A 600-kg car is going over a curve with a radius of 120 meters that is banked at an angle of 25 degrees with a speed of 30 meters per second. The coefficient of static friction between the car and the road is 0.3. What is the normal force exerted by the road on the car? a) 7240 N b) 1590 N c) 5330 N d) 3430 N e) 3620 N Note: The speed is not Vmax (roughly 32 m/s). I've been working under the assumption that mgcos(theta) will be increased by the frictional component acting downward. I simply can't seem to get the solution.

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http://i.imgur.com/kkaO7.png the triangles that the dots/circles/particles are on are cross sections of the curve in the road. the triangles coming from the dots/circles/particles are force vectors and their components parallel and perpendicular to the ramp. Fn = normal force Fc = centripetal force Fg = force of gravity Fgp = force of gravity's component perpendicular to the road Fc = centripetal force's component perpendicular to the road Fn = Fgp - Fcp Fgp = cos(25)Fg Fcp = -sin(25)Fc Fg = mg -> (600kg)(9.81m/s/s) Fc = (mv^2)/r -> (600kg)(30m/s)(30m/s)/(120m) Fn = cos(25)(600kg)(9.81m/s/s) - sin(25)(600kg)(30m/s)(30m/s)/(120m) Fn is approximately 3432.7N, so choice D. the centripetal acceleration pulls the particle (car) towards the center of the circle. this decreases the normal force from the road, because the car is not pushing as hard into the ramp. you can also use vector projection ( http://en.wikipedia.org/wiki/Vector_projection#Vector_projection_2 ) to project the centripetal acceleration along a unit vector perpendicular to the road. to do this, you would define: (note: ^ is denoting a unit vector, ~ is denoting a vector) ^r = the unit vector along the ramp ^p = the unit vector perpendicular to the ramp ~Fc = the vector of centripetal acceleration. points in -x direction. Fc = magnitude of ~Fc, which is just (mv^2)/r ||~A|| = magnitude of vector A. ^r = (cos(25))^i + (sin(25))^j ^p = (sin(25))^i + (-cos(25))^j ~Fc = (-Fc)^i projecting vector ~A along ~B: ( (~A dot ~B)/(||~B||^2) )*~B projecting vector ~Fc along ^p ( (~Fc dot ^p)/(||^p||^2) )*^p note: unit vectors are defined to have a magnitude of 1. therefore ||^p|| = 1 (~Fc dot ^p) = -sin(25)(Fc) therefore the centripetal force subtracts a force from the net force "pushing" into the road. the force of friction only acts tangential to the curve of the road, and therefore the dot product of this frictional force with the unit vector that is PERPENDICULAR to the ramp results in 0, meaning it does not effect the normal force. The final answer is D.

@neugauss This is not the right answer. Sorry, but I have no time right now to point exactly where your equations are incorrect. This very question has already been answered on OS by viniterranova in the question by jgonzales8.

i'm struggling to see how my answer is incorrect. whenever you have the time to tell me what exactly is wrong with it, i'd love to hear. viniterranova's solution is in broken english and it insists that the value of gravity is unknown. however, the idea is silly because gravity near the surface of the earth is constant?? i don't see how the gravity would be different at all? and then he seems to solve for the "gravity" by assuming the answer to the question is 7240N? I don't see at all how he comes to that number? please tell me where and how i'm incorrect, because the solution posted by viniterranova makes no sense to me at all.

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