Recent content by demente182

I tried again starting from zero, and this time I set the sum of the forces in the Y direction to zero, but now I get a negative angle. So: F_N = -ma_c / cos (90 + "theta") Replaced into : sin (90 + "theta") * F_N = mg and I got "theta" = -126.825869

Yes, my v^2/r is 13.08 m/s^2. If the car slides down the slope friction will act in the opposite direction the car is sliding, so friction acts up the slope. Force of friction = F_N x uk. Am I right?

This is what I found out after doing the diagram: *theta = angle at which the curve is banked. F_N = -ma_c / cos ("theta" +90) so I replaced into: mg + sin ("theta" +90) x F_N = ma_c mg + sin ("theta" +90) x ( (-ma_c) / (cos("theta" +90) ) = ma_c (-9.8) + tan ("theta"+90) x ( v^2...

Homework Statement A curve of radius 65m is banked for a design speed of 105 km/h. If the coefficient of static friction is 0.34 (wet pavement), at what range of speeds can a car safely make the curve? I am not sure how can I get range of velocities at which the car can travel without...