Question

The components of vector A are Ax and Ay (both positive), and the angle that it makes with respect to the positive x axis is θ. Find the angle θ if the components of the displacement vector A are: a. Ax = 11 m and Ay = 11 m. b. Ax = 19 m and Ay = 11 m. c. Ax = 11 m and Ay = 19 m.

Answer 1

(a) theta = arctan (Ay/Ax) = arc tan (11/11) = 45°

(b) theta = arctan (Ay/Ax) = arc tan (11/19) = 30.07°

(c) theta = arctan (Ay/Ax) = arc tan (19/11) = 59.93°

To find the angle a vector makes with the positive axis, one needs to know the components of the given vector. Then the arctan of the ratio of the component of the vector along the y-axis and the component along the x-axis is computed. This gives the angle the vector makes with the positive x-axis.

Step-by-step explanation:

This angle is always measured from the positive x-axis in a counter clockwise direction towards the positive y-axis. The values of theta range from 0 - 360°.

The component vectors Ay and Ax are simply representing of the vector A along the respective axis. In terms of force, they represent the effect that force will have along the x- and y-axis respectfully. Mathematically they can be calculated using the formulas

Ax = Acos(theta)

Ay = Asin(theta)

Where theta is measured from the positive X-axis counter clock wisely

A = magnitude of the vector A and mathematically

A = squareroot(Ay² + Ax²)

Answer 2

a)
`\theta = tan^(-1) ((11)/(11))= tan^(-1) (1)=45` degrees

b)
`\theta = tan^(-1) ((11)/(19))= tan^(-1) (0.579)=30.07` degrees

c)
`\theta = tan^(-1) ((19)/(11))= 59.93` degrees

Step-by-step explanation:

If we have a vector
`A= (A_x ,A_y)` in a two dimensional space. The angle respect the x axis can be founded from the following expression:

`tan (\theta)= (A_y)/(A_x)`

And then the angle is given by:

`\theta = tan^(-1) ((A_y)/(A_x))`

Part a

Ax = 11 m and Ay = 11 m

For this case the angle would be:

`\theta = tan^(-1) ((11)/(11))= tan^(-1) (1)=45` degrees

Part b

Ax = 19 m and Ay = 11 m

For this case the angle would be:

`\theta = tan^(-1) ((11)/(19))= tan^(-1) (0.579)=30.07` degrees

Part c

Ax = 11 m and Ay = 19 m

For this case the angle would be:

`\theta = tan^(-1) ((19)/(11))= 59.93` degrees