Question

Two equal forces F have resultant of 1.5F .Find the angle between the forces​

Answer 1

Answer:

The angle between the two forces will be "120°".

Explanation:

Given,

The resultant of force = 1.5F

Now,

By using formula for two resultant of two equal forces, we get

$R^2=A^2+B^2+2ABCos\theta$

On putting the values, we get

⇒  $1.5^2=1.5^2+1.5^2+2(1.5)(1.5)Cos\theta$

⇒  $2.25=2.25+2.25+4.5Cos\theta$

⇒  $2.25=4.5+4.5Cos\theta$

⇒  $4.5Cos\theta=2.25-4.5$

⇒  $4.5Cos\theta=-2.25$

⇒  $Cos\theta=\frac{-2.25}{4.5}$

⇒  $Cos\theta=-\frac{1}{2}$

⇒  $\theta=120^{\circ}$

Answer 2

Answer:

Angle between the forces​ $= 82.82$ degrees

Explanation:

The resultant force value $= 1.5$  times of Force F

$|R| = |F1| = |F2| =|F|$

Where R is the resultant force

$R^2 = A^2 + B^2 + 2AB cos X\\(1.5F)^2 = F^2 + F^2 + 2F^2 cos X\\2.25F^2 = 2F^2 + 2F^2 cos X$

Cos X is the angle between the two forces.

On further simplifying the equation we get

$2.25F^2 = 2F^2(1 + cos X)$

$(1 + cos X) = \frac{2.25F^2}{2F^2} \\(1 + cos X) = \frac{2.25}{2} \\(1 + cos X) = 1.125\\cos X = 1.125 - 1\\Cos X = 0.125\\X = Cos^{-1}0.125$

$X = 82.82$ degrees

Angle between the forces​ $= 82.82$ degrees