Physics, consider two vectors

Question

Consider two vectors F~ 1 and F~ 2 with magnitude F1 = 80 N and F2 = 76 N and where θ1 = 275 ◦ and θ2 = 203 ◦ . The angles are measured from the positive x axis with the counter-clockwise angular direction as positive. What is the magnitude of the resultant vector kF~ k, where F~ = F~ 1 + F~ 2 ? Answer in units of N.

Answer 1

Here is a picture:

Vector F₂ can be broken into 2 components:

F_2x = 76cos(23°) = 69.958 but it is in the negative x-direction so F_2x = -69.958

F_2y = 76sin(23°) = 29.696 but it is in the negative y-direction so F_2y = -29.696

Vector F₁ can be broken into 2 components:

F_1x = 80cos(85°) = 6.9725

F_1y = 80sin(85°) = 79.696 but it is in the negative y-direction so F_2y = -79.696

Adding the two x-components gives the resultant x component of:

R_x = -69.958 + 6.9725 = -62.986

Adding the two y-components gives the resultant y component of:

R_y = -29.696 + -79.696 = -109.391

Using the Pythagorean Theorem to put these components back together we get the magnitude of the resultant as: √(-62.986)² + (-109.391)²) = 126 N

The direction of this vector is not asked for, but it is in Quadrant 3. To find the angle (θ_R) we can say tan(θ_R) = (-109.391)/(-62.986) so θ_R = tan^-1((-109.391)/(-62.986)) = 60° however, since this is in Q3, we need to add 180° to it to get the angle from the positive x-axis so θ_R = 240°