<p>The resultant of two forces <em>P</em> N and 3 N is a force of 7 N. If the direction of 3 N force were reversed, the resultant would be \(\sqrt{19}\) N. The value of <em>P</em> is</p>
Step-by-Step Solution
Key Concept: Use the parallelogram law of vector addition: R² = A² + B² + 2AB cos θ for the original configuration, then apply it again with the reversed direction (angle becomes 180° - θ) to create two equations in two unknowns (P and θ).
Step 1: Let the angle between forces P and 3 be θ. Using R^2 = A^2 + B^2 + 2AB cos θ for the first case: 7^2 = P^2 + 3^2 + 2(P)(3) cos θ 49 = P^2 + 9 + 6P cos θ 40 = P^2 + 6P cos θ ... (1) Step 2: When 3 N force is reversed, the angle becomes (180° - θ), so cos(180° - θ) = -cos θ (√19)^2 = P^2 + 3^2 + 2(P)(3)(-cos θ) 19 = P^2 + 9 - 6P cos θ 10 = P^2 - 6P cos θ ... (2) Step 3: Add equations (1) and (2): 40 + 10 = P^2 + 6P cos θ + P^2 - 6P cos θ 50 = 2P^2 P^2 = 25 P = 5 N Verification: From (1): 40 = 25 + 6(5) cos θ ⟹ cos θ = 1/2, so θ = 60° Check (2): 10 = 25 - 6(5)(1/2) = 25 - 15 = 10 ✓ ∴ Answer: A (P = 5 N)
Correct Answer: A