Math Problem Statement

For the semi-circular arc shown with radius of 5 m, supported by 7.5 m long cable, determine the tension in cable BD.

Solution

To solve this problem, we need to analyze the forces acting on the semi-circular arc and determine the tension in cable BD. Here's a step-by-step approach:

Given:

  • Radius of the semi-circular arc, R=5mR = 5 \, \text{m}
  • Length of cable BD, L=7.5mL = 7.5 \, \text{m}
  • Distance between points A and D, AD=10.0mAD = 10.0 \, \text{m}
  • Force acting at point B, F=18kNF = 18 \, \text{kN}

Assumptions:

  • The arc is symmetrical.
  • The tension in the cable is uniform.
  • The system is in static equilibrium.

Solution:

  1. Identify Points and Angles:

    • Points A and D are at the base of the semi-circular arc.
    • Point B is at the top of the arc.
    • Point C is where the cable BD meets the arc.
    • The angle subtended by the arc (from A to D) is 180 degrees.
  2. Analyze Geometry:

    • The semi-circular arc has a radius of 5 m. The distance from B to the arc's center is also 5 m.
    • Since AD=10mAD = 10 \, \text{m}, the center of the arc is at the midpoint between A and D.
    • The length of the cable BD is 7.5 m.
  3. Calculate the Angles:

    • Use the Law of Cosines in triangle BCD to find the angle θ\theta between BD and the vertical.

    cosθ=BC2+CD2BD22BCCD\cos \theta = \frac{BC^2 + CD^2 - BD^2}{2 \cdot BC \cdot CD} Where:

    • BC=5mBC = 5 \, \text{m} (radius)
    • CD=5mCD = 5 \, \text{m} (radius)
    • BD=7.5mBD = 7.5 \, \text{m}
  4. Set Up Equilibrium Equations:

    • Resolve the forces vertically and horizontally:
    • Vertical forces: Tsinθ=18kNT \sin \theta = 18 \, \text{kN}
    • Horizontal forces: Tcosθ=horizontal component of the reaction at A and DT \cos \theta = \text{horizontal component of the reaction at A and D}
  5. Solve for Tension TT:

Let's calculate these values now.The tension TT in cable BD is approximately 18.14 kN. The angle θ\theta between cable BD and the vertical is about 97.18 degrees.

This tension accounts for balancing the vertical force of 18 kN applied at point B while considering the geometry of the arc and cable.

Would you like more details, or do you have any questions?

Here are 5 related questions that might help expand your understanding:

  1. How would the tension change if the length of cable BD was different?
  2. What would happen to the equilibrium if the force at B was not vertical?
  3. How can we compute the horizontal components of forces at points A and D?
  4. What would be the impact on tension if the arc radius changed?
  5. How do we use vector resolution for complex force systems like this?

Tip: When solving problems involving tension and angles, always consider the geometry and use trigonometric identities for accurate calculations.

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Math Problem Analysis

Mathematical Concepts

Statics
Geometry
Trigonometry

Formulas

Law of Cosines: cos(θ) = (BC^2 + CD^2 - BD^2) / (2 * BC * CD)
Vertical equilibrium: T*sin(θ) = Applied force
Horizontal equilibrium: T*cos(θ) = Horizontal reaction forces

Theorems

Law of Cosines
Equilibrium of Forces

Suitable Grade Level

College-level Engineering or Physics