Two forces 8n and 12n act at 120 the force required to keep the body in equilibrium is ?
Answers
Answer ⇒ Magnitude of opposite force is 40.89°. The direction is just opposite to the resultant of given two vectors.
SEE THE ATTACHMENT.
Explanation ⇒ The Force which can keep both the body in equilibrium is the Resultant Force but the resultant force will be just opposite in direction to the resultant of these two vectors.
Let us first find the resultant of these 2 given vectors,
R² = A² + B² + 2ABCosФ
In given question,
A = 8 N, B = 12 N and Ф = 120.
Thus,
R² = 8² + 12² + 2 × 8 × 12 × Cos120
∴ R² = 64 + 144 + 16 × 12 × -1/2
∴ R² = 208 - 96
∴ R² = 112
∴ R = √112 N.
Now, The angle of the Resultant with an Force B,
tanθ = ASinΦ/( B + ACosΦ)
∴ tanθ = 8 × Sin120/(12 + 8 × Cos120)
∴ tanθ = (8 × √3/2)/(12 - 8 × 1/2)
∴ tanθ = (4√3)/(12 - 4)
∴ tanθ = √3/2
or θ = 40.89°
Thus, Resultant opposite force that should be applied to keep them in equilibrium is opposite to there resultant which is making angle of 40.89°.
Hope it helps.
Answer:
Explanation:Let us first find the resultant of these 2 given vectors,
R² = A² + B² + 2ABCosФ
In given question,
A = 8 N, B = 12 N and Ф = 120.
Thus,
R² = 8² + 12² + 2 × 8 × 12 × Cos120
∴ R² = 64 + 144 + 16 × 12 × -1/2
∴ R² = 208 - 96
∴ R² = 112
∴ R = √112 N.
Now, The angle of the Resultant with an Force B,
tanθ = ASinΦ/( B + ACosΦ)
∴ tanθ = 8 × Sin120/(12 + 8 × Cos120)
∴ tanθ = (8 × √3/2)/(12 - 8 × 1/2)
∴ tanθ = (4√3)/(12 - 4)
∴ tanθ = √3/2
or θ = 40.89°