Below is a partially filled table of vectors and their magnitudes, directions, a

Question

Below is a partially filled table of vectors and their magnitudes, directions, and components in an x-y coordinate system. All magnitudes are in meters, and all angles are in degrees measured counter-clock-wise from the +x direction.

Table I

Vector

Magnitude

Angle

x-component

y-component

A

8

6

B

7

100

C

300

3

D=2XB

Table II

Vector

Magnitude

Angle

x-component

y-component

E=A+B+C

F= -A+B-C

G=E+F

(8 pts) Calculate all the missing information in Table I.

(4 pts) Sketch the vectors A, B, C and D as arrows with their tails at the origin of the x-y coordinate system. (The sketch doesn’t have to be to precise, but as much as possible, try to properly represent the magnitudes and directions.)

(4 pts) Sketch how to graphically find the vectors E and F in Table II.

(8 pts) Using the component method, calculate the components, magnitudes, and directions of the vectors E and F.

(1 pt) What is the magnitude & direction of the vector G?

Table I

Vector

Magnitude

Angle

x-component

y-component

A

8

6

B

7

100

C

300

3

D=2XB

Table II

Vector

Magnitude

Angle

x-component

y-component

E=A+B+C

F= -A+B-C

G=E+F

Explanation / Answer

here,

for vector A

Ax = 8 m

Ay = 6 m

tan(theta) = Ay/Ax

theta = arctan( 6/8)

theta = 36.87 degree

A = sqrt(Ax^2 + Ay^2)

|A| = 10 m

for vector B

Bx = B * cos(100)

Bx = 7 * cos(100)

Bx = - 1.21 m

By = B * sin(theta)

By = 7 * sin(100)

By = 6.89 m

for vector C

theta = 300 degree

Cx = C * cos(theta)

3 = C * cos(300)

C = 6 m

Cy = C * sin(theta)

Cy = 6 * sin(300)

Cy = - 5.2 m

for vector D

D = 2 * B

|D| = 14 m

theta = 100 degree

Dx = - 2.42 m

Dy = 13.78 m

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