the resultant of two forces 3P and 2P is R. If the first force is doubled then the resultant is also doubled. then the angle between the two forces us. plz xplain with the soln..

Dear Nandini,

In 1st case, R²=(3P)²+(2P)²+2*3*2*P²*cosΘ = 9P²+4P²+12P²cosΘ = 13P²+12P²cosΘ ----->(1)

In 2nd case, (2R)²=(6P)²+(2P)²+2*6*2*P²*cosΘ ; 4R²=36P²+4P²+24P²cosΘ = 40P²+24P²cosΘ ------->(2)

Multiply (1) by 4; 4R²=52P²+48P²cosΘ ------>(3)

(3)-(2) gives, 0=12P²+24P²cosΘ

Therefore, cosΘ = (-1/2) => Θ = cos^-1(-1/2) = 120°

case 1, R²=(3P)²+(2P)²+2*3*2*P²*cosΘ = 9P²+4P²+12P²cosΘ = 13P²+12P²cosΘ ----->(1)

case 2, (2R)²=(6P)²+(2P)²+2*6*2*P²*cosΘ ; 4R²=36P²+4P²+24P²cosΘ = 40P²+24P²cosΘ ------->(2)

Multiply (1) by 4; 4R²=52P²+48P²cosΘ ------>(3)

(3)-(2) gives, 0=12P²+24P²cosΘ

Therefore, cosΘ = (-1/2) => Θ = cos^-1(-1/2) = 120°

r²=9p²+4p²+2.6p²cosθ....................1

4r²=(6p)²+4p²+2.12cosθ..................2

dividing 2 from 1 

4=(36+4+24cosθ)/(9+4+12cosθ)

9+4+12cosθ=10+6cosθ

6cosθ=-3

cosθ=-0.5

θ=120degree