Question

The x- and y -components of several unknown forces are listed below 1Fx, Fy2. For each force, draw on an x, y coordi- nate system the components of the force vectors. Determine the magnitude and direction of each force:

(a) 1+100 N, -100 N2,
(b) 1-300 N, -400 N2, and
(c) 1-400 N, +300 N2.

Answer 1

For each force vector, we can determine the magnitude and direction using the x- and y-components. For force (a), the magnitude is 141.4 N and it is directed 45° below the negative x-direction. For force (b), the magnitude is 500 N and it is directed 53.1° below the negative x-direction. For force (c), the magnitude is 500 N and it is directed 36.9° above the negative x-direction.

Step-by-step explanation:

1. For force (a), we have F1x = +100 N and F1y = -100 N. To determine the magnitude and direction of the force, we can use the Pythagorean theorem. The magnitude is given by √(Fx^2 + Fy^2), which in this case is √((100 N)^2 + (-100 N)^2) = 141.4 N. The direction can be found using the tangent function: tan^(-1)(|F1y|/|F1x|) = tan^(-1)(|-100 N|/|100 N|) = 45° below the negative x-direction.
2. For force (b), we have F1x = -300 N and F1y = -400 N. Using the Pythagorean theorem, the magnitude is √((-300 N)^2 + (-400 N)^2) = 500 N. And the direction is tan^(-1)(|-400 N|/|-300 N|) = tan^(-1)(4/3) = 53.1° below the negative x-direction.
3. For force (c), we have F1x = -400 N and F1y = +300 N. Using the Pythagorean theorem, the magnitude is √((-400 N)^2 + (300 N)^2) = 500 N. And the direction is tan^(-1)(|300 N|/|-400 N|) = tan^(-1)(3/4) = 36.9° above the negative x-direction.