Find the sum of the following four vectors ((a) and (b) for x and y components respectively), and as (c) a magnitude and (d) an angle relative to +x (including sign plus or minus).

Question

Find the sum of the following four vectors ((a) and (b) for x and y components respectively), and as (c) a magnitude and (d) an angle relative to +x (including sign plus or minus).: 

17.0 m, at 31.0o counterclockwise from +x

16.0 m, at 12.0o counterclockwise from +y

15.0 m, at 17.0o clockwise from -y

13.0 m, at 47.0o counterclockwise from -y

Answer 1

Let's first start by finding the x and y components of each vector.

Vector 1: 17.0m at 31.0-degrees counterclockwise from +x

1. The 17.0m line will be pointing 31-degrees counterclockwise from +x, meaning the vector will be in the first quadrant of the xy-plane (meaning both the x and y components will be positive).
2. We can create a right triangle and use trigonometry to find the legs of the right triangle (which will also be the x and y components of the vector). Using the 31-degree angle and the known hypotenuse of 17m, we can use the sine function to find the y-component and the cosine function to find the x-component.

sin(θ) = opposite/hypotenuse :: hypotenuse*sin(θ) = opposite (y-component)

cos(θ) = adjacent/hypotenuse :: hypotenuse*cos(θ) = adjacent (x-component)

17sin(31-degrees) = y-component = 8.76

17cos(31-degrees) = x-component = 14.57

Vector 1 is 14.57i+8.76j

Vector 2: 16.0m at 12.0-degrees counterclockwise from +y

1. The 16.0m line will be pointing 12-degrees counterclockwise from +y, meaning the vector will be in the second quadrant of the xy-plane (meaning the x component will be negative and the y component will be positive).
2. We can create a right triangle and use trigonometry to find the legs of the right triangle (which will also be the x and y components of the vector). Using the 12-degree angle and the known hypotenuse of 16m, we can use the sine function to find the x-component and the cosine function to find the y-component.

sin(θ) = opposite/hypotenuse :: hypotenuse*sin(θ) = opposite (x-component)

cos(θ) = adjacent/hypotenuse :: hypotenuse*cos(θ) = adjacent (y-component)

16sin(12-degrees) = x-component = 3.33

16cos(12-degrees) = y-component = 15.65

Finally, let's include the negative sign on the x-component since we know the vector is in the second

quadrant.

Vector 2 is -3.33i+15.65j

Vector 3: 15.0m at 17.0-degrees clockwise from -y

1. The 15.0m line will be pointing 17-degrees clockwise from -y, meaning the vector will be in the third quadrant of the xy-plane (meaning both the x and y components will be negative).
2. We can create a right triangle and use trigonometry to find the legs of the right triangle (which will also be the x and y components of the vector). Using the 17-degree angle and the known hypotenuse of 15m, we can use the sine function to find the x-component and the cosine function to find the y-component.

sin(θ) = opposite/hypotenuse :: hypotenuse*sin(θ) = opposite (x-component)

cos(θ) = adjacent/hypotenuse :: hypotenuse*cos(θ) = adjacent (y-component)

15sin(17-degrees) = x-component = 4.39

15cos(17-degrees) = y-component = 14.34

Finally, let's include the negative sign on the x and y-components since we know the vector is in the third

quadrant.

Vector 3 is -4.39i-14.34j

Vector 4: 13.0m at 47.0-degrees counterclockwise from -y

1. The 13.0m line will be pointing 47-degrees counterclockwise from -y, meaning the vector will be in the fourth quadrant of the xy-plane (meaning the x component will be positive and the y component will be negative).
2. We can create a right triangle and use trigonometry to find the legs of the right triangle (which will also be the x and y components of the vector). Using the 47-degree angle and the known hypotenuse of 13m, we can use the sine function to find the x-component and the cosine function to find the y-component.

sin(θ) = opposite/hypotenuse :: hypotenuse*sin(θ) = opposite (x-component)

cos(θ) = adjacent/hypotenuse :: hypotenuse*cos(θ) = adjacent (y-component)

13sin(47-degrees) = x-component = 9.51

13cos(47-degrees) = y-component = 8.87

Finally, let's include the negative sign on the y-component since we know the vector is in the fourth

quadrant.

Vector 4 is 9.51i-8.87j

Vector 1 is 14.57i+8.76j

Vector 2 is -3.33i+15.65j

Vector 3 is -4.39i-14.34j

Now we can add all four vectors together:

(a) x-component: 14.57-3.33-4.39+9.51 = 16.36

(b) y-component: 8.76+15.65-14.34-8.87 = 1.2

Sum Vector: 16.36i + 1.2j

From here, we can find the magnitude of the vector by summing the squaring its components and taking the square root of that sum (Note: sqrt is meant to represent the square root)

(c) Magnitude: sqrt(16.36²+1.2²) = 4.19

We can also use the tangent trigonometric function to find the angle relative to +x axis. (Note that since the Sum Vector has a positive x and y-component, the vector is in the first quadrant and will have a positive angle) Let's say the angle between the vector and the +x axis is θ.

tan(θ) = opposite/adjacent = y-component/x-component

tan(θ) = 1.2/16.36

θ = tan^-1(1.2/16.36)

(d) θ = 4.20-degrees

Sanity check: The 4.20-degrees makes sense as the Sum Vector has a large x-component and small y-component, suggesting that the vector will be close to the x-axis and make an acute angle with the x-axis.