- #1
Joshuarr
- 23
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Homework Statement
Curved portions of highways are always banked (tilted) to prevent cars from sliding off the highway. When a highway is dry, the frictional force between the tires and the road surface may be enough to prevent sliding. When the highway is wet, however, the frictional force may be negligible, and banking is then essential. Figure 6-13a represents a car of mass m as it moves at a constant speed v of 20 m/s around a banked circular track of radius R = 190 m. If the frictional force from the track is negligible, what bank angle θ prevents sliding?
Homework Equations
Uniform circular motion: a = v^2/R
Magnitude of centripetal force: F = m*a = m*v^2/R
I used this identity in solving the problem: sin(2θ) = 2*sin(θ)*cos(θ)
The Attempt at a Solution
So, I managed to solve this problem, but the answer I got seems to have a conceptual difference than the book. So I'm wondering how I can correct my thinking.
I started by drawing a similar diagram in 6-13b, EXCEPT, I assumed that Fg has the largest magnitude, and when decomposing the vectors, I chose it as the hypotenuse.
This lead me to the relationships: F_N = mg*cos(θ) and F_Nr = mg*cos(θ)*sin(θ) = mg*sin(2θ)/2
I labeled F_Nr as the centripetal force, so: F_Nr = mg*sin(2θ)/2 = m*v^2/R,
which gives: θ = Acos( 2*v^2/(R*g))/2 = Acos(2*(20)^2/(190*9.8))/2 = 12.722°
The book got: Atan(v^2/(g*R)) = Atan(20^2/(9.8*190)) = 12.124°
They are similar, and normally I'd just ignore the difference, but there seems to be a difference in the formulas themselves [for example, they get F_n = mg*(1/cos(θ)) as opposed to my F_n = mg*cos(θ)], which leads me to believe that my thinking was wrong, and I just got a very close answer by serendipity.
I don't understand why F_N is larger than F_g. It seems to me the force that "causes" all the other forces is the force of gravity, so it should be the largest.
How do I know that that isn't the case?
The diagram the book gives, including their free-body diagram, is attached.
Thanks in advance!