Resolution of forces/vectors:

The process of splitting up vectors into their component forces is called the resolution of forces.

OR

Splitting up a force into two mutually perpendicular components is called the resolution of that force. Vector resolution is reverse from vector addition.

Perpendicular component/rectangular components:

Consider a force F represented by line OA making an angle with an x-axis. Draw a perpendicular AB on the x-axis from A. According to the head-to-tail rule, OA is the resultant vector represented by OB and BA.

Thus

OA=OB+BA………………. (1)

From figure

F= $F_{x}+ F_{y}$ ………………… (2)

Magnitude of horizontal component:

In right angled triangle OBA

$\cos \theta=\frac{base}{hypotenuse}=\frac{OB}{OA}$

$\frac{Fx}{F}= \cos\theta$

$Fx=F \cos\theta$ …………………….. (3)

Magnitude of vertical component (F_y):

$\sin\theta=\frac{perpendicular}{hypotenuse}=\frac{BA}{OA}$

$\frac{Fy}{F}= \sin\theta$ 

F_y= F Sin ……………… (4)

Equations (3) and (4) give the magnitude of horizontal and rectangular components.

Trigonometric Table

Mini Exercise

In a right-angled triangle length of the base is 4 cm and its perpendicular is 3 cm. Find:

(i) Length of hypotenuse  (ii) sin θ

(iii)     cos θ                     (iv)     tan θ

Solution:

(i) Length of hypotenuse:

Pythagoras theorem:

$\left(Hypotenuse\right)^{2}= \left(Base\right)^{2}+\left(Perpendicular\right)^{2}$

$\left(Hypotenuse\right)^{2}= \left(4\right)^{2}+\left(3\right)^{2}$

$\left(Hypotenuse\right)^{2}= 16+9$

$\left(Hypotenuse\right)^{2}= 25$ by taking square root on both sides

Hypotenuse = 5 cm

(ii) $\sin\theta$:

$\sin\theta= \frac{Perpendicular}{Hypotenuse}= \frac{3}{5}$

(iii) $\cos\theta$:

$\cos\theta= \frac{Base}{Hypotenuse}= \frac{4}{5}$

(iv) $\tan\theta$:

$\tan\theta= \frac{Perpendicular}{Base}= \frac{3}{4}$