Consider an object kept at the centre, in the $X Y$-plane, on which five coplanar forces act as shown in the figure. The resultant force on the object is

• Correct answer
  A

  $6.5 \mathrm{~N}, 330^{\circ}$
• B

  $6.5 \mathrm{~N}, 300^{\circ}$
• C

  $6.5 \mathrm{~N}, 30^{\circ}$
• D

  $5.7 \mathrm{~N}, 331^{\circ}$

Solution

The total force in $X$-direction, $\begin{aligned} & \Sigma F_x=15 \cos 60^{\circ}+19-11 \cos 30^{\circ}-16 \cos 45^{\circ} \\ & \Sigma F_x=5.66 \mathrm{~N} \end{aligned}$ The total force in $Y$-direction, $\begin{aligned} & \Sigma F_y=-22+15 \sin 60^{\circ}+16 \cos 45^{\circ}-11 \sin 30^{\circ} \\ & \Sigma F_y=-3 \cdot 196 \mathrm{~N} \end{aligned}$ Then, resultant force $\begin{aligned} \Rightarrow \quad R & =\sqrt{F_x^2+F_y^2}=\sqrt{(5 \cdot 66)^2+(3 \cdot 196)^2} \\ R & =6 \cdot 5 \mathrm{~N} \end{aligned}$ Direction of resultant, $\begin{aligned} \Rightarrow & \tan \theta & =F_y / F_x=\frac{-3 \cdot 196}{5 \cdot 66} \\ \Rightarrow & \theta & =330^{\circ} \end{aligned}$

Hence, the correct answer is option: 'A'.