultant force help in determining other unknowns?
4. How does vector addition work when forces are at different angles?
5. What are some common mistakes when solving equilibrium problems?

Tip: When solving problems with multiple forces, always ensure the units are consistent and double-check angle directions (clockwise vs counterclockwise) for accuracy.