Question

3. Two forces 7 Newton and 9 Newton are acting on a point at an angle of 60 degree. Find the magnitude and direction of the resultant force?​

Answer 1

$\large{\red{\frak{Given }}}\begin{cases}\bullet\textsf{ There are two forces of 7N and 9N .}\\\bullet\textsf{ They are acting at angle of 60 ^{\circ} .}\end{cases}$

$\large{\red{\frak{To \ Find }}}\begin{cases}\bullet\textsf{The magnitude of the resultant force.}\\\bullet\textsf{ The direction of the resultant force. }\end{cases}$

So here we need to find the magnitude and the direction of the resultant force . Here we can find out magnitude by Vector Law of Addition .

$\underline{\dag\textsf{\textbf{ Finding the resultant force :- }}}$

$\sf:\implies\pink{F_{(net)}= \sqrt{F_1^2+F_2^2+2F_1F_2 cos\theta }}\\\\\sf:\implies F_{(net)}= \sqrt{ (9N)^2+(7N)^2+2(7N)(9N)cos 60^o }\\\\\sf:\implies F_{(net)}= \sqrt{49N^2+81N^2+126N^2\times\dfrac{1}{2}}\\\\\sf:\implies F_{(net)}= \sqrt{130N^2+63N^2}\\\\\sf:\implies F_{(net)}= \sqrt{193N^2} \\\\\sf:\implies \underset{\blue{\sf Resultant \ Magnitude }}{\underbrace{\boxed{\pink{\frak{ Force_{(net)}= 13.89 \ Newtons }}}}}$

$\rule{200}2$

$\underline{\dag\textsf{\textbf{ Finding the direction of Resultant :- }}}$

$\sf:\implies \pink{ \phi = tan^{-1}\bigg( \dfrac{B \ sin\theta}{A+B \ cos \theta } \bigg) }\\\\\sf:\implies \phi = tan^{-1}\bigg( \dfrac{9 sin60^o}{7+9 cos\theta }\bigg)\\\\\sf:\implies \phi = tan^{-1}\left(\dfrac{9\times\dfrac{\sqrt3}{2}}{7+9\times\dfrac{1}{2}}\right) \\\\\sf:\implies \phi =tan^{-1}\left( \dfrac{\dfrac{9\sqrt3}{2}}{7+\dfrac{9}{2}}\right) \\\\\sf:\implies \phi =tan^{-1} \left(\dfrac{\dfrac{9\sqrt3}{2}}{\dfrac{23}{2}}\right)\\\\\sf:\implies\underset{\blue{\sf Required\ Angle }}{\underbrace{\boxed{\pink{\frak{ \phi =tan^{-1}\bigg(\dfrac{9\sqrt3}{23}\bigg)}}}}}$