Question

Three forces F1=3 i* +6j*, F2= 3i*+2j*, and F3 = -6i* + b j* are in equilibrium then the value of b=......

Answer 1

Question :

Three forces $\vec{F}_{1} = 3i + 6j$ , $\vec{F}_{2} = 3i + 2j$ ,

$\vec{F}_{3} = -6i + bj$ are in equilibrium then find the value of b ?

Theory :

• Equlilbrum of a particle

Translatory equlilbrum:

when coherent and coplanar forces act on a body or a system then to be in equilibrium

${\purple{\boxed{\large{\bold{ \sum \vec{F} = 0}}}}}$

Solution :

★ Given forces :

$\vec{F}_{1} = 3i + 6j$

$\vec{F}_{2} = 3i + 2j$

and $\vec{F}_{3} = -6i + bj$

_________________

$\sum \vec{f}_{net} = f _{1} + f _{2} + f_{3}$

$= 3i + 6j + 3 i + 2j - 6i + bj$

$= 6i + 8j - 6i + bj$

$= 8j + bj$

_________________________

For equlilbrum :

$\sum \vec{f} = 0$

$\implies \: 8j + bj = 0$

$\implies \: (8 + b)j = 0$

j is a unit vector , it's magnitude is 1

$\implies \: b = - 8$

${\purple{\boxed{\large{\bold{b = -8 }}}}}$