Question

The vector sum of three forces having magnitudes $|{\overrightarrow{F}}_1|=100N$, $|{\overrightarrow{F}}_2|=80N$ & $|{\overrightarrow{F}}_3|=60N$ acting on a particle is zero. the angle between ${\overrightarrow{F}}_1$ & ${\overrightarrow{F}}_2$ is nearly:-

• A. ${53}^0$
• B. ${143}^0$
• C. ${37}^0$
• D. ${127}^0$

Answer 1

The correct option is C ${37}^0$

$\text{For equilibrium:}\begin{array}{c}|{\overrightarrow{F}}_1+{\overrightarrow{F}}_2|=\left|{\overrightarrow{F}}_3\right|\end{array}$

$\begin{array}{c}\text{Squaring both side}\\ \Rightarrow {|{\overrightarrow{F}}_1+{\overrightarrow{F}}_2|}^2={\left|{\overrightarrow{F}}_3\right|}^2\\ \Rightarrow {\left|{\overrightarrow{F}}_1\right|}^2+{\left|{\overrightarrow{F}}_2\right|}^2+2\left|F_1\right|\left|F_2\right|\cos \theta \\ ={\left|{\overrightarrow{F}}_3\right|}^2\\ \Rightarrow {100}^2+{80}^2+2\times 100\times 80\times \cos \theta \\ ={60}^2\end{array}$

$\begin{array}{cc}\Rightarrow \cos \theta & =\frac{12800}{16000}\\ \Rightarrow \cos \theta & =\frac{4}{5}\end{array}$

$\theta ={37}^{\circ }$