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IrishTwins

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I've never understood how to tell when x and y vector components need to be "switched" to correspond to sin and cos respectively. Example:

In setting up the vectors for an inclined plane problem, it is "more helpful to express the components of W in terms of the angle theta, which the inclined surface makes with the horizontal. In terms of theta, the components of W are:

Wx = W sin theta = mg sin theta (W = mg)
Wy = W cos theta = mg cos theta"


I really can't visually or even conceptually understand why the x now corresponds to sin, and y now corresponds to cos.

Any help would be great, thanks!

(Reference: Page 205 of Physics review notes, Kaplan.)
 

vsl5

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Well assuming that you have a block on an inclined plane, the normal force is mgcos(theta) and the force down the incline is mgsin(theta). To arrive at this, you start with mg pointing straight down and draw the normal force which is perpendicular to the surface of the incline plane. Connect these ends and you get the incline of the plane (mgsin(theta)). So in this case, mg is actually your hypotenuse and theta is the angle between mg and the normal force. Since the triangle formed here is similar to the triangle formed by the incline plane, it is the same.
 

IrishTwins

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Okay, drew the triangle and can see that it is similar to the one that forms the inclined plane. However, I'm still unsure how we know when to use the other angle, which reverse our sin/cos. Oh well ... I suppose I can just know to do this when I see an inclined plane problem. Hopefully there aren't any other circumstances where that would help, because for some reason I just can't wrap my mind around it. It's probably incredibly easy, but there's just a mental block there. Thanks for trying to help anyway.
 

IrishTwins

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Okay, I spent some more time with it and figured out what you were saying. The adjacent angle ends up being "Wy", so that's why we use cos*theta*hypotenuse (W) to find it.

Thanks again!
 
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mdgirl01

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I've always had trouble with inclined plane problems (especially with the angle and trig thing). However, this link posted by Khan Academy, helped clarify everything for me. Apparently, while sitting in my college physics class, I never could access what I learned in 8th grade geometry.

So if you're a highly visual learner like I am, then hopefully this link/explanation will help you out.

http://www.youtube.com/watch?v=TC23wD34C7k&feature=player_embedded#at=601
 
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