Understanding Incline Planes & Vector Issues

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SUMMARY

This discussion addresses the complexities of resolving vectors in incline plane problems, specifically in frictionless scenarios. The participant grapples with the correct application of sign conventions when decomposing the weight vector into its x and y components. The consensus emphasizes the importance of defining positive directions clearly, suggesting that both the x-component and y-component should be treated with consistent sign conventions to avoid confusion. The discussion concludes that using parallel and perpendicular components rather than traditional x and y axes simplifies the analysis.

PREREQUISITES
  • Understanding of vector decomposition in physics
  • Familiarity with Newton's second law (w=ma)
  • Knowledge of trigonometric functions (sine and cosine)
  • Concept of coordinate systems in physics
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  • Research vector decomposition techniques in physics
  • Study the effects of friction on incline planes
  • Learn about coordinate system transformations in mechanics
  • Explore advanced applications of Newton's laws in multi-dimensional motion
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Students and educators in physics, particularly those focusing on mechanics and vector analysis, as well as anyone seeking to clarify the application of sign conventions in incline plane problems.

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This is just a basic incline plane problem. I know how to do it, but these sign conventions and vector issues are what confuse me. I know that I align my coordinate system so the x-axis is basically the surface the block is resting/moving on, the net weight vector of the block will be directed down at some angle, and the normal force will be directed perpendicular to the surface the block is resting on (so that its x-component is zero), etc.

But here's my issue, in particular if the set up looks like this http://www.cs.utah.edu/~zachary/isp/applets/SlidingBlock/image1.gif, the incline is FRICTIONLESS!, and my coordinate system is set up in such a way that the block will be moving in the POSITIVE direction on the x-axis (when it gets moving that is). Say the angle here is 30 degrees. When resolving the weight vector into the x and y components, how do you account for the fact that gravity/acceleration is a vector (has direction and magnitude)? I understand that w=ma. The way I reason is that if you're trying to find the y-component of vector w, this is w=m(-9.8)cos(30), which makes sense because the y-component vector is directed in the negative direction in relation to the y-axis. But (HERES MY PROBLEM) if you resolve the weight vector into its x-component the same way, w=m(-9.8)sin(30), you get a negative number obviously indicating that the block is moving in the opposite direction that it should be. And this doesn't make sense. Other books simply don't account for the sign on the acceleration vector but rather look at the vector as a whole: example
y-comp is w=-m(9.8)cos(30)
x-comp is w=m(9.8)sin(30)
which gives the correct direction of movement for the x-component. While this gives the correct magnitude and direction, my issue is why doesn't such an x-component equation account for the negative sign for acceleration/gravity, because acceleration/gravity is a vector. This is driving me nuts.

Thanks
 
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You can choose which directions to make positive. It makes sense to eliminate as many minus signs as possible, as it's easy to 'lose' them!

If you had a falling object, the displacement, velocity and acceleration vectors all point down, so it's easiest to make "down" the positive direction.

It's the same here. There's no need to get confused with signs. For horizontal and vertical, make down and right your positive directions. And for parallel and perpendicular, make down the slope and into the slope your positive directions.

PS. It's probably best to think of your components as parallel and perpendicular rather than x and y, as x and y strongly implies horizontal and vertical.
 

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