Banked curve min/max velocities

AI Thread Summary
A banked curve with a radius of 15 m allows a 930 kg car to travel at 48.0 km/h without skidding, even on icy conditions with negligible friction. The problem requires determining the range of speeds for safe navigation around the curve, factoring in a static friction coefficient of 0.300. The equations of motion involve balancing gravitational and centripetal forces, leading to a derived angle of 50.38 degrees. Attempts to incorporate friction into the calculations resulted in incorrect velocity estimates of 37.50 km/h and 56.58 km/h. The discussion highlights the complexity of combining forces in a banked curve scenario, especially when friction is considered.
ndoc
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Homework Statement



A curve of radius 15 m is banked so that a 930 kg car traveling at 48.0 km/h can round it even if the road is so icy that the coefficient of static friction is approximately zero. You are commissioned to tell the local police the range of speeds at which a car can travel around this curve without skidding. Neglect the effects of air drag and rolling friction. If the coefficient of static friction between the road and the tires is 0.300, what is the range of speeds you tell them?

Homework Equations



Nsinø = (mv^2)/r
N cos ø = mg

The Attempt at a Solution



Combining these and simplifying produces rg tan ø = v^2
Solving for theta I get 50.38

I then return to the equations, but instead add friction force

mgtanø = (mv^2)/r + μ(mg/cosø) and mgtanø = (mv^2)/r - μ(mg/cosø)
solving these for velocity, but this produces the wrong answer (I get 37.50 km/h and 56.58 km/h)

Thanks in advance for the help!
 
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Welcome to PF, ndoc.
It seems to me the situation is more complicated for the case with friction. I'm thinking of mg down and the CENTRIFUGAL force horizontal to my right. So the normal force is mg*cos(A) + mv^2/r*sin(A). Certainly it makes sense that as the car goes faster, it gets pressed harder against the banked road, increasing friction.
 
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