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I'm trying to work out the forces acting on the left and the right rail if the track has curvature and cant (i.e. is an inclined plane)

I've come this far as you see in the picture but I'm struggling with working out the effective gravity acting on the left and the right rail. I know the center of mass of the train (it's about the middle between left and right but somewhat closer to the bottom than the top).

Now I managed to work out the point of the track where gravity is acting on which is Δ (orange on the picture) from the center of the track. But I can't figure out how to split the mass between the left and the right rail.

Is it correct to just assume that (where m is the mass of the train)

m₁ = m/2 * ((⍵/2+Δ)/(⍵/2-Δ))

m₂ = m/2 * ((⍵/2-Δ)/(⍵/2+Δ))

enter image description here

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enter image description here

you can solve your problem like this:

we put the coordinate system at the center of mass.

$$\sum{F}_y={\it F2}+{\it F1}-\sin \left( \alpha \right) m{\omega}^{2}R-\cos \left( \alpha \right) mg=0\tag 1$$

and torque sum at the center of mass

$$\sum{\tau}_{cm}=F1\,a-F2\,b\tag 2=0$$

Thus:

$$F1={\frac {m \left( \sin \left( \alpha \right) {\omega}^{2}R+\cos \left( \alpha \right) g \right) b}{a+b}} $$ $$F2={\frac {ma \left( \sin \left( \alpha \right) {\omega}^{2}R+\cos \left( \alpha \right) g \right) }{a+b}} $$

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  • $\begingroup$ Thank you so much!!! $\endgroup$ – rhedak May 28 at 23:23

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