Adding 2 vectors. One is 20 degrees below x-axis

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The discussion revolves around adding two vectors, with one vector pointing 20 degrees below the x-axis and the other 43 degrees above it. The user is confused about how to correctly represent the angle of the first vector, considering both -20 degrees and 340 degrees as potential options. It is clarified that either representation is valid, provided the calculator is set to degree mode. The user is guided on calculating the x and y components using the cosine and sine functions, respectively. The conversation emphasizes the importance of understanding vector direction and component calculation methods.
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I've done many other problems like this and understood them all but when it comes to 20 degrees below the x-axis it threw me off. I tried putting the angle to 340 degrees but that was incorrect. I know something has to be negative, i just can't keep guessing at getting it wrong. Hope someone can help!

Homework Statement



A vector A has a magnitude of 51.0 m and points in a direction 20.0° below the x axis. A second vector, B, has a magnitude of 70.0 m and points in a direction 43.0° above the x axis. Using the component method of vector addition, find the x- and y- components of the vector C.

Homework Equations



Vector A + Vector B = New Vector C

Cx = Ax + Bx
Cy = Ay + By


The Attempt at a Solution



I'm really not sure. In the book it says to put vector B at the head of vector A. I just drew each vector as a separate triangle and solved for Ax, Ay and so on. Like i said before, i just need to know what to do with the "20 degrees below the X-axis. Is it -20? 340? I've tried both and they came out wrong so help! Thanks again.
 
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"I'm really not sure. In the book it says to put vector B at the head of vector A."

This is one way of finding C, but is not the method you are asked to use

"I just drew each vector as a separate triangle and solved for Ax, Ay and so on."

This sounds like what they are asking you to do.



"Like i said before, i just need to know what to do with the "20 degrees below the X-axis. Is it -20? 340? I've tried both and they came out wrong so help! Thanks again."

You should be able to use either -20 deg or +340 deg. Make sure your calculator is i degree mode, if that's how you chose to do the calculation.

In any case

Ax = 51 cos(-20) = 51 cos(340) = 49.9

Ay = 51 sin(-20) = 51 sin(340) = -17.4
 
Thanks a lot for the help! This is a great forum. I'm glad I found it.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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