Static friction on banked curve

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To determine the minimum coefficient of static friction for a car traveling at 91.8 km/hr on a banked curve with a radius of 89.0 m, the angle of banking (theta) was calculated to be 24.03 degrees. The equations involving normal force and gravitational force were used, but the initial attempt to find the coefficient of static friction was incorrect. The static friction's role was clarified, emphasizing its horizontal and vertical components in relation to centripetal force and vertical force balance. A detailed drawing was requested to illustrate the calculations, leading to further clarification of the problem. Ultimately, the correct approach was confirmed by a participant in the discussion.
runner2392
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If a curve with a radius of 89.0m is perfectly banked for a car traveling 71.0km/hr, what is the minimum coefficient of static friction for a car not to skid when traveling at 91.8km/hr?

I figured out theta = 24. 03degs from the equations F(normal)*cos(theta) = mg and F(normal)*sin(theta) = m(v^2/r).

Then to find mus, I tried: m(v^2/r) = F(normal)sin(theta) -F(static f). For F(normal) I substituted mg/cos(theta) but ultimately I got the incorrect answer. Can someone please help?
Thanks
 
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Make a drawing and show your work in detail, please. The static friction acts along the load: It contributes with its horizontal component to the centripetal force, and its vertical component has to be taken into account in the equation for the vertical force components.

ehild
 
How did you calculate with the friction?

ehild
 
figured it out
 
Last edited:

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