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# acceleration on inclined plane

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i have a problem to become the accelaration for a object on a inclined plane ... i have a plane and their normal vector, more i have the grafity of e.g. 9.81N now i searching the acceleration along this plane... please help me ! have tryed alot but nothing was right !

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No need to cross post in other forums. I''ve deleted the other two.

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F=mg*sin(theta)
F/m=a=g*sin(theta)

I belive this is correct.

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also how do you do friction? I knew once but i forgot. I know it has to do with cosine (the normal force still able to act on the object). but, well,yeah.

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jep this formel i know ... my proble is to become tha acceleration
allong the axis ! i have formels which work correct if the plane is
inclined in only one axis

e.g. 45Â° to the x->axis the acceleration for ->x = 4.905 and ->y = 4.905

but all this formels fails by a plane inclined in two axis !

here a picture to the problem ... the result (x,y,z) is frong it should be allong the plane !
http://www.aacm.ch/data/project/opengl/winkel.gif

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theta is the angle between the gravity vector (let''s call it G) and the normal N of the plane you''re on.

theta = ArcCos(Dot(-G,N)/(Length(N)*Length(G)))

As G is probably (0,-g,0) (a downward vector of a length of the gravity force)

the force will be

f=mg sin (theta) = mg sin( ArcCos(Dot(-G,N)/(Length(N)*Length(G)))) = mg sin(ArcCos(Dot(-G,N)/Length(N))) = m sin(ArcCos(Ny/Length(N))) = mg sqr(1- Ny*Ny/(Nx*Nx+Ny*Ny+Nz*Nz))

Plus, if N is normalized, you have

F=mg sqr(1 - Ny*Ny)

Cool huh?

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Just a precision: Ny is the y component of the N vector

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good math ! but what is the forces direction !! this is my big problem

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Oh!

Ok. The vector V (your "force direction" you want to find is the following:
you want it to be normal to the normal of the plane () Dot(N,V)=0 and it must have to be the most "vertical" possible, that is with the smallest length possible for a fixed y component (maybe not very intuitive). Let''s say Vy=1 (or whatever) You have NxVx + Ny + NzVz = 0 where Nx,y,z is fixed
=> Vz = -(NxVx + Ny)/Nz
And you want to minimize Vz^2+Vx^2 (in order to minimize vector''s length)

We have Vz^2+Vx^2 = (-(NxVx + Ny)/Nz )^2+Vx^2 = ((Nx/Nz)^2+1)*Vx^2 + 2*NyNx/Nz*Vx + (some constant)

taking the first derivative, we got:

2*((Nx/Nz)^2+1)*Vx + 2*NyNx/Nz = 0 (to get the min. of the length of V)

thus Vx = - (NyNx)/(Nz(Nx/Nz)^2 + Nz)

Vz is given by Vz = -(NxVx + Ny)/Nz

Then you have your vector V which you would normalize and then multiply by the force I gave before. Just a last thing: dont forget to take the opposite... (As Ny was set at 1 rather than -1)

I hope I didnt do mistake... but that should be correct...

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(sorry for "my" english :D)

Well I''ve just seen sth:
if Nz is 0 then just replace every z with x in the above equations, otherwise div by 0. if x and z are 0 then the force is 0 and there isn''t any force.

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