Problem Statement
Two forces of magnitudes 3P and 2P acting at a point have resultant R. When the first force is doubled, the resultant also doubles. Find the angle between the forces.
Given Parameters
Initial Forces
3P and 2P
Modified Force
6P (doubled first force)
Resultant Relationship
2R = New resultant
Step-by-Step Solution
Original Configuration
Step 1: Initial resultant calculation:
R² = (3P)² + (2P)² + 2·3P·2P·cosθ
R² = 9P² + 4P² + 12P²cosθ
R² = 13P² + 12P²cosθ (Equation 1)
R² = 9P² + 4P² + 12P²cosθ
R² = 13P² + 12P²cosθ (Equation 1)
Modified Configuration
Step 2: New resultant after doubling:
(2R)² = (6P)² + (2P)² + 2·6P·2P·cosθ
4R² = 36P² + 4P² + 24P²cosθ
4R² = 40P² + 24P²cosθ (Equation 2)
4R² = 36P² + 4P² + 24P²cosθ
4R² = 40P² + 24P²cosθ (Equation 2)
Solving Equations
Step 3: Substitute R² from Equation 1 in Equation 2:
4(13P² + 12P²cosθ) = 40P² + 24P²cosθ
52P² + 48P²cosθ = 40P² + 24P²cosθ
52P² + 48P²cosθ = 40P² + 24P²cosθ
Step 4: Simplify equation:
12P² + 24P²cosθ = 0 ⇒ 24P²cosθ = −12P²
cosθ = −½ ⇒ θ = 120°
cosθ = −½ ⇒ θ = 120°
Angle Between Forces: θ = 120°
Physical Interpretation
Force Vector Analysis
The 120° angle creates a specific force balance where doubling one force exactly doubles the resultant. This occurs because the negative cosine value (cos120° = −½) introduces a cancellation effect that maintains proportional relationships between the forces.
Verification Table
| Original Resultant: | √[13P² + 12P²(−½)] = √7P ≈ 2.645P |
| Modified Resultant: | √[40P² + 24P²(−½)] = √28P ≈ 5.291P = 2×2.645P |
Engineering Significance
- Demonstrates non-linear relationships in vector systems
- Important for tension calculations in cable systems
- Fundamental for understanding force equilibrium conditions
