Vector A, a wind velocity vector, has a magnitude of 20 miles per hour and is blowing toward the North. A second wind velocity vector, Vector B, has a magnitude of 50 miles per hour and is blowing toward the Southeast. A third wind velocity vector, Vector C, has a magnitude of 10 miles per hour and is blowing toward the Southwest These three winds emerge (come out of) three separate canyons and merge (meet) at a flagpole carrying a cloth flag. What direction (angle with respect to zero degrees) will the flag be pointing as a result of the three winds? Also, what is the magnitude of the resulting wind velocity vector (expressed in m/s)?
This is how far I got... and is it correct?????
R=20Cos0 N + 50cos135N+50sin135E+10Cos225N+10sin225(-E)
Do I just find the angles to all of them?
Most of us say a North wind is FROM the North, but oh well I will pretend to be a mathematician.
North:
A 20
B -50 *.707
C -10 *.707
East:
A 0
B 50*.707
C -10*.707
total N = -22.42
total E = +28.28
so in quadrant 4
tan angle S of E = 22.42/28.28
angle S of E = 38.4
That is all I can say. Being a navigator for me zero is north but in math it is east :) You decide
v = sqrt (22.42^2 + 28.28^2)
Calculate resultant in mi/h and convert to m/s. All angles are CCW from +x-axis.
R = 20[90o] + 50[315o] + 10[225].
R = 20i + 35.36-35.36i + -7.07-7.07i
R = 28.29 - 22.43i = 36.1 mi/h[-38.4o] = 36.1 mi/h[38.1o] S. of E.
R = 36.1mi/h * 1600m/mi * 1h/3600s = 16m/s[38.4o] S. of E.
To find the resulting wind direction and magnitude, we need to add the three wind velocity vectors together.
First, let's break down each wind velocity vector into its horizontal (x-axis) and vertical (y-axis) components.
Vector A:
Magnitude = 20 mph
Direction = North (0 degrees)
Horizontal component (Ax) = 20 * cos(0) = 20 * 1 = 20 mph
Vertical component (Ay) = 20 * sin(0) = 20 * 0 = 0 mph
Vector B:
Magnitude = 50 mph
Direction = Southeast (135 degrees)
Horizontal component (Bx) = 50 * cos(135) = 50 * (-0.7071) ≈ -35.3553 mph
Vertical component (By) = 50 * sin(135) = 50 * 0.7071 ≈ 35.3553 mph
Vector C:
Magnitude = 10 mph
Direction = Southwest (225 degrees)
Horizontal component (Cx) = 10 * cos(225) = 10 * (-0.7071) ≈ -7.0711 mph
Vertical component (Cy) = 10 * sin(225) = 10 * (-0.7071) ≈ -7.0711 mph
Next, we can add up the horizontal and vertical components separately to get the resultant (combined) horizontal (Rx) and vertical (Ry) components.
Rx = Ax + Bx + Cx = 20 mph + (-35.3553 mph) + (-7.0711 mph) ≈ -22.4264 mph
Ry = Ay + By + Cy = 0 mph + 35.3553 mph + (-7.0711 mph) ≈ 28.2842 mph
Now, we can use the horizontal and vertical components to find the magnitude and direction of the resulting wind velocity vector.
Magnitude of the resulting vector (R) = sqrt(Rx^2 + Ry^2) ≈ sqrt((-22.4264 mph)^2 + (28.2842 mph)^2) ≈ 36.0555 mph
To find the direction (angle with respect to zero degrees), we can use the inverse tangent function (tan^-1) to find the angle between the resultant vector and the positive x-axis.
Direction (θ) = tan^-1(Ry / Rx) ≈ tan^-1((28.2842 mph) / (-22.4264 mph)) ≈ tan^-1(-1.2629) ≈ -51.34 degrees
Note: The negative sign indicates that the resultant vector is pointing in the opposite direction (southwest) from the positive x-axis.
Therefore, the flag will be pointing approximately 51.34 degrees with respect to zero degrees in the southwest direction. The magnitude of the resulting wind velocity vector is approximately 36.0555 mph.