Question

The resultant of two forces at right angle is 5 N. When the angle between them is
120, the resultant is 13 N. Then the forces are
​

Answer 1

Answer

The two forces are 17.25 N and -0.25 N

Solution

Given -

$\rm \vec R = 5 N when \theta = 90\degree$

$\rm \vec R = 13 N when \theta = 120\degree$

To find -

The forces A and B (let)

Formula used -

$\rm R = \sqrt{A^2 + B^2 + 2ABcos\theta}$

When the vectors are perpendicular -

$\implies$$\rm \vec R = 5 N$

$\implies$$\rm \theta = 90\degree$

Substituting the value in formula -

$\implies$$\rm R = \sqrt{A^2 + B^2 + 2ABcos\theta}$

$\implies$$\rm 5 = \sqrt{A^2 + B^2 + 2ABcos90}$

$\implies$$\rm 5^2 = A^2 + B^2 + 2AB \: \: [cos90\degree = 0]$

$\implies$$\rm 5^2 = A^2 + B^2$

$\implies$$\rm A^2 + B^2 = 5^2$ eq i

When angle between them is 120°

$\implies$$\rm \vec R = 13 N$

$\implies$$\rm \theta = 120\degree$

Substituting the value in formula -

$\implies$$\rm R = \sqrt{A^2 + B^2 + 2ABcos\theta}$

$\implies$$\rm 13 = \sqrt{A^2 + B^2 + 2ABcos120}$

$\implies$$\rm 13^2 = A^2 + B^2 + 2AB (\frac{-1}{2})$

$\implies$ $\rm 169 = A^2 + B^2 - AB$ eq ii

By substituting the value of $\rm A^2 + B^2$ from equation i in equation ii -

$\implies$$\rm 169 = A^2 + B^2 - AB$

$\implies$$\rm 169 = 25 - AB$

$\implies$$\rm AB = -144$

$\implies$$\rm A ^2 + B^2 + 2AB = (A + B)^2$

$\implies$$\rm 25 - 288 = (A + B)^2$

$\implies$$\rm 263 = (A + B)^2$

$\implies$$\rm A + B = \sqrt{263} = 17.1$

$\implies$$\rm A^2 + B^2 - 2AB = (A - B)^2$

$\implies$$\rm 25 - 2 ( - 144 ) = (A - B)^2$

$\implies$$\rm (A - B)^2 = 313$

$\implies$$\rm A - B = \sqrt{313} = 17.6$

$\implies$$\rm A + B = 17.1$

$\implies$$\rm A - B = 17.6$

Solving the equation -

$\rm 2A = 34.7$

$\implies$$\rm A = 17.35$

$\implies$$\rm B = -0.25$

The two forces are 17.25 N and -0.25 N.