Two forces of unequal magnitude simultaneously act on a particle making an angle `theta (=120^(@))` with each other.If one of them is reversed ,the accelaration of the particle is become `sqrt(3)` times. Calculate the ratio of the magnitude of the forces.

Explanation

Let the two forces be `vec(F)_(1)` and `vec(F)_(2)` that are inclined at angle `theta`.
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The resultant of these forces is `vec(R)_(1)=vec(F)_(1)+ vec (F)_(2)`. Then `|vec(R)_(1)|=|vec(F)_(1)+vec(F)_(2)|`.
Using parallelogram law of vector addition,we have
`|vec(R)_(1)|=sqrt(F_(1)^(2)+F_(2)^(2)+2F_(1)F_(2)costheta)`....(i)
If the direction of `vec(F)_(2)`is reversed,new
resultant force `vec(R)_(2)=vec(F)_(1)+(-vec(F)_(2))`.
Using parallelogram law of vectors, the magnitude of new resultant is `|vec(R)_(2)|=sqrt(F_(1)^(2)+F_(2)^(2)+2F_(1)F_(2)cos theta)`..(ii)

since force is directly propotional to acceleration `vec(F)prop vec(a)`.
Hence, `|vec(R)_(2)|/(|vec(R)_(1)|)=a_(2)/(a_(1))=sqrt(3/1)`
From Eqs.(i) (ii) ,
`sqrt(F_(1)^(2)+F_(2)^(2)-2F_(1)F_(2)costheta)/sqrt(F_(1)^(2)+F_(2)^(2)+2F_(1)F_(2)costheta)=sqrt(3)`
`rArrF_(1)^(2)+F_(2)^(2)+4F_(1)F_(2)cos theta=0`..(iii)
Here `theta=120^(@)` and dividing Eq.(iii) by `F_(2)^(2)`,
`((F_(1))/(F_2))^(2)+1-2((F_2)/(F_1))=0`..(iv)
Let `F_(1)/F_(2)=x`.Equation (iv)becomes `(x)^(2)+1-2(x)=0`
after solving ,`x=1 "or" F_(1)/F_(2)=1`.