Math Problem Statement

  1. 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?

Solution

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 0.00.0^\circ, it lies entirely along the x-axis.
    • F1x=12.5Ncos(0)=12.5NF_{1x} = 12.5 \, \text{N} \cdot \cos(0^\circ) = 12.5 \, \text{N}
    • F1y=12.5Nsin(0)=0.0NF_{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.
    • F2x=17.3Ncos(34.0)=17.3N0.829=14.34NF_{2x} = 17.3 \, \text{N} \cdot \cos(-34.0^\circ) = 17.3 \, \text{N} \cdot 0.829 = 14.34 \, \text{N}
    • F2y=17.3Nsin(34.0)=17.3N0.559=9.67NF_{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:

Fx=0andFy=0\sum F_x = 0 \quad \text{and} \quad \sum F_y = 0

  • For the x-direction: F1x+F2x+F3x=012.5+14.34+F3x=0F_{1x} + F_{2x} + F_{3x} = 0 \quad \Rightarrow \quad 12.5 + 14.34 + F_{3x} = 0 Solving for F3xF_{3x}: F3x=(12.5+14.34)=26.84NF_{3x} = -(12.5 + 14.34) = -26.84 \, \text{N}

  • For the y-direction: F1y+F2y+F3y=00+(9.67)+F3y=0F_{1y} + F_{2y} + F_{3y} = 0 \quad \Rightarrow \quad 0 + (-9.67) + F_{3y} = 0 Solving for F3yF_{3y}: F3y=9.67NF_{3y} = 9.67 \, \text{N}

Step 3: Calculate the magnitude and direction of the third force

The magnitude of the third force F3F_3 is given by the Pythagorean theorem:

F3=F3x2+F3y2=(26.84)2+(9.67)2=720.27+93.48=813.75=28.53NF_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 θ3\theta_3 of the third force is found using the inverse tangent function:

θ3=tan1(F3yF3x)=tan1(9.6726.84)=tan1(0.36)=19.8\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?


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.

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Math Problem Analysis

Mathematical Concepts

Vector Addition
Equilibrium
Trigonometry

Formulas

Vector components: F_x = F * cos(θ), F_y = F * sin(θ)
Equilibrium conditions: ΣF_x = 0, ΣF_y = 0
Pythagorean theorem: F = √(F_x² + F_y²)
Inverse tangent: θ = tan⁻¹(F_y / F_x)

Theorems

Newton's First Law of Motion (Equilibrium of Forces)

Suitable Grade Level

Grades 11-12