To compute the x - and y -components of the vectors A ⃗ \vec{A} A , B ⃗ \vec{B} B , C ⃗ \vec{C} C , and D ⃗ \vec{D} D , we can use trigonometric functions based on the magnitudes of the vectors and the angles they make with the coordinate axes.
Components of Vector A ⃗ \vec{A} A :
Magnitude of A ⃗ = 8.0 m \vec{A} = 8.0 \, \text{m} A = 8.0 m
Angle with negative y y y -axis = 0 ∘ 0^\circ 0 ∘ (since it's directly downward)
A x = 8.0 m × cos ( 270 ∘ ) = 8.0 m × 0 = 0 m A_x = 8.0 \, \text{m} \times \cos(270^\circ) = 8.0 \, \text{m} \times 0 = 0 \, \text{m} A x = 8.0 m × cos ( 27 0 ∘ ) = 8.0 m × 0 = 0 m
A y = 8.0 m × sin ( 270 ∘ ) = 8.0 m × ( − 1 ) = − 8.0 m A_y = 8.0 \, \text{m} \times \sin(270^\circ) = 8.0 \, \text{m} \times (-1) = -8.0 \, \text{m} A y = 8.0 m × sin ( 27 0 ∘ ) = 8.0 m × ( − 1 ) = − 8.0 m
Components of Vector B ⃗ \vec{B} B :
Magnitude of B ⃗ = 15.0 m \vec{B} = 15.0 \, \text{m} B = 15.0 m
Angle with positive x x x -axis = 30 ∘ 30^\circ 3 0 ∘
B x = 15.0 m × cos ( 30 ∘ ) = 15.0 m × 3 2 ≈ 12.99 m B_x = 15.0 \, \text{m} \times \cos(30^\circ) = 15.0 \, \text{m} \times \frac{\sqrt{3}}{2} \approx 12.99 \, \text{m} B x = 15.0 m × cos ( 3 0 ∘ ) = 15.0 m × 2 3 ≈ 12.99 m
B y = 15.0 m × sin ( 30 ∘ ) = 15.0 m × 1 2 = 7.5 m B_y = 15.0 \, \text{m} \times \sin(30^\circ) = 15.0 \, \text{m} \times \frac{1}{2} = 7.5 \, \text{m} B y = 15.0 m × sin ( 3 0 ∘ ) = 15.0 m × 2 1 = 7.5 m
Components of Vector C ⃗ \vec{C} C :
Magnitude of C ⃗ = 12.0 m \vec{C} = 12.0 \, \text{m} C = 12.0 m
Angle with negative x x x -axis = 25 ∘ 25^\circ 2 5 ∘ (which is equivalent to 180 ∘ − 25 ∘ = 155 ∘ 180^\circ - 25^\circ = 155^\circ 18 0 ∘ − 2 5 ∘ = 15 5 ∘ )
C x = 12.0 m × cos ( 180 ∘ − 25 ∘ ) = 12.0 m × ( − cos ( 25 ∘ ) ) ≈ − 12.0 m × 0.9063 ≈ − 10.88 m C_x = 12.0 \, \text{m} \times \cos(180^\circ - 25^\circ) = 12.0 \, \text{m} \times (-\cos(25^\circ)) \approx -12.0 \, \text{m} \times 0.9063 \approx -10.88 \, \text{m} C x = 12.0 m × cos ( 18 0 ∘ − 2 5 ∘ ) = 12.0 m × ( − cos ( 2 5 ∘ )) ≈ − 12.0 m × 0.9063 ≈ − 10.88 m
C y = 12.0 m × sin ( 180 ∘ − 25 ∘ ) = 12.0 m × sin ( 25 ∘ ) ≈ 12.0 m × 0.4226 ≈ 5.07 m C_y = 12.0 \, \text{m} \times \sin(180^\circ - 25^\circ) = 12.0 \, \text{m} \times \sin(25^\circ) \approx 12.0 \, \text{m} \times 0.4226 \approx 5.07 \, \text{m} C y = 12.0 m × sin ( 18 0 ∘ − 2 5 ∘ ) = 12.0 m × sin ( 2 5 ∘ ) ≈ 12.0 m × 0.4226 ≈ 5.07 m
Components of Vector D ⃗ \vec{D} D :
Magnitude of D ⃗ = 10.0 m \vec{D} = 10.0 \, \text{m} D = 10.0 m
Angle with negative x x x -axis = 53 ∘ 53^\circ 5 3 ∘ (which is equivalent to 180 ∘ − 53 ∘ = 127 ∘ 180^\circ - 53^\circ = 127^\circ 18 0 ∘ − 5 3 ∘ = 12 7 ∘ )
D x = 10.0 m × cos ( 127 ∘ ) = 10.0 m × ( − cos ( 53 ∘ ) ) ≈ 10.0 m × ( − 0.6018 ) ≈ − 6.02 m D_x = 10.0 \, \text{m} \times \cos(127^\circ) = 10.0 \, \text{m} \times (-\cos(53^\circ)) \approx 10.0 \, \text{m} \times (-0.6018) \approx -6.02 \, \text{m} D x = 10.0 m × cos ( 12 7 ∘ ) = 10.0 m × ( − cos ( 5 3 ∘ )) ≈ 10.0 m × ( − 0.6018 ) ≈ − 6.02 m
D y = 10.0 m × sin ( 127 ∘ ) = 10.0 m × sin ( 53 ∘ ) ≈ 10.0 m × 0.7986 ≈ 7.99 m D_y = 10.0 \, \text{m} \times \sin(127^\circ) = 10.0 \, \text{m} \times \sin(53^\circ) \approx 10.0 \, \text{m} \times 0.7986 \approx 7.99 \, \text{m} D y = 10.0 m × sin ( 12 7 ∘ ) = 10.0 m × sin ( 5 3 ∘ ) ≈ 10.0 m × 0.7986 ≈ 7.99 m
Summary of Components:
A ⃗ \vec{A} A : A x = 0 m A_x = 0 \, \text{m} A x = 0 m , A y = − 8.0 m A_y = -8.0 \, \text{m} A y = − 8.0 m
B ⃗ \vec{B} B : B x ≈ 12.99 m B_x \approx 12.99 \, \text{m} B x ≈ 12.99 m , B y = 7.5 m B_y = 7.5 \, \text{m} B y = 7.5 m
C ⃗ \vec{C} C : C x ≈ − 10.88 m C_x \approx -10.88 \, \text{m} C x ≈ − 10.88 m , C y ≈ 5.07 m C_y \approx 5.07 \, \text{m} C y ≈ 5.07 m
D ⃗ \vec{D} D : D x ≈ − 6.02 m D_x \approx -6.02 \, \text{m} D x ≈ − 6.02 m , D y ≈ 7.99 m D_y \approx 7.99 \, \text{m} D y ≈ 7.99 m
Would you like a more detailed breakdown of any of the components? Here are some related questions that might interest you:
How can we determine the resultant vector of all four vectors?
What is the magnitude of the resultant vector of A ⃗ \vec{A} A and B ⃗ \vec{B} B ?
How would the components change if vector B ⃗ \vec{B} B were rotated by an additional 15 ∘ 15^\circ 1 5 ∘ counterclockwise?
What is the angle between vectors A ⃗ \vec{A} A and D ⃗ \vec{D} D ?
Can you compute the dot product of vectors C ⃗ \vec{C} C and D ⃗ \vec{D} D ?
Tip: When dealing with vector components, always ensure you accurately resolve the angles with respect to the coordinate axes to avoid sign errors.