Understanding Incline Planes & Vector Issues

[korters]

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

 

[heth]

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.

 

[thomate1]

for reaching out with your question about incline planes and vector issues. It can be confusing to understand how to properly set up and solve problems involving incline planes, especially when considering the direction and magnitude of vectors like gravity and acceleration. Let's break down the issue you're having and see if we can come to a better understanding.

First, let's start with the basics. Incline planes involve a block or object resting or moving on an angled surface. In order to properly set up our coordinate system, we align the x-axis with the surface the object is resting on, as you mentioned. This allows us to easily break down the weight vector, which is the force of gravity acting on the object, into its x and y components.

Next, we need to consider the normal force, which is the force exerted by the surface on the object and is always perpendicular to the surface. In a frictionless incline plane, the normal force will be equal to the weight vector's y-component, as you correctly stated.

Now, let's move on to the issue you're having with the sign conventions and resolving the weight vector into its components. When we resolve the weight vector into its x and y components, we are essentially breaking it down into two separate vectors that act in those respective directions. This means that the x-component of the weight vector will act in the positive x-direction, while the y-component will act in the negative y-direction.

Now, let's consider the equation w=ma. This equation represents Newton's Second Law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the weight vector is acting as the net force on the object. So, when we solve for the y-component of the weight vector, we get w=m(-9.8)cos(30) or w=-m(9.8)cos(30), depending on whether we use positive or negative values for gravity/acceleration. This is because the y-component is acting in the negative y-direction in relation to the y-axis.

However, when we solve for the x-component of the weight vector, we get w=m(-9.8)sin(30) or w=m(9.8)sin(30). This may seem confusing because the x-component is acting in the positive x-direction, but remember that we are breaking down the weight vector into two separate vectors, and the x-component is acting in the