q) Radius of a curved road is 'R' and the width of the road is 'b'. The outer edge of the road is raised by the height of 'h' wrt the inner edge so that the car with velocity 'v' can pass safe over it. What is the value of 'h'?

h =  V² b /  R g

Just make a right angle triangle ...take the values of the sides as given in the question .....now as u shall know ....



tan theta = V(sq) / R g for a safe turn



=> h / b = V(sq) / R g



=> h = V(sq) b / R g

I'm getting

q) Radius of a curved road is 'R' and the width of the road is 'b'. The outer edge of the road is raised by the height of 'h' wrt the inner edge so that the car with velocity 'v' can pass safe over it. What is the value of 'h'?

ANSWER :

mv² / R = mg sin@

v² / R = g sin@

BUT sin@ = h/b

So -

v² / R = g h / b

h = v²b / Rg

shouldn't it be

edit

didnt see the above ans

Yeah,thats right...b is technically speaking,the hypotenuse.

And for tan@ =v²/rg, draw the FBD and divide the two components of normal reaction..thats Rsin@/Rcos@..you get it...

so tan@ =v²/rg..

From figure, base =( b²-h²)^0.5

tan@ =h/( b²-h²)^0.5...

Usually h<<b..so that simplifies to tan@ =h/b..

Hope that clears it ! :-)