Multiple Concept Example 9 provides background pertinent to this problem. The magnitudes of the four displacement vectors shown in the drawing are A = 17.0 m, B = 10.0 m, C = 11.0 m, and D = 23.0 m. Determine the (a) magnitude and (b) direction for the resultant that occurs when these vectors are added together. Specify the direction as a positive (counterclockwise) angle from the +x axis.
This cannot be commented on without knowing the directions of each vector.
Here is my suggestion. Add all the x components MEASURED FROM THE X AXIS COUNTER CLOCKWISE:
Rx=A*cosTheta+B*cosTheta+....+E*cosTheta
Ry=A*sinTheta+B*sinTheta+...+EsinTheta.
Now for magnitude:
R=sqrt(Rx^2 + Ry^2)
Angle=arcTan(Ry/Rx). Remember that is the principle angle, there is at least one other angle in the other quadrants that will work. Take a look at the vectors,especially if either Rx or Ry are negative
To determine the magnitude and direction of the resultant vector, we need to add the given displacement vectors A, B, C, and D.
(a) Magnitude of the Resultant:
To find the magnitude of the resultant, we can use the Pythagorean theorem. The magnitude (R) of the resultant vector is given by:
R = √(Ax² + Ay²)
where Ax and Ay are the x and y components of the resultant vector.
To find the components of the resultant vector, we need to decompose the given displacement vectors into their x and y components.
The x components (Rx) can be found by adding the x components of each individual vector:
Rx = Ax + Bx + Cx + Dx
Similarly, the y components (Ry) can be found by adding the y components of each individual vector:
Ry = Ay + By + Cy + Dy
Now, we can substitute the values of the displacement vectors A, B, C, and D:
Rx = 0 + B + C + D
Ry = A + 0 + 0 + 0
Substituting the values:
Rx = 0 + 10.0 m + 11.0 m + 23.0 m = 44.0 m
Ry = 17.0 m + 0 + 0 + 0 = 17.0 m
Now, we can find the magnitude of the resultant using the Pythagorean theorem:
R = √(Rx² + Ry²)
R = √((44.0 m)² + (17.0 m)²) = √(1936.0 m² + 289.0 m²) = √2225.0 m² = 47.2 m (approx)
Therefore, the magnitude of the resultant vector is approximately 47.2 m.
(b) Direction of the Resultant:
To find the direction of the resultant vector, we can use trigonometry. The direction is specified as a positive (counterclockwise) angle from the +x axis.
The angle (θ) can be found using the inverse tangent (arctan) function:
θ = arctan(Ry/Rx)
Substituting the values:
θ = arctan(17.0 m/44.0 m) ≈ 22.1°
Therefore, the direction of the resultant vector is approximately 22.1° counterclockwise from the +x axis.