# Vector Addition By Rectangular Components (F.Sc-Physics-Chapter 2)

Let A and B be two vectors which are represented by two directed lines OM and ON respectively. The vector B is added to A by the head to tail rule of vector addition (fig 2.9). thus the resultant vector R = A + B is give, in direction and magnitude, by the vector OP.

In the Fig 2.9 , and are the x components of the vectors A, B and R and their magnitudes are given by the lines OQ, MS, and OR respectively, but

Which means that the sum of the magnitudes of x-components of two vectors which are to be added, is equal to the x-component of the resultant. Similarly the sum of the magnitudes of y-components of two vectors is equal to the magnitude of y-component of the resultant, that is

Since and are the rectangular components of the resultant vector R, hence

The magnitude of the resultant vector R is thus given as

And the direction of the resultant vector is determined from

Similarly for any number of coplanar vectors A, B, C…….. we can write

**Do you know?**

The Chinese acrobats in this incredible balancing act are in equilibrium.

The vector addition by rectangular components consists of the following steps.

i) Find x and y components of all given vectors.

ii) Find x-component of the resultant vector by adding the x-component of all the vectors.

iii) Find y-component of the resultant vector by adding the y-components of all the vectors.

iv) Find the magnitude of resultant vector R using

v) Find the direction of resultant vector R by using

where is the angle, which the resultant vector makes with positive x-axis. The signs of and determine the quadrant in which resultant vector lies. For that purpose proceed as given below.

Irrespective of the sign of , determine the value of from the calculator or by consulting trigonometric tables. Knowing the value of ф, angle θ is determined as follows.

a) If both are positive, then the resultant lies in the first quadrant and its direction is θ = ф.

b) If is – ive and is +ive, the resultant lies in the second quadrant and its direction is θ = 180⁰ – ф.

c) If both are –ive, the resultant lies in the third quadrant and its direction is θ = 180⁰ – ф.

d) If is positive and is negative, the resultant lies in the fourth quadrant and its direction is θ = 360⁰ – ф.

**Example 2.2:** two forces of magnitude 10 N and 20 N act on a body in direction making angles 30⁰ and 60⁰ respectively with x-axis. Find the resultant force.

**Solution:**

**Step (i) x-components **

The x-component of the first force : cos 30⁰

= 10 N x 0.866 = 8.66 N

The x-components of second force =

= 20 N x 0.866 = 17.32 N

**Step (ii)**

The magnitude of y component of the resultant force F

**Step (iii) **

The magnitude of y component of the resultant force F

**Step (iv)**

The magnitude F of the resultant force F

**Step (v)**

F the resultant force F makes and angle θ with the x-axis then

**Example: 2.3:** find the angle between two forces of equal magnitude when the magnitude of their resultant is also equal to the magnitude of either of these forces.

**Solution:** let θ be the angle between two forces F1 is along x-axis. Then x-component of their resultant will be

And y-component of their resultant is

**Point to ponder **

Why do you keep your legs far apart when you have to stand in the aisle of a bumpy-riding bus?