Displacement vectors A,B,and C add up to a total of zero. Vector A has a magnitude of 1550 m and a direction of 24.1° north of east. Vector B has a direction of 41.0° east of south, and vector C has a direction of 33.4° north of west. Find the magnitudes of vector B and vector C.
To find the magnitudes of vectors B and C, we can use the concept of vector addition and trigonometry.
First, let's break down vector A into its x and y components.
The x-component of vector A can be found by using the formula:
Ax = A * cos(θ)
where A is the magnitude of vector A and θ is the angle it makes with the positive x-axis.
Substituting the given values:
Ax = 1550 m * cos(24.1°)
Similarly, the y-component of vector A can be found using the formula:
Ay = A * sin(θ)
where A is the magnitude of vector A and θ is the angle it makes with the positive x-axis.
Substituting the given values:
Ay = 1550 m * sin(24.1°)
Now, let's move on to vector B. Vector B has a direction of 41.0° east of south. To find its components, we first need to determine the angle it makes with the positive x-axis.
Since it is east of south, we need to subtract the given angle from 90° to get the angle with the positive x-axis:
θB = 90° - 41.0°
Now, we can find the x and y components of vector B using the formulas:
Bx = B * cos(θB)
By = B * sin(θB)
Similarly, for vector C, which has a direction of 33.4° north of west, we need to find the angle it makes with the positive x-axis.
Since it is north of west, we need to add the given angle to 180° to get the angle with the positive x-axis:
θC = 33.4° + 180°
Now, we can find the x and y components of vector C using the formulas:
Cx = C * cos(θC)
Cy = C * sin(θC)
Since the sum of all three vectors adds up to zero, we can write the following equations:
Ax + Bx + Cx = 0
Ay + By + Cy = 0
Now, substitute the calculated values for Ax, Ay, Bx, By, Cx, and Cy into the equations above and solve for B and C.