Question

Help me with this problem

Vector E is 0.111 m long in a 90.0 direction.Vector F is 0.234 m long in a 300 direction. What is the magnitude and direction of their vector sum?

Answer 1

0.149 m,
`321.8^(\circ)`

Step-by-step 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=√(R_x^2 +R_y^2 )=√((0.117)^2+(-0.092)^2)=0.149 m`

And the direction is

`\theta=tan^(-1) ((|R_y|)/(R_x))=-tan^(-1) ((0.092)/(0.117))=-38.2^(\circ)=321.8^(\circ)`