Find the magnitude of two forces, such that if they act at right angles, their resultant is √10 N, but if they act at 60' their resultant is √13 N.
Answers
Answer:
1 N and 3 N
Explanation:
The resultant of two vectors is given by,
R² = P² + Q² + 2PQcosθ
where
• P, Q denotes the magnitude of vectors
• θ denotes the angle between the two vectors
Let the two forces be F₁ and F₂
Case - (i) :
If two forces act at right angles, their resultant is √10 N
right angle = 90°
√(10)² = F₁² + F₂² + 2F₁F₂cos 90°
10 = F₁² + F₂² + 2F₁F₂ (0)
10 = F₁² + F₂² + 0
10 = F₁² + F₂² -- (1)
Case - (i) :
If two forces act at 60°, their resultant is √13 N
√(13)² = F₁² + F₂² + 2F₁F₂cos 60°
13 = F₁² + F₂² + 2F₁F₂ (1/2)
13 = F₁² + F₂² + F₁F₂
Put F₁² + F₂² = 10 [eqn. (1) ]
13 = 10 + F₁F₂
F₁F₂ = 13 – 10
F₁F₂ = 3
we know,
(a + b)² = a² + b² + 2ab
Similarly,
(F₁ + F₂)² = F₁² + F₂² + 2F₁F₂
(F₁ + F₂)² = 10 + 2(3)
(F₁ + F₂)² = 10 + 6
(F₁ + F₂)² = 16
(F₁ + F₂) = √16
(F₁ + F₂) = 4 –– [2]
Also,
(a – b)² = a² + b² – 2ab
Similarly,
(F₁ – F₂)² = F₁² + F₂² – 2F₁F₂
(F₁ – F₂)² = 10 _ 2(3)
(F₁ – F₂)² = 10 – 6
(F₁ – F₂)² = 4
(F₁ – F₂) = √4
(F₁ – F₂) = 2 –– [3]
Adding both equations (2) and (3),
(F₁ + F₂) + (F₁ – F₂) = 4 + 2
2F₁ = 6
F₁ = 3 N
Substitute in equation (2),
(F₁ + F₂) = 4
(3 + F₂) = 4
F₂ = 4 – 3
F₂ = 1 N
The magnitudes of two forces are 1 N and 3 N