Three displacement vectors of a croquet ball are shown in figure below, where |A| = 20.0 units, |B| = 40.0 units, and |C| = 30.0 units. Calculate
(a) the resultant in unit-vector notation and
(b) the magnitude and direction of the resultant displacement
To calculate the resultant displacement, we can add the three displacement vectors together.
(a) To find the resultant in unit-vector notation, we need to add the x-components and the y-components of the vectors separately.
Let's assume the x-direction is represented by î and the y-direction is represented by ĵ.
The x-component of vector A can be calculated using cosθA = Ax / |A|:
Ax = |A| * cosθA = 20 * cos(0) = 20
The y-component of vector A can be calculated using sinθA = Ay / |A|:
Ay = |A| * sinθA = 20 * sin(0) = 0
Since the x-component of vector A is 20 and the y-component is 0, vector A in unit-vector notation is A = 20î.
Similarly, we can calculate the x and y components for vectors B and C:
Bx = |B| * cosθB = 40 * cos(0) = 40
By = |B| * sinθB = 40 * sin(0) = 0
Cx = |C| * cosθC = 30 * cos(0) = 30
Cy = |C| * sinθC = 30 * sin(0) = 0
Therefore, B = 40î and C = 30î.
Now we can add the x-components and y-components of the vectors together to find the resultant vector:
Rx = Ax + Bx + Cx = 20 + 40 + 30 = 90
Ry = Ay + By + Cy = 0 + 0 + 0 = 0
The resultant vector in unit-vector notation is R = 90î.
(b) To find the magnitude and direction of the resultant displacement, we can use Pythagoras' theorem and simple trigonometry.
The magnitude of the resultant displacement, |R|, can be found using |R| = √(Rx² + Ry²):
|R| = √(90² + 0²) = √(8100) = 90 units.
The direction of the resultant displacement, θ, can be found using tanθ = Ry / Rx:
θ = tan⁻¹(Ry / Rx) = tan⁻¹(0 / 90) = 0°.
Therefore, the magnitude of the resultant displacement is 90 units, and the direction is 0°.
To calculate the resultant displacement, we need to add the three displacement vectors together.
(a) Resultant in Unit-Vector Notation:
To represent the vectors in unit-vector notation, we need to determine their individual components.
Let's assume vector A is represented as A = A_x i + A_y j + A_z k, where i, j, and k are the unit vectors in the x, y, and z directions. Similarly, we'll represent vectors B and C as B = B_x i + B_y j + B_z k and C = C_x i + C_y j + C_z k.
Given that |A| = 20.0 units, |B| = 40.0 units, and |C| = 30.0 units, we can express them in unit-vector notation as:
A = 20.0 A_x i + 20.0 A_y j + 20.0 A_z k
B = 40.0 B_x i + 40.0 B_y j + 40.0 B_z k
C = 30.0 C_x i + 30.0 C_y j + 30.0 C_z k
Now, let's add these vectors together to calculate the resultant in unit-vector notation. The resultant vector R is given by:
R = A + B + C
Add the component vectors of A, B, and C separately:
R = (20.0 A_x + 40.0 B_x + 30.0 C_x) i + (20.0 A_y + 40.0 B_y + 30.0 C_y) j + (20.0 A_z + 40.0 B_z + 30.0 C_z) k
Therefore, the resultant displacement in unit-vector notation is R = (20.0 A_x + 40.0 B_x + 30.0 C_x) i + (20.0 A_y + 40.0 B_y + 30.0 C_y) j + (20.0 A_z + 40.0 B_z + 30.0 C_z) k.
(b) Magnitude and Direction of the Resultant Displacement:
To find the magnitude of the resultant displacement, we can use the Pythagorean theorem. The magnitude |R| is given by:
|R| = sqrt((20.0 A_x + 40.0 B_x + 30.0 C_x)^2 + (20.0 A_y + 40.0 B_y + 30.0 C_y)^2 + (20.0 A_z + 40.0 B_z + 30.0 C_z)^2)
To determine the direction of the resultant displacement, we need to find the angles it makes with the x, y, and z axes. The direction can be expressed in terms of angles or as a direction cosine.
Please provide the values of A_x, A_y, A_z, B_x, B_y, B_z, C_x, C_y, and C_z so that I can calculate the resultant magnitude and direction for you.
Resultant-69.49i+27.07i
magnitude=74.58
theta=21.3