Is the Normal Force Component in Superelevated Curves the Centripetal Force?

AI Thread Summary
In superelevated curves, the normal force has a component directed towards the center of the curve, which contributes to the centripetal force required for a car's circular motion. This component of the normal force is indeed responsible for providing the necessary centripetal acceleration. The relationship can be expressed as centripetal acceleration equating to the centripetal force divided by mass. Understanding this dynamic is crucial for analyzing vehicle motion on curved paths. The discussion highlights the importance of free body diagrams in visualizing these forces.
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A question about when a car take a superelevated curve. I don't know how to be more precise in English. Take a look at this photo, maybe you'll understand better : http://www.fmciclismo.com/noticias/bmx/imagenes/peralte%20bmx%20copia.jpg
My question is : when you draw the free body diagram, you see that the normal force (N) has a component in direction of the center of the radius of the curve that describe the movement of the car. Does this component is precisely the centripetal force? That would mean that the centripetal acceleration is worth \frac{F_c}{m}.
 
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Yes, that component of the normal force creates the centripetal acceleration. Nice picture, by the way.
 
Thanks for your answer. And for the picture, thanks to google :rolleyes:.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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