# How does Normal reaction work to provide Centrifugal force in a banked road when some of force die to gravity is already there?

I came across the problem-

A turn of radius 20m is banked for the vehicles going at a speed 36km/hr. If the coefficient of static friction between road and the is 0.4, what are the possible speeds of a vehicle so that it neither slips down or skids up?

While solving it, I come across resolving forces along the road as frictional force, sine of force due to gravity and also Sine of Normal Reaction force which just complicates the solution

What is the way to solve this question as the solution I have just says ignore normal reaction and then solve the question, and it works. How? I mean if Normal reaction is giving the force, why is it not in the Free Body Diagram?

• – sammy gerbil Jan 22 '18 at 19:05
• Whoa that was so really different that what I asked – Ashish Shukla Jan 22 '18 at 19:07
• -1 The problem is almost identical to yours. It does not answer your particular question about normal reaction. It does show how to solve the problem. Car on a frictionless banked curve asks about the normal force. Normal reaction is needed to provide centripetal force, so it should be included in the Free Body Diagram. You have not provided the solution which you are asking about so we cannot explain why normal reaction is not in the FBD. – sammy gerbil Jan 22 '18 at 19:12
• Gonna do that quickly – Ashish Shukla Jan 22 '18 at 19:18
• The normal reaction is labelled $R$ in the diagrams. – sammy gerbil Jan 22 '18 at 19:29