Forces Question (Object on Ramp)

[Lucky_69]

Hello!

This is my first post so please forgive me if I'm doing this wrong. Thanks!

http://img4.imageshack.us/img4/3922/godfdb.jpg

Ok it has always stumbled me why we use the angle the ramp makes with the surface it rests on when resolve gravity into it's components

I understand why and how to resolve vectors into it's components

the force of gravity, x component = mg sin theta
the force of gravity, y component = mg cos theta

the force of gravity is clearly the hypotenuse of the triangle
the force of gravity, the x component, is clearly the opposite side of the triangle
the force of gravity, the y component, is clearly the adjacent side of the triangle

ok but it's very clear that the angle the ramp makes with the surface that it sits on is not the anlge being used in those formulas...

So I've tried to find the anlge that I believe is being used. I drew the triangle (in a dark red) in the diagram above with the angle that I believe is the angle that is being used in those formulas.

Ok I have always been told that you can use the angle the ramp makes with the surface it sits on. This lead me to believe that the two angles (indicated in the diagram I drew) are equal to each other...

Ok, how do I prove that these angles are equal to each other?

It has been a while sense I have taken geometry so my skills are a little bit rusy in proofs. I always remeber doing them in statement reason format and we would make a table... Please tell me step by step because I'm a little bit rusty in geometry it's been a while, just telling me that the angles are complementary won't help because I have forgten how to prove two different angels are complementary in different tirangles when the triangles are seperate. I do remeber what complementary means. So if you could please give both the statements and reasons for why the statements are true that would be great... =]

also it seems a little bit odd to me drawing the x component of the force of gravity down there to complete the triangle (the red one) because it's no longer sitting on the x axis... If we place it down there don't we also have to place the x-axis down there? I've always been told to put the x-axis through the center of the object and parrallel to the ramp that way when you do problems like trying to find how far it'll fall down you would have to do less work... so it just seems a bit odd removing from the x component from the x-axis and placing it randomly on the diagram to complete a triangle...

Thanks so much!

 

[dx]

This is how I think of it: first imagine theta to be zero. In this case it is clear that the two angles in question are equal. Now, since the actual situation is just a rigid rotation of this case, the angles must still be equal.

 

[rcgldr]

If the angle between two lines is θ, then the angle on the opposite side of that intersection that is also θ. The angles on the adjacent sides = π - θ. If lines are drawn perpendicular to those two lines, then the angles between those 2 lines will also be θ and π - θ.

Assume base of plane is horizontal, there angle between plane and base is θ. Now draw a line perpendicular to the base, which is the direction of gravity, and draw a line perpendicular to the plane, which is the "normal" line to the plane. The angles between gravity and the normal line will also be θ and π - θ. Effectively you're just rotating both lines by π/2.

I don't recall the geometric names for these relationships.