The three displacement vectors in the drawing have magnitudes of A= 5.00 m B= 5.00 m and C= 4.00 m. Find the resultant( magnitude and directional angle) of three vectors by mean of the component method. express the directional angle as an angle above the positive or negative x-axis
The Correct Answer and Explanation is:
To find the resultant displacement vector of three given vectors A, B, and C, using the component method, we follow these steps:
Step 1: Understand the vectors and their directions
Assume the vectors are oriented as follows (a common setup unless stated otherwise):
- Vector A is along the positive x-axis
- Vector B is at a 45° angle above the positive x-axis
- Vector C is along the negative y-axis
Given:
- A = 5.00 m
- B = 5.00 m
- C = 4.00 m
Step 2: Break each vector into components
Vector A:
- Ax = 5.00 m
- Ay = 0.00 m
Vector B:
- Bx = B × cos(45°) = 5.00 × 0.7071 ≈ 3.54 m
- By = B × sin(45°) = 5.00 × 0.7071 ≈ 3.54 m
Vector C:
- Cx = 0.00 m
- Cy = –4.00 m (since it is directed downward)
Step 3: Add components
Total x-component (Rx):
Rx = Ax + Bx + Cx
Rx = 5.00 + 3.54 + 0.00 = 8.54 m
Total y-component (Ry):
Ry = Ay + By + Cy
Ry = 0.00 + 3.54 – 4.00 = –0.46 m
Step 4: Find magnitude of the resultant vector
R=(Rx)2+(Ry)2=(8.54)2+(−0.46)2≈72.96+0.21≈73.17≈8.56 mR = \sqrt{(Rx)^2 + (Ry)^2} = \sqrt{(8.54)^2 + (-0.46)^2} \approx \sqrt{72.96 + 0.21} \approx \sqrt{73.17} \approx 8.56\ \text{m}R=(Rx)2+(Ry)2=(8.54)2+(−0.46)2≈72.96+0.21≈73.17≈8.56 m
Step 5: Find the directional angle
θ=tan−1(RyRx)=tan−1(−0.468.54)≈tan−1(−0.0539)≈−3.1∘\theta = \tan^{-1}\left(\frac{Ry}{Rx}\right) = \tan^{-1}\left(\frac{-0.46}{8.54}\right) \approx \tan^{-1}(-0.0539) \approx -3.1^\circθ=tan−1(RxRy)=tan−1(8.54−0.46)≈tan−1(−0.0539)≈−3.1∘
Since Rx is positive and Ry is negative, the vector lies slightly below the positive x-axis.
Final Answer:
- Magnitude = 8.56 m
- Direction = 3.1° below the positive x-axis
Or equivalently:
Angle = 356.9° measured counterclockwise from the positive x-axis
