In the sum A+B=C, vector A has a magnitude of 10.2 m and is angled 48.0° counterclockwise from the +x direction, and vector C has a magnitude of 15.0 m and is angled 22.0° counterclockwise from the -x direction. What are (a) the magnitude and (b) the angle (relative to +x) of Vector B? State your angle as a positive number.
To find the magnitude and angle of vector B, we can use vector addition and trigonometry.
(a) Magnitude of vector B:
Let's find the x and y components of the vectors A and C.
For vector A:
Magnitude = 10.2 m
Angle = 48.0° counterclockwise from the +x direction
Ax = 10.2 * cos(48.0°)
Ay = 10.2 * sin(48.0°)
For vector C:
Magnitude = 15.0 m
Angle = 22.0° counterclockwise from the -x direction
Cx = 15.0 * cos(-22.0°)
Cy = 15.0 * sin(-22.0°)
Now, let's find the components of vector B:
Bx = Cx - Ax
By = Cy - Ay
Magnitude of B = sqrt(Bx^2 + By^2)
Calculate the values:
Ax = 10.2 * cos(48.0°) = 6.598 m
Ay = 10.2 * sin(48.0°) = 7.536 m
Cx = 15.0 * cos(-22.0°) = 14.015 m
Cy = 15.0 * sin(-22.0°) = -5.541 m
Bx = 14.015 m - 6.598 m = 7.417 m
By = -5.541 m - 7.536 m = -13.077 m
Magnitude of B = sqrt((7.417 m)^2 + (-13.077 m)^2) = 14.956 m
Therefore, the magnitude of vector B is 14.956 m.
(b) Angle of vector B:
The angle can be found using the arctan function:
Angle of B = atan(By/Bx)
Calculate the value:
Angle of B = atan((-13.077 m) / (7.417 m)) = -59.770°
Since we want the angle relative to the +x direction, we need to add 180° to the angle:
Angle of B = -59.770° + 180° = 120.230°
Therefore, the angle of vector B relative to +x is 120.230°.
To find the magnitude and angle of vector B, we can use vector addition and trigonometry.
(a) Magnitude of vector B:
Since the sum of vectors A and B is equal to vector C, we can write their components as:
Ax + Bx = Cx
Ay + By = Cy
Given that vector A has a magnitude 10.2 m, we can break it down into its x and y components using trigonometry:
Ax = 10.2 × cos(48°)
Ay = 10.2 × sin(48°)
Similarly, we can express vector C with its components using its magnitude and angle:
Cx = 15.0 × cos(180° - 22°) (since it is counterclockwise from -x)
Cy = 15.0 × sin(180° - 22°)
Now, we can solve for the x and y components of vector B by subtracting the corresponding x and y components of vector A and C:
Bx = Cx - Ax
By = Cy - Ay
To find the magnitude of vector B, use the Pythagorean theorem:
Magnitude of B = √(Bx² + By²)
(b) Angle of vector B:
To find the angle of vector B relative to the +x direction, we can use trigonometry:
Angle of B = atan(By/Bx)
Let's calculate the values:
Ax = 10.2 × cos(48°) ≈ 6.648 m
Ay = 10.2 × sin(48°) ≈ 7.643 m
Cx = 15.0 × cos(180° - 22°) ≈ -12.906 m
Cy = 15.0 × sin(180° - 22°) ≈ -5.342 m
Bx = Cx - Ax ≈ -12.906 m - 6.648 m ≈ -19.554 m
By = Cy - Ay ≈ -5.342 m - 7.643 m ≈ -12.985 m
Magnitude of B = √(Bx² + By²) ≈ √((-19.554 m)² + (-12.985 m)²) ≈ √(383.098 m² + 168.768 m²) ≈ √(551.866 m²) ≈ 23.49 m
Angle of B = atan(By/Bx) ≈ atan((-12.985 m)/(-19.554 m)) ≈ atan(0.664) ≈ 34.26°
Therefore, (a) the magnitude of vector B is approximately 23.49 m and (b) the angle of vector B relative to the +x direction is approximately 34.26°.
10.2m[48o] + B = 15m[15o]
10.2*cos48+i10.2*sin48+B=15^cos22+i15*sin22.
6.83 + i7.58 + B = 13.91 + i5.62
B=13.91-6.83 + i5.62 - i7.58=7.08-i1.96
B = sqrt(7.08^2 + (-1.96)^2) = 7.35 m.
b. Tan A = Y/X = -1.96/7.08 = -0.27684
A = -15.47o S of E = 344.5o, CCW.