Question

Forces F1,F2, and F3, act at O as shown. If F2 = 5√3 N and 'O' is in equilibrium, value of F1 and F3 are​

Answer 1

Answer:

Refer the image for the answer :)

Answer 2

Given : Forces $\[{{F_1}}\]$, $\[{{F_2}}\]$  and $\[{{F_3}}\]$, act at O, 'O' is in equilibrium

     $\[{{F_2}}\]$ = 5√3 N.

To Find : Value of $\[{{F_1}}\]$ and $\[{{F_3}}\]$

Solution :

Angle between $\[{{F_2}}\]$  and  $\[{{F_1}}\]$ are $\[90^\circ \]$

Angle between $\[{{F_2}}\]$  and $\[{{F_3}}\]$ are $\[90^\circ \]$ + $\[60^\circ \]$= $\[150^\circ \]$

Angle between   $\[{{F_1}}\]$  and $\[{{F_3}}\]$, are  $\[90^\circ \]$ + $\[30^\circ \]$ =$\[120^\circ \]$

thus , by Lami Theorem

$\frac{A}{\sin \alpha}=\frac{B}{\sin \beta}=\frac{C}{\sin \gamma}$

where A, B and C are the magnitudes of the three coplanar, concurrent and non-collinear vectors,which keep the object in static equilibrium, and α, β and γ are the angles directly opposite to the vectors.

Using Lami theorem formulae ,

$\[\frac{{{F_1}}}{{\sin 150^\circ }} = \frac{{{F_2}}}{{\sin 120^\circ }} = \frac{{{F_3}}}{{\sin 90^\circ }}\]$                         (Given :$\[{{F_2}}\]$  = $\[{5\sqrt 3 }\]$)

$\[\frac{{{F_1}}}{{\sin 150^\circ }} = \frac{{5\sqrt 3 }}{{\sin 120^\circ }} = \frac{{{F_3}}}{{\sin 90^\circ }}\]$

by this,

$\[\frac{{{F_1}}}{{\sin 150^\circ }} = \frac{{5\sqrt 3 }}{{\sin 120^\circ }}\]$

$\[{{F_1}}\]$ = $\[\frac{{15}}{3}\]$

and by this

$\[{{F_3}}\]$ = $\[\frac{{5\sqrt 3 }}{2}\]$

Hence, value of $\[{{F_1}}\]$ is $\[\frac{{15}}{3}\]$ and value of $\[{{F_2}}\]$ is  $\[\frac{{5\sqrt 3 }}{2}\]$ .