Question

two forces of unequal magnitude simultaneously act on a particle making an angle theta is equal to 150 degree with each other. if one of them is river the net force on the particle is doubled .calculate the ratio of the magnitude of the forces​

Answer 1

Given that,

Angle = 150°

Let the forces acting on the particle be F₁ and F₂.

The resultant force is

$R^2=F_{1}^2+F_{2}^2+2F_{1}F_{2}\cos\theta$

Put the value of angle

$R^2=F_{1}^2+F_{2}^2+2F_{1}F_{2}\cos150$

$R^2=F_{1}^2+F_{2}^2-2F_{1}F_{2}\times\dfrac{\sqrt{3}}{2}$

$R^2=F_{1}^2+F_{2}^2-\sqrt{3}F_{1}F_{2}$......(I)

If one of them is reversed the net force on the particle is doubled.

If one of the force is reversed than the angle between the forces becomes 30°.

The resultant force is

$(S)^2=F_{1}^2+F_{2}^2+2F_{1}F_{2}\cos\theta$

Put the value into the formula

$(2R)^2=F_{1}^2+F_{2}^2+2F_{1}F_{2}\cos30$

$4R^2=F_{1}^2+F_{2}^2+2F_{1}F_{2}\times\dfrac{\sqrt{3}}{2}$

$4R^2=F_{1}^2+F_{2}^2+\sqrt{3}F_{1}F_{2}$

$R^2=\dfrac{F_{1}^2+F_{2}^2+\sqrt{3}F_{1}F_{2}}{4}$....(II)

We need to calculate the ratio of the magnitude of the forces​

From equation (I) and (II)

$F_{1}^2+F_{2}^2-\sqrt{3}F_{1}F_{2}=\dfrac{F_{1}^2+F_{2}^2+\sqrt{3}F_{1}F_{2}}{4}$

$4F_{1}^2+4F_{2}^2-4\sqrt{3}F_{1}F_{2}=F_{1}^2+F_{2}^2+\sqrt{3}F_{1}F_{2}$

$3F_{1}^2+3F_{2}^2-5\sqrt{3}F_{1}F_{2}=0$

On dividing by $F_{2}^2$ on both side

$\dfrac{3F_{1}^2}{F_{2}^2}+3-\dfrac{5\sqrt{3}F_{1}}{F_{2}}=0$

Let $\dfrac{F_{1}}{F_{2}}=x$

So, $3x^2-5\sqrt{3}x+3=0$

After solving

$x = 0.40, 2.48$

Hence, The ratio of the magnitude of the forces is either 0.40 and 2.48.