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# PLEASE GIVE PROPER EXPLANATION.

10 years ago

This question will be faulty until you give some relation between radius of curvatures of pt. A,B and C.

Now,if radius of curvatures at all 3 pts are assumed to be equal,then:

For A,

N+(mv^2/r)=mg;           because centrifugal for will act along the radial vector joining A and centre of curvature and in opposite           N=mg-(mv^2/r)             direction,so will get added to normal reaction force(N) which is along the same direction.

For B,

N=mg+(mv^2/r);           because N will act in up direction while centrifugal force and weight will act in forward direction.

For C,

Same as A.

Therefore,N(A)=N(C)<N(B).

NOTE:This question has been solved assuming the plane of motion to be vertical.

If the plane of motion is horizontal,N(A)=N(B)=N(C)=mg.

10 years ago

in this question if you consider your reference frame on the car and then apply the concept of centrifugal force.

now,at point A,

mg=n+centrifugal

at B,

mg+centrifugal=n

so the normal at b is definately more than at a.for c we have to note that the centrifugal depends on radius and hence more the radius less the centrifugal.