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bleedblue1234
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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 ...