Safe Speed of Frictionless Banked Turn

[bleedblue1234]

Homework Statement

On banked turns, the horizontal component of the normal force exerted by the road reduces the need for friction to prevent skidding. In fact, for a turn banked at a certain angle, there is one speed for which no friction is required! The horizontal component of the normal force can provide all of the needed centripetal force. Find the so-called "safe speed" of a turn of radius 40.3 meters banked at 20.4 degrees. Hint: Draw a free-body diagram for a car on a frictionless banked turn. Then consider the net vertical force and the net horizontal force.

Homework Equations

2(pi)r/T = V
a = V^2/r
Fc = mac
w = mg
weight parallel = 9.8 * x * sin(20.4)
weight perp = 9.8 * x * cos(20.4)

The Attempt at a Solution

So what I start to do is I first find the weight
I assign m = 100kg to the car

so w = (100)(9.8)
so w = 980 N

then I find weight parallel to road

so 980 * sin (20.4)
= 341.6006064 N

so weight parallel to road will be equal in magnitude to parallel component of the normal force

so N = 341.6006064

which will need to equal centripital force to keep it from slipping

so Fc = 341.6006064

so Fc = ma
so 341.6006064N/100 kg = a

a = 3.416006064

and a = V^2.r
so 3.416006064*40.3m = V^2

so V^2 = 137.7750444 m/s
so V = 11.7 m/s

but its wrong ...

 

[Delphi51]

I think it is a mistake to find the component of mg parallel to the road.
The question suggests you find the x and y components of the normal force. Draw N perpendicular to the road and show its components N*sinθ and N*cosθ.
In the vertical direction, the forces on the car are one of those components and the force of gravity. Their total must be zero because the car does not accelerate upward or downward. That gives you equation [1].

In the horizontal direction, the question says the x component of N provides the centripetal force. Equation [2]. You can put the two equations together to eliminate N and find a nice simple expression for Fc which you can use to find the velocity.