In the sum Vector A plus Vector B equals vector C, vector A has a magnitude of 11.6 m and is angled 36.5° counterclockwise from the +x direction, and vector C has a magnitude of 14.0 m and is angled 17.1° 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.
First, let's break down the given information:
Vector A: magnitude = 11.6 m, angle = 36.5° counterclockwise from +x direction
Vector C: magnitude = 14.0 m, angle = 17.1° counterclockwise from -x direction
To find Vector B, we need to use vector addition. The sum of Vector A and Vector B is equal to Vector C:
Vector A + Vector B = Vector C
To add vectors, we need to break them down into their x and y components. Let's consider the x-component first.
The x-component of Vector A is given by: A_x = A * cos(angle of A)
A_x = 11.6 m * cos(36.5°) = 9.379 m (rounded to three decimal places)
The x-component of Vector C is given by: C_x = C * cos(angle of C)
C_x = 14.0 m * cos(17.1°) = 13.367 m (rounded to three decimal places)
Since Vector B is the unknown vector, let's assume its magnitude as B_m and its angle as θ.
Now, let's calculate the x-component of Vector B, denoted as B_x:
Vector A_x + Vector B_x = Vector C_x
9.379 m + B_x = 13.367 m
B_x = 13.367 m - 9.379 m
B_x = 3.988 m (rounded to three decimal places)
Now, let's find the y-components of Vector A, Vector B, and Vector C.
The y-component of Vector A is given by: A_y = A * sin(angle of A)
A_y = 11.6 m * sin(36.5°) = 6.769 m (rounded to three decimal places)
The y-component of Vector C is given by: C_y = C * sin(angle of C)
C_y = 14.0 m * sin(17.1°) = 4.038 m (rounded to three decimal places)
Now, let's calculate the y-component of Vector B, denoted as B_y:
Vector A_y + Vector B_y = Vector C_y
6.769 m + B_y = 4.038 m
B_y = 4.038 m - 6.769 m
B_y = -2.731 m (rounded to three decimal places)
Since vector B is in the second quadrant (y-component is negative), its angle will be greater than 90°.
Using the x and y components of vector B, we can calculate its magnitude and angle using the Pythagorean theorem and trigonometry.
(a) The magnitude of Vector B can be found using the Pythagorean theorem:
|B| = sqrt(B_x^2 + B_y^2)
|B| = sqrt((3.988 m)^2 + (-2.731 m)^2)
|B| = sqrt(15.904 m^2 + 7.468 m^2)
|B| = sqrt(23.372 m^2)
|B| = 4.834 m (rounded to three decimal places)
(b) The angle of Vector B relative to the +x direction can be calculated using the inverse tangent (tan^(-1)) function:
θ_B = tan^(-1)(B_y / B_x)
θ_B = tan^(-1)(-2.731 m / 3.988 m)
θ_B = tan^(-1)(-0.684)
θ_B = -34.894° (rounded to three decimal places)
Since the angle is given in the counterclockwise direction, we can convert it to a positive angle relative to the +x direction:
Positive angle = 360° - (-34.894°)
θ_B = 360° + 34.894°
θ_B = 394.894° (rounded to three decimal places)
Therefore, the magnitude of Vector B is 4.834 m and the angle (relative to +x) is 394.894°.