Question

Five forces $2N,\sqrt{3}N,5N,\sqrt{3}Nand2N$ respectively act at a particle $P$ as shown in the figure.
The resultant force on the particle $P$ is

• A. None of these
• B. $10N$ making angle ${60}^{\circ }$ with $X-$axis
  
• C. $20N$ along $Y-$axis
  
• D. $10N$ making angle ${60}^{\circ }$ with $Y-$axis
  

Answer 1

The correct option is B $10N$ making angle ${60}^{\circ }$ with $X-$axis


Find the net force along $x-$axis.
$\sum F_x=\text{sum of}x-\text{component of forces}$
$=2+\sqrt{3}\cos {30}^{\circ }+5\cos {60}^{\circ }-2\sin {30}^{\circ }$
$=2+\frac{3}{2}+\frac{5}{2}-1=5N$

Find the net force along $y-$axis.
$\sum F_y=\text{sum of}y-\text{component of forces}$
$=\sqrt{3}\sin {30}^{\circ }+5\sin {60}^{\circ }+\sqrt{3}+2\cos {30}^{\circ }$
$\frac{\sqrt{3}}{2}+\frac{5\sqrt{3}}{3}+\sqrt{3}+\sqrt{3}=5\sqrt{3}N$
$\therefore F=\sqrt{(\sum F_x{)}^2+(\sum F_y{)}^2}$
$=\sqrt{5^2+(5\sqrt{3}{)}^2}$
$=\sqrt{100}=10N$

Find the direction of force on the particle.
Formula Used: $\tan \theta =\frac{\sum F_y}{\sum F_x}$
$\therefore \tan \theta =\frac{\sum F_y}{\sum F_x}=\frac{5\sqrt{3}}{5}=\sqrt{3}\Rightarrow \theta ={60}^{\circ }$
Final Answer: (𝑨)