7. A force table is set up in a physics laboratory and
three spring scales are hooked to the central ring
as shown above. The first scale reads 12.5 N at an
angle of 0.0° and the second scale reads 17.3 N at
an angle of -34.0°. If the three forces are balanced,
what is the reading and angle of the third spring
scale?

To solve this problem, we need to balance the forces acting on the central ring, which means the vector sum of the three forces must equal zero. Since two of the forces are given, we can determine the magnitude and direction of the third force using vector components.

### Step 1: Resolve the given forces into components
We will break down the forces into their **x** (horizontal) and **y** (vertical) components using trigonometry.

1. **Force 1 (12.5 N at 0.0°):**
   - Since the angle is \( 0.0^\circ \), it lies entirely along the x-axis.
   - \( F_{1x} = 12.5 \, \text{N} \cdot \cos(0^\circ) = 12.5 \, \text{N} \)
   - \( F_{1y} = 12.5 \, \text{N} \cdot \sin(0^\circ) = 0.0 \, \text{N} \)

2. **Force 2 (17.3 N at -34.0°):**
   - We need to resolve this force into its x and y components.
   - \( F_{2x} = 17.3 \, \text{N} \cdot \cos(-34.0^\circ) = 17.3 \, \text{N} \cdot 0.829 = 14.34 \, \text{N} \)
   - \( F_{2y} = 17.3 \, \text{N} \cdot \sin(-34.0^\circ) = 17.3 \, \text{N} \cdot -0.559 = -9.67 \, \text{N} \)

### Step 2: Set up equations for equilibrium
In equilibrium, the sum of the forces in the **x** and **y** directions must be zero:

\[
\sum F_x = 0 \quad \text{and} \quad \sum F_y = 0
\]

- For the **x**-direction:
  \[
  F_{1x} + F_{2x} + F_{3x} = 0 \quad \Rightarrow \quad 12.5 + 14.34 + F_{3x} = 0
  \]
  Solving for \( F_{3x} \):
  \[
  F_{3x} = -(12.5 + 14.34) = -26.84 \, \text{N}
  \]

- For the **y**-direction:
  \[
  F_{1y} + F_{2y} + F_{3y} = 0 \quad \Rightarrow \quad 0 + (-9.67) + F_{3y} = 0
  \]
  Solving for \( F_{3y} \):
  \[
  F_{3y} = 9.67 \, \text{N}
  \]

### Step 3: Calculate the magnitude and direction of the third force
The magnitude of the third force \( F_3 \) is given by the Pythagorean theorem:

\[
F_3 = \sqrt{F_{3x}^2 + F_{3y}^2} = \sqrt{(-26.84)^2 + (9.67)^2} = \sqrt{720.27 + 93.48} = \sqrt{813.75} = 28.53 \, \text{N}
\]

The direction \( \theta_3 \) of the third force is found using the inverse tangent function:

\[
\theta_3 = \tan^{-1} \left( \frac{F_{3y}}{F_{3x}} \right) = \tan^{-1} \left( \frac{9.67}{-26.84} \right) = \tan^{-1}(-0.36) = -19.8^\circ
\]

### Final Answer:
The third spring scale reads **28.5 N** at an angle of **-19.8°**.

Would you like more details or have any questions?

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Here are 5 related questions to explore this topic further:
1. How do you determine the resultant force when two forces act at an angle?
2. What is the difference between resolving a force and finding its resultant?
3. How does the concept of equilibrium apply to real-life scenarios like suspension bridges?
4. What role does vector addition play in balancing forces in physics?
5. Can equilibrium be achieved with more than three forces acting on an object?

**Tip**: When working with vectors, always break them into components along perpendicular axes to simplify calculations.

A force table is set up in a physics laboratory and three spring scales are hooked to the central ring as shown above. The first scale reads 12.5 N at an angle of 0.0° and the second scale reads 17.3 N at an angle of -34.0°. If the three forces are balanced, what is the reading and angle of the third spring scale?

Solve a physics problem involving balancing three forces on a force table. Learn to resolve forces into vector components and apply equilibrium conditions to find the magnitude and direction of the third force. This problem covers vector addition and trigonometry, suitable for students in Grades 11-12.

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