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

Answer:

a) $\theta = tan^{-1} (\frac{11}{11})= tan^{-1} (1)=45$ degrees

b) $\theta = tan^{-1} (\frac{11}{19})= tan^{-1} (0.579)=30.07$ degrees

c) $\theta = tan^{-1} (\frac{19}{11})= 59.93$ degrees

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:

$tan (\theta)= \frac{A_y}{A_x}$

And then the angle is given by:

$\theta = tan^{-1} (\frac{A_y}{A_x})$

Part a

Ax = 11 m and Ay = 11 m

For this case the angle would be:

$\theta = tan^{-1} (\frac{11}{11})= tan^{-1} (1)=45$ degrees

Part b

Ax = 19 m and Ay = 11 m

For this case the angle would be:

$\theta = tan^{-1} (\frac{11}{19})= tan^{-1} (0.579)=30.07$ degrees

Part c

Ax = 11 m and Ay = 19 m

For this case the angle would be:

$\theta = tan^{-1} (\frac{19}{11})= 59.93$ degrees

Answer 2

Answer:

(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.

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)