## Composition Of Several Vectors By Resolution

Three or more vectors can be added by the law of polygon of vectors. This method is geometrically convenient but is inconvenient mathematically. However, when vectors lie on a plane, an elegant method is to resolve each of the vectors into two mutually perpendicular components while performing addition.

**Composition Of Several Vectors By Resolution Process:** The vectors are drawn on a plane from a common initial point. Two mutually perpendicular axes X and Y are drawn from that point and each vector is resolved into components along these two axes.

Next, algebraic sums of the x-components and of the y -y-components of the vectors with proper sign (+ or -) are determined separately. Now, vector addition of the resultant x and y components gives the resultant of all the vectors.

**Composition Of Several Vectors By Resolution Calculation:** Let \(\vec{P}, \vec{Q}, \vec{R}\) be three coplanar vectors having a common initial point O. To find the resultant, axes OX and OY are drawn, coplanar with the vectors.

Let \(\vec{P}, \vec{Q}, \vec{R}\) make angles α, β, γ respectively with the x-axis. Thus their respective components along the x-axis are \(P_x=P \cos \alpha, Q_x=Q \cos \beta, and R_x=R \cos \gamma\).

Similarly \(P_y=P \sin \alpha, Q_y=Q \sin \beta and R_y=R \sin \gamma\).

Hence, the resultants (sums) F_{x} and F_{y} along x and y axes are \(F_x=P \cos \alpha+Q \cos \beta+R \cos \gamma\)….(1)

and \(F_y=P \sin \alpha+Q \sin \beta+R \sin \gamma\)….(2)

**Read and Learn More: Class 11 Physics Notes**

For performing the summation, positive or negative values of sine and cosine functions for the angles α, β, γ should be taken into consideration. \(\vec{F}\) represents the resultant of F_{x} and F_{y}, according to the triangle or the parallelogram law of vector addition.

F = \(\sqrt{F_x^2+F_y^2}\)…(3)

and the inclination θ of \(\vec{F}\) with the x-axis is given by \(\tan \theta=\frac{F_y}{F_x}\)…(4)

Equations (3) and (4) give the magnitude and direction respectively of the resultant of the three vectors \(\vec{P}, \vec{Q}, \vec{R}\). The resultant of any number of coplanar vectors can be found using this method.

## Composition Of Several Vectors By Resolution Numerical Examples

**Example 1. Compute the resultant of three coplanar vectors P, 2P and 3P inclined at 120° with one another.**

**Solution:**

Let the x-axis be taken along vector P. Then the vectors 2P and 3P will make angles 120° and 240° respectively with the x-axis. Resolving the vectors into components along the x and y axes and adding, we get the stuns R_{x} and R_{y} respectively along the x and y axes.

⇒ \(R_x=P \cos 0^{\circ}+2 P \cos 120^{\circ}+3 P \cos 240^{\circ}\)

∴ \(R_x=P+2 P\left(-\frac{1}{2}\right)+3 P\left(-\frac{1}{2}\right)=P-P-\frac{3}{2} P=-\frac{3}{2} P\)

⇒ \(R_y=P \sin 0^{\circ}+2 P \sin 120^{\circ}+3 P \sin 240^{\circ}\)

= \(0+2 P \times \frac{\sqrt{3}}{2}+3 P\left(-\frac{\sqrt{3}}{2}\right)=(2-3) \frac{\sqrt{3}}{2} P=-\frac{\sqrt{3}}{2} P\)

Hence, the value of the resultant \(\vec{R}\) is, R = \(\sqrt{R_x^2+R_y^2}=\sqrt{\frac{9}{4} P^2+\frac{3}{4} P^2}=\sqrt{3} P\)

∴ \(\vec{R}\) makes an angle θ with the x-axis such that,\(\tan \theta=\frac{R_y}{R_x}=\frac{-\frac{\sqrt{3}}{2} P}{-\frac{3}{2} P}=\frac{1}{\sqrt{3}}=\tan 30^{\circ} \text { or, } \tan 210^{\circ}\)

As \(R_x\) and \(R_y\) are both negative, R must be in the third

As R_{x} and R_{y} are both negative, R must be in the third quadrant, and θ = 210°.

**Example 2. Five coplanar forces, each of magnitude F, are acting on a particle. Each force is inclined at an angle of 30° with the previous one. Find out the magnitude and direction of the resultant force on the particle.**

**Solution:**

Let us choose the x-axis along the first force, and y-axis along the perpendicular direction on the plane of the forces

The angles made by the five forces with the x-axis are 0°, 30°, 60°, 90° and 120°.

∴ Component of resultant force along x-axis, F_{x} = Fcos0° + Fcos30° + Fcos60° + Fcos90° + Fcos120°

= \(F\left[1+\frac{\sqrt{3}}{2}+\frac{1}{2}+0-\frac{1}{2}\right]\)

= \(F\left(1+\frac{\sqrt{3}}{2}\right)\)

and component of resultant force along y-axis, Fy = Fsin0° + Fsin30° + Fsin60° + Fsin90° + Fsinl20°

= \(F\left[0+\frac{1}{2}+\frac{\sqrt{3}}{2}+1+\frac{\sqrt{3}}{2}\right]=F\left(\frac{3}{2}+\sqrt{3}\right)\)

= \(\sqrt{3} F\left(1+\frac{\sqrt{3}}{2}\right)\)

So, the magnitude of the resultant force, \(F^{\prime}=\sqrt{F_x^2+F_y^2}=F\left(1+\frac{\sqrt{3}}{2}\right) \sqrt{1^2+(\sqrt{3})^2}\)

= \(2 F\left(1+\frac{\sqrt{3}}{2}\right)=(2+\sqrt{3}) F\)

The angle of inclination of F’ with the x-axis, \(\theta=\tan ^{-1} \frac{F_y}{F_x}=\tan ^{-1} \sqrt{3}=60^{\circ} \text { (along the third force) }\)