## Homework Statement

calculate max volocity of car with road banking and no friction

## Homework Equations

The normal reaction N.
Resolving in the vertical component,
Ncosθ=mg (which make N >=mg)

Resolving horizontal component,
Nsinθ=mv^2/r

My question is if we resolve the forces perpendicular to the road:
N = mgcosθ (which would make N < =mg) , it would give a completely different N
. We use N=mgcosθ in inclined plane problems, why can't we use it here?

collinsmark
Homework Helper
Gold Member
Hello mike168,

Welcome to Physics Forums!
My question is if we resolve the forces perpendicular to the road:
N = mgcosθ (which would make N < =mg) ,
That's not true for this problem. :tongue2:

But it would be true if the car was just siting there motionless on the inclined road, and it was held in place by friction; or if the car was allowed to slide down the side (but still not moving forward). That would be just like an incline plane problem.

But this problem is quite different.
it would give a completely different N
. We use N=mgcosθ in inclined plane problems, why can't we use it here?
The simplest way to see this is to draw a free body diagram (FBD).

Draw the FBD for this problem (speeding car along a banked, frictionless curve), where there are three forces involved. There is the force of gravity, the normal force, and horizontal resultant force [Edit: where this horizontal, resultant force is the same thing as the centripetal force].

Compare that to an inclined plane problem FBD where a block slides down an incline. In a problem like that there are also three forces. There is the gravitational force, the normal force, and the tangential resultant force that is directed along the plane.

In each problem, the forces can be put together to form a triangle. But the triangles will be differently shaped in each FBD. For each FBD, ask yourself, "Which forces are perpendicular to each other?" and "Which vector is the hypotenuse?"

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