Problem Statement
A wheel has five equally spaced radial spokes, all in tension. If the tensions of three consecutive spokes are 500 N, 700 N and 600 N respectively, find the tensions in the other two spokes.
Step-by-Step Solution
Key Geometric Properties
- Five spokes are equally spaced at 72° angles
- The three consecutive known tensions are 500 N, 700 N, and 600 N (spokes 1, 2, and 3)
- We need to find tensions P₁ and P₂ in spokes 4 and 5
- For equilibrium, the sum of all force components must equal zero in both horizontal and vertical directions
Step 1: Resolve Forces Along Horizontal Direction
For horizontal equilibrium: ΣH = 0
500 + 700·cos72° – 600·cos36° – P₁·cos36° + P₂·cos72° = 0
Calculate the trigonometric values:
- cos72° = 0.309
- cos36° = 0.809
Substituting these values:
500 + 700 × 0.309 – 600 × 0.809 – P₁ × 0.809 + P₂ × 0.309 = 0
500 + 216.3 – 485.4 – 0.809P₁ + 0.309P₂ = 0
0.809P₁ – 0.309P₂ = 230.9
Dividing both sides by 0.809:
P₁ – 0.38P₂ = 285.4 … (i)
Step 2: Resolve Forces Along Vertical Direction
For vertical equilibrium: ΣV = 0
700·sin72° + 600·sin36° – P₁·sin36° – P₂·sin72° = 0
Calculate the trigonometric values:
- sin72° = 0.951
- sin36° = 0.588
Substituting these values:
700 × 0.951 + 600 × 0.588 – P₁ × 0.588 – P₂ × 0.951 = 0
665.7 + 352.8 – 0.588P₁ – 0.951P₂ = 0
0.588P₁ + 0.951P₂ = 1018.5
Dividing both sides by 0.588:
P₁ + 1.62P₂ = 1732.1 … (ii)
Step 3: Solve the Simultaneous Equations
Subtract equation (i) from (ii):
(P₁ + 1.62P₂) – (P₁ – 0.38P₂) = 1732.1 – 285.4
2.0P₂ = 1446.7
P₂ = 723.3 N
Substitute the value of P₂ back into equation (i):
P₁ – 0.38 × 723.3 = 285.4
P₁ – 274.9 = 285.4
P₁ = 560.3 N
Final Result
Tension in spoke 5 (P₂) = 723.3 N
Conceptual Explanation & Applications
Core Concepts:
- Static Equilibrium: The sum of all forces must equal zero in both horizontal and vertical directions for the wheel to be in equilibrium
- Vector Resolution: Breaking each tension force into horizontal and vertical components using trigonometric functions
- Radial Symmetry: The five equally spaced spokes create a regular pentagonal pattern with 72° between each spoke
- Simultaneous Equations: Using horizontal and vertical equilibrium to create two equations with two unknowns
Real-World Applications:
- Mechanical Engineering: Analysis of tension in bicycle or vehicle wheel spokes
- Structural Engineering: Designing tensioned radial support systems in structures
- Materials Science: Understanding load distribution in symmetric structures
- Aerospace Engineering: Analysis of satellite dish support structures or radar arrays
Why It Works:
This problem demonstrates the principle of static equilibrium in a radial system. By resolving forces into their orthogonal components and applying the condition that their sums must equal zero, we can determine unknown forces in the system. The angular spacing provides the necessary trigonometric relationships to resolve each force accurately. The resulting tensions ensure the wheel remains in perfect equilibrium despite the uneven distribution of forces among the spokes.
