the magnitude of resultant od two forces of magnitudes 3P and 2P is R. If the first force is doubled,the magnitude of the resultant is also doubled. Find the angle between the two forces.
the magnitude of resultant od two forces of magnitudes 3P and 2P is R. If the first force is doubled,the magnitude of the resultant is also doubled. Find the angle between the two forces.
The resultant of two vectors A and B is given as √(A^2 + B^2 + 2ABcosΦ), where Φ is the angle between them (what I 've shown in 'phi ', not 'theta ' but the naming of the angle doesn 't really matter; theta cannot be properly represented here)
Therefore √(9p^2 + 4p^2 + 12p^2.cosΦ) is their resultant C (let)
3p doubled -> 6p; therefore new resultant = √(36p^2 + 4p^2 + 24p^2.cosΦ) = D (let)
Now it 's given that, D = 2C
So, squaring on both sides, 4(13p^2 + 12p^2.cosΦ) = 40p^2 + 24p^2.cosΦ
Equating, and cancelling p^2, cosΦ = 1/2 => Φ = 60 degrees or π/3 radians
The resultant of two vectors A and B is given as √(A^2 + B^2 + 2ABcosΦ), where Φ is the angle between them (what I 've shown in 'phi ', not 'theta ' but the naming of the angle doesn 't really matter; theta cannot be properly represented here)
Therefore √(9p^2 + 4p^2 + 12p^2.cosΦ) is their resultant C (let)
3p doubled -> 6p; therefore new resultant = √(36p^2 + 4p^2 + 24p^2.cosΦ) = D (let)
Now it 's given that, D = 2C
So, squaring on both sides, 4(13p^2 + 12p^2.cosΦ) = 40p^2 + 24p^2.cosΦ
Equating, and cancelling p^2, cosΦ = 1/2 => Φ = 60 degrees or π/3 radians
