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.