What are the Magnitude and Angle of Vectors in Terms of their Components?

AI Thread Summary
The discussion focuses on converting vectors into their magnitude and angle based on their components. For the velocity vector with components -75 m/s and 35 m/s, the magnitude is calculated using the square root of the sum of the squares of the components, while the angle requires adding 180 degrees to the arctangent result due to the negative x-component. In the case of the force vector in the third quadrant with a magnitude of 50 lb and an x-component of 40 lb, the angle is determined similarly, but the magnitude is already provided. Participants are advised to visualize the vectors on an x-y coordinate plane to better understand the angles. Emphasizing the importance of understanding the geometric relationships between the components and the hypotenuse helps clarify the calculations.
Susanem7389
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Rewrite the following vectors in terms of their magnitude and angle (counterclockwise from the +x direction)
a) A velocity vector with an x component of -75 m/s and a y component of 35 m/s
- I found the magnitude by using A= square root of ( Ax squared plus Ay squared ), however I did not get the correct answer for the angle. I used the formula angle = tan -1 ( Ay/ Ax)

b) A force vector with a magnitude of 50 lb that is in the third quadrant with an x component whose magnitude is 40 lb.
- I could not find the correct magnitude and angle with the same formula used in part A.
 
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Could you show your angel for part a) as well as magnitude and angel for part b)?
 
Note that your calculator cannot distinguish arctan(1/1) from arctan(-1/-1).
Rule of thumb... if the x-component is negative, add 180 degrees to what your calculator tells you when using arctan. (Some calculators may have something like an atan2 function.)
 
For part A, adding 180 degrees to what the calculator gave me, I got the correct answer, however it did not work for part b.
 
Susanem7389 said:
For part A, adding 180 degrees to what the calculator gave me, I got the correct answer, however it did not work for part b.

You'll have to show your work...
 
I would strongly suggest drawing out each of these vectors on an x-y coordinate plane so you can see exactly what the angles the formulas are giving you. The vector angle is always going to be measured with respect to the +x direction, in a counter-clockwise fashion; in other words, a vector in this +x direction would have an angle of 0 degrees.

Try to not just remember formulas, look at the right triangles the vector magnitude, x-component, and y-component are forming. The magnitude is going to be the hypotenuse of the right triangle that is formed.

for b) They give the the magnitude of the vector already, you just need to find the angle. Remember that the angle will be measure from the +x direction rotating counter-clockwise until it meets the vector in question.
 
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