Problem Statement
Find the components of a force of 150 N into two directions inclined at 45° and 30° with the force.
Step-by-Step Solution
Given Values
- Total force P = 150 N
- Angles: α = 45°, β = 30°
Step 1: Law of Sines Formula
P1 = P · [sin β / sin(α + β)]
P2 = P · [sin α / sin(α + β)]
P2 = P · [sin α / sin(α + β)]
Step 2: Calculate P₁ Component
P1 = 150 × [sin 30° / sin(75°)]
= 150 × [0.5 / 0.9659]
= 150 × 0.5175
= 77.63 N
= 150 × [0.5 / 0.9659]
= 150 × 0.5175
= 77.63 N
Step 3: Calculate P₂ Component
P2 = 150 × [sin 45° / sin(75°)]
= 150 × [0.7071 / 0.9659]
= 150 × 0.7319
= 109.78 N
= 150 × [0.7071 / 0.9659]
= 150 × 0.7319
= 109.78 N
Final Components
P1 = 77.63 N
P2 = 109.78 N
P2 = 109.78 N
Conceptual Explanation
This solution uses the Law of Sines for force resolution:
- The total force forms a triangle with its components
- Angles between components sum to 75° (45° + 30°)
- Sine values determine force distribution:
- sin 30° = 0.5
- sin 45° ≈ 0.7071
- sin 75° ≈ 0.9659
- Components are inversely proportional to their angles from resultant
The smaller angle component (30°) gets less force, while the larger angle component (45°) receives more force.
