Question

Help me with this problem

Answer 1

Answer:

0.149 m, $321.8^{\circ}$

Explanation:

Let's start by resolving each vector into its components along the x- and y- direction:

$E_x = E cos \theta = (0.111) cos 90^{\circ}=0\\E_y = E sin \theta = (0.111) sin 90^{\circ}=0.111 m$

And

$F_x = F cos \theta = (0.234) cos 300^{\circ} = 0.117 m\\F_y = F sin \theta = (0.234) sin 300^{\circ} = -0.203 m$

So the components of the vector sum are

$R_x = E_x + F_x = 0+0.117 = 0.117 m\\R_y = E_y + F_y = 0.111 -0.203 = -0.092 m$

The magnitude of the vector sum is

$R=\sqrt{R_x^2 +R_y^2 }=\sqrt{(0.117)^2+(-0.092)^2}=0.149 m$

And the direction is

$\theta=tan^{-1} (\frac{|R_y|}{R_x})=-tan^{-1} (\frac{0.092}{0.117})=-38.2^{\circ}=321.8^{\circ}$