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...