The problem presented is about vector addition and equilibrium, where three teams are pulling on a metal ring with forces such that the total force must add up to zero. Here's how to approach the problem:

Problem Breakdown:

• Team 1 pulls with a force of 1200 N horizontally (positive x-direction).
• Team 2 pulls with 1000 N of force at an angle of 80° from Team 1's force (measured counterclockwise from the x-axis).
• Team 3's force is unknown, but we know that the total of all three forces must sum to zero because the ring doesn't move (equilibrium condition).

Part (a) - Decomposing Forces into Components:

We need to find the x and y components of the forces from Team 1 and Team 2, and then solve for the components of Team 3 (since the total force in both directions must be zero).

Force Components:

For each force vector, we can decompose into x and y components using trigonometry:

• Team 1 (Force $F_A = 1200 \, \text{N}$):

  • $F_{Ax} = 1200 \, \text{N}$ (since it's entirely along the x-axis).
  • $F_{Ay} = 0 \, \text{N}$ (no component in the y-direction).

• Team 2 (Force $F_B = 1000 \, \text{N}$ at 80°):

  • $F_{Bx} = 1000 \cos(80^\circ)$
  • $F_{By} = 1000 \sin(80^\circ)$

Let’s calculate these values.

Calculations for Team 2:

Let’s compute these components:

$F_{Bx} = 1000 \cos(80^\circ), \quad F_{By} = 1000 \sin(80^\circ)$

I'll calculate this now.The components of Team 2's force are as follows:

• $F_{Bx} \approx 173.65 \, \text{N}$ (x-component).
• $F_{By} \approx 984.81 \, \text{N}$ (y-component).

Part (b) - Solving for Team 3's Force:

Since the ring is in equilibrium, the sum of forces in both the x-direction and y-direction must equal zero. Therefore:

$F_{Ax} + F_{Bx} + F_{Cx} = 0$ $F_{Ay} + F_{By} + F_{Cy} = 0$

Using this, we can solve for the x and y components of Team 3's force:

1. $F_{Cx} = -(F_{Ax} + F_{Bx})$
2. $F_{Cy} = -(F_{Ay} + F_{By})$

Let me compute the values for $F_{Cx}$ and $F_{Cy}$.The components of Team 3's force are:

• $F_{Cx} \approx -1373.65 \, \text{N}$ (x-component),
• $F_{Cy} \approx -984.81 \, \text{N}$ (y-component).

Magnitude and Direction of Team 3's Force:

We can now calculate the magnitude and direction of Team 3's force.

• Magnitude: $F_C = \sqrt{F_{Cx}^2 + F_{Cy}^2}$
• Direction (angle $\theta$ measured counterclockwise from the positive x-axis): $\theta = \text{atan2}(F_{Cy}, F_{Cx})$

Let me compute these values for you.The magnitude and direction of Team 3's force are as follows:

• Magnitude: $F_C \approx 1690.19 \, \text{N}$
• Direction: $\theta \approx -144.36^\circ$ (measured counterclockwise from the positive x-axis, indicating a clockwise direction from the positive x-axis).

Would you like more details or further explanation on any part? Here are a few related questions to deepen your understanding:

1. How would the equilibrium condition change if one of the forces increased by 50%?
2. Can you visualize how the vectors would look if drawn to scale?
3. What trigonometric identities are crucial for solving vector components?
4. How does the concept of torque relate to this problem?
5. How would you apply this method to more than three forces?

Tip: Always check the direction of angles when resolving forces to ensure proper vector decomposition.