Consider two vectors F~ 1 and F~ 2 with magnitude F1 = 52 N and F2 = 78 N and where θ1 = 263 ◦ and θ2 = 336 ◦

Question

1.Consider two vectors F~ 1 and F~ 2 with magnitude F1 = 52 N and F2 = 78 N and where θ1 = 263 ◦ and θ2 = 336 ◦ . 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.

2.What is the direction of this resultant vector F~ ? Use counterclockwise as the positive angular direction, between the limits of −180◦ and +180◦ as measured from the positive x axis. Answer in units of ◦ .

Answer 1

Step 1: Draw a picture. Since I can't do that for you, I will describe it. You will have your coordinate axis. F1 will be in the third quadrant pointing down and left (neg. x and neg y) since θ1 = 263^o.WRT the positive x-axis = 263 - 180 = 83^o WRT the negative x-axis. F2 is in the 4th quadrant (since 270 < θ2 = 336^o.< 360, so points down (neg y) and to the right (pos x) at an angle of 360 - 336 = 24^o in the clockwise direction from the positive x-axis.

Step 2: Use the angles that are <90 to find the components of F1 and F2, making sure to keep the signs correct based on the quadrant/drawing.

F1 = - F1 * cos(83) i - F1 * sin(83) j

F2 = F2 * cos(24) i - F2 * sin(24) j

(where i is a unit vector in the x-direction and j is a unit vector in the y direction)

The vector, F = F1 + F2; add by components.

F = [- F1 * cos(83) + F2 * cos(24)] i + [ - F1 * sin(83) - F2 * sin(24) ] j

= [ - 52*cos(83) + 78 * cos(24)] i + [ - 52 * sin(83) - 78 * sin(24) ] j

F = 65 i - 83 j

If you draw this vector on your coordinate system, it will be in the 4th quadrant

(1) The magnitude of a vector is the sum of the squares of the components:

|F| = √[ (65)² + (83)²] = 105 N (or 106 N, depending on rounding)

(2) The direction of the vector is found using the inverse tangent function

θ = tan^-1 (opp/adj) = tan^-1 (-83/52) = -52^o = 360 - 52 = 308^o