The displacement vectors A ? and B ? shown in Figure P3.11 both have magnitudes of 3.00 m. The direction of vector A ? is ? 5 30.0°. Find graphically (a) A ? + B ?, (b) A ? – B ? ,(c) B ? – A , and (d) A ? – 2 B ? .(Report all angles counterclockwise from the positive x axis.)
Answer
Graphical Analysis of Displacement Vectors A and B
Given: Vectors A and B both have magnitudes of 3.00 m. Direction of A is 30.0° from the positive x-axis. Assume B is oriented along the positive x-axis unless otherwise specified.
(a) A + B
Using graphical vector addition, place vector B tail-to-tip with A.
Component form:
Ax = 3 cos(30°) = 2.598 m, Ay = 3 sin(30°) = 1.500 m
Bx = 3.00 m, By = 0.00 m
Component form:
Ax = 3 cos(30°) = 2.598 m, Ay = 3 sin(30°) = 1.500 m
Bx = 3.00 m, By = 0.00 m
Resultant R = A + B:
Rx = 2.598 + 3.00 = 5.598 m, Ry = 1.500 + 0 = 1.500 m
Rx = 2.598 + 3.00 = 5.598 m, Ry = 1.500 + 0 = 1.500 m
|R| = √(5.598² + 1.5²) ≈ 5.796 m, θ ≈ tan⁻¹(1.5 / 5.598) ≈ 15.0°
(b) A – B
Subtract B from A by reversing B and adding:
Bx = -3.00 m, By = 0.00 m
Bx = -3.00 m, By = 0.00 m
Rx = 2.598 – 3.00 = -0.402 m, Ry = 1.500 m
|R| ≈ √((-0.402)² + 1.5²) ≈ 1.553 m, θ ≈ tan⁻¹(1.5 / -0.402) ≈ 105.2°
(c) B – A
Reverse A and add to B:
Ax = -2.598 m, Ay = -1.500 m
Ax = -2.598 m, Ay = -1.500 m
Rx = 3.00 – 2.598 = 0.402 m, Ry = 0.000 – 1.500 = -1.500 m
|R| ≈ √(0.402² + 1.5²) ≈ 1.553 m, θ ≈ tan⁻¹(-1.5 / 0.402) ≈ 284.8°
(d) A – 2B
2Bx = 6.00 m, 2By = 0.00 m
Rx = 2.598 – 6.00 = -3.402 m, Ry = 1.500
Rx = 2.598 – 6.00 = -3.402 m, Ry = 1.500
|R| ≈ √(3.402² + 1.5²) ≈ 3.723 m, θ ≈ tan⁻¹(1.5 / -3.402) ≈ 156.4°