(a) An object is subjected to two displacements. The displacement vector of the first, V1 , has a magnitude of 186 m at an angle 23° North of East. The second displacement vector, V2 , has a magnitude of 327 m at an angle 43° West of North. find the magnitude and direction of the resultant vector.
i got 502.02 m and 38.25 degrees but have no idea if this is anywhere near correct ! Help I split each into x and y components and added but had to use 47 degrees for the angle for vector two as subracted the 43 degrees from 90 because it said it is west of north... have i done this correctly ?
In terms of x & y,
186@E23°N = 171.21,72.68
327@N43°W = -223.01,239.15
Add them up: -51.80,311.83
That is 361.10 at 99.43°
or, as a heading, [email protected]°W
Vr = 186m[23o] + 327m[133o]
X = 186*cos23 + 327*cos133 = -51.8 m/s.
Y = 186*sin23 + 327*sin133 = 311.8 m/s.
Tan Ar = Y/X = 311.8/-51.8 = -6.01986
Ar = -80.6o = Reference angle.
A = -80.6 + 180 = 99.4o CCW = 9.4o W of
N.
Vr = Y/sin A = 311.8/sin99.4)=316 m/s
[99.4o] = 316 m/s[9.4o]W of N.
thank you ! got it now
To find the magnitude and direction of the resultant vector, you need to correctly add the components of the two displacement vectors V1 and V2. Let's break down the steps:
1. Convert the given angles to directional angles in the Cartesian coordinate system, where East is the positive x-axis and North is the positive y-axis. Note that in this system, angles are measured counterclockwise from the positive x-axis.
- For V1: Angle = 23° + 90° = 113° counterclockwise from the positive x-axis.
- For V2: Angle = 360° - 43° = 317° counterclockwise from the positive x-axis.
2. Find the x and y components of each displacement vector:
For V1:
- x-component (V1x) = magnitude (V1) * cos(angle (V1))
- y-component (V1y) = magnitude (V1) * sin(angle (V1))
For V2:
- x-component (V2x) = magnitude (V2) * cos(angle (V2))
- y-component (V2y) = magnitude (V2) * sin(angle (V2))
3. Add the x-components and y-components separately:
Resultant x-component (Rx) = V1x + V2x
Resultant y-component (Ry) = V1y + V2y
4. Calculate the magnitude of the resultant vector (R):
R = √(Rx^2 + Ry^2)
5. Calculate the direction of the resultant vector (θ):
θ = tan^(-1)(Ry / Rx)
Now let's calculate the magnitude and direction using the provided values:
1. V1:
- Magnitude (V1) = 186 m
- Angle (V1) = 113°
2. V2:
- Magnitude (V2) = 327 m
- Angle (V2) = 317°
3. Calculate the x and y components of each vector:
V1x = 186 * cos(113°)
V1y = 186 * sin(113°)
V2x = 327 * cos(317°)
V2y = 327 * sin(317°)
4. Calculate the resultant x and y components:
Rx = V1x + V2x
Ry = V1y + V2y
5. Calculate the magnitude of the resultant vector:
R = √(Rx^2 + Ry^2)
6. Calculate the direction of the resultant vector:
θ = tan^(-1)(Ry / Rx)
Now, performing the calculations:
V1x ≈ 186 * cos(113°) ≈ -99.29 m
V1y ≈ 186 * sin(113°) ≈ 158.41 m
V2x ≈ 327 * cos(317°) ≈ 53.25 m
V2y ≈ 327 * sin(317°) ≈ -299.15 m
Rx ≈ -99.29 m + 53.25 m ≈ -46.04 m
Ry ≈ 158.41 m - 299.15 m ≈ -140.74 m
R ≈ √((-46.04)^2 + (-140.74)^2) ≈ 150.98 m (rounded to two decimal places)
θ ≈ tan^(-1)(-140.74 / -46.04) ≈ 70.52° (rounded to two decimal places)
Therefore, the magnitude of the resultant vector is approximately 150.98 m, and the direction is approximately 70.52° counterclockwise from the positive x-axis.