Components of Weight on an Inclined Plane

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nDever
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Hey,

Imagine a frictionless inclined plane situation where the ramp goes up and to the right.

If we put an object on it with any mass m, would it be fair to say that both of the components of the weight are negative? My reasoning is that the x component causes acceleration to the left and the y component presses the object against the ramp (which is downward with respect to the x axis).

The components of the weight would therefore be computed as follows.

Wy= -mgcos(θ)
Wx= mgsine(-θ)

Correct?
 
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You can get those results if you choose a particular sign convention. The sign convention is arbitrary, but you need to be consistent throughout the problem. Your sign convention is perfectly natural-- the x-axis is positive upward and to the right, the y-axis is positive upward and to the left. Any other sign convention would be equally correct, as long as it was clear and consistent.

Note you can also use the traditional vertical y-axis and horizontal x axis, in which case gravity is all in y, but the other forces will need to be analyzed in components. Your way is more convenient.
 
I have gotten used to using cosine to find the component along the x axis and sine along the y. I was wondering if there were any other methods of finding the components (with the correct sign) of weight on an inclined plane.
 
Sure, there's any way you want. Just draw arbitrary axes and you're off, if your conventions are consistent then the problem does not care how you draw your axes (although the trigonometric functions you encounter will). Still, there are only two convenient situations-- one where the x-axis is along the inclined plane, and the other where it is along the horizontal (or you can exchange the x and y axes but that's not going to change anything but the labels). Your way is most convenient, because you only have to break up gravity into components, and you usually know gravity already. The other way means you don't have to break up gravity into components, but you do have to break up the forces you don't necessarily know yet, so that's less convenient.