Supose that the motor cycle is at position P while ascending the overbridge from A to B . The component of its weight ` mg cos theta ` is ditected ` along `PO.
`therefore ` IF R is the normal reaction at P, then C.P force `(mv^(2))/(r ) `
` = mg cos theta - R `
` therefore R= mg cos theta + (mv^(2))/(r )`
As he moves upwards , ` theta ` decreases and ` cos theta ` increases
` therefore ` the normal reaction ( increases ).
Finally at `B, theta = 0^(@) and R= m g + (mv^(2))/(r )`