Understanding weight on an inclined plane I'm trying to solve a problem where I have an object resting on an inclined plane, with the angle of the plan being alpha, and the weight being w. I'm having trouble figuring out how I can calculate the component of the weight parallel to the plane. I also want to find out the weight component perpendicular to the plane.
I don't want an outright answer, more of an explanation to help me understand.
Thanks!
 A: Here's a quick sketch:

Gravity is the vector $u$.  Its components in the plane and against the plane are $v$ and $w$ respectively.  You want to find $v$.
The angle between the plane and horizontal is the same as the angle between $w$ and $u$, which allows you to find a simple trig relation to solve the problem.
A: Imagine the inclined plane, and imagine an arrow pointing straight down, representing your weight (the force of gravity). Then imagine rotating that picture so that the plane is horizontal, with the arrow rotated sideways by alpha. Imagine now a horizontal x axis, and a vertical y axis, like they always draw in the textbooks. The x and y axes should cross (intersect) at the tail of the weight arrow. Now imagine a horizontal line that starts at the tip of the arrow, and ends at the (vertical) y axis. That line is the component of the weight that is parallel to the inclined plane.
Then use what you know about sines and cosines (trigonometry) to calculate the amount of weight in the directions parallel and perpendicular to the incline. 
A: Here's an animation which shows how to keep the triangle correct when working incline plane problems. Hope this helps... The red lines are parallel.
Inclined Plane Problem
