Question

Two forces of $3\text{N}$ and $4\text{N}$ are acting at a point such that the angle between them is $60\text{degrees}$. Find the resultant force.

• A. Mangnitude of $R=8\text{N}$ and angle $q={\tan }^{-1}0.472$
• B. Mangnitude of $R=6\text{N}$ and angle $q={\tan }^{-1}0.572$
• C. Mangnitude of $R=6.08\text{N}$ and angle $q={\tan }^{-1}0.472$
• D. Mangnitude of $R=6.08\text{N}$ and angle $q={\tan }^{-1}0.872$

Answer 1

The correct option is C Mangnitude of $R=6.08\text{N}$ and angle $q={\tan }^{-1}0.472$

magnitude of resultant $R$ is given as
= $\sqrt{(3{)}^2+(4{)}^2+2\times 3\times 4\cos {60}^{\circ }}$
$=\sqrt{37}=6.08\text{N}$
Direction of R is given by finding the angle $q$
Thus, $\tan q=\frac{3\sin \left({60}^{\circ }\right)}{4+3\cos \left({60}^{\circ }\right)}=0.472$
$\Rightarrow q={\tan }^{-1}0.472$