# Question 89967

May 23, 2016

Magnitude of resultant force $R = \sqrt{{A}^{2} + {B}^{2} + 2 A B \cos \theta}$
Angle $\alpha$ of resultant force with $x$ axis $= {\tan}^{-} 1 \left(\frac{B \sin \theta}{A + B \cos \theta}\right)$

#### Explanation:

We need to use the Law of parallelogram of vectors.

The law states that when two vectors (forces) act on a particle at the same time are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point their resultant vector is represented by the diagonal of the parallelogram in magnitude and direction drawn from the same point.

As given in the above figure two forces $\vec{A} \mathmr{and} \vec{B}$, inclined at angle θ#, act on a particle simultaneously. Let these be represented in magnitude and direction by two adjacent sides $O P$ and $P Q$ of parallelogram drawn from origin $O$ (other two sides omitted for simplicity).

According to parallelogram law of vectors, Magnitude and Direction of Resultant $\vec{R}$, obtained by joining tail of $\vec{A}$ with tip of $\vec{B}$ is found as follows.
Let $\vec{R}$ make an angle $\alpha$ with $x$ axis.
Draw perpendicular $Q E$ on $O P$ extended. Considering only the scalar part of the vectors
1. In right $\Delta$ $Q P E$
$P E = B \cos \theta$
$E Q = B \sin \theta$
2. In right $\Delta$ $O Q E$
$O E = O P + P E = A + B \cos \theta$
${R}^{2} = {\left(A + B \cos \theta\right)}^{2} + {\left(B \sin \theta\right)}^{2}$, using Pythagoras theorem.
$\implies {R}^{2} = \left({A}^{2} + {B}^{2} {\cos}^{2} \theta + 2 A B \cos \theta\right) + \left({B}^{2} {\sin}^{2} \theta\right)$

$\implies {R}^{2} = \left({A}^{2} + {B}^{2} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right) + 2 A B \cos \theta\right)$,
simplifying and taking square root of both sides we get
$R = \sqrt{{A}^{2} + {B}^{2} + 2 A B \cos \theta}$
3. $\tan \alpha = \frac{Q E}{O E} = \frac{B \sin \theta}{A + B \cos \theta}$
$\implies \alpha = {\tan}^{-} 1 \left(\frac{B \sin \theta}{A + B \cos \theta}\right)$