# Normal reaction in banked road circular motion

ΣF=0

## The Attempt at a Solution

Basically I don't agree with this solution, I think it is R1 + R2 - mgcos5 + mv^2sin5/r = 0. Can you help me please?
thanks!

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Basically, I don't like that for ΣF it takes the fictional inertial force opposite to centripetal force, instead of taking the actual centripetal force. Isn't this wrong?
Second thing I don't know is the direction of the centripetal force in such situation: isn't it from the centre of gravity of the vehicle towards the centre of the circular motion? Ie. isn't it PARALLEL to the surface of the banked road? Because here, he has it parallel to the horizon, instead of parallel to the banked road!

BvU
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2019 Award
isn't it from the centre of gravity of the vehicle towards the centre of the circular motion? Ie. isn't it PARALLEL to the surface of the banked road? Because here, he has it parallel to the horizon, instead of parallel to the banked road!
Fom the centre of gravity of the vehicle to the axis of the circular motion. And when banking a road, they usually have a verical axis (otherwise you would just have a sloping plane).
for ΣF it takes the fictional inertial force opposite to centripetal force
You should see is as: $R_1+R_2-mg\cos 5^\circ$ contribute a fraction $\sin 5^\circ$ to the actual centripetal force $mv^2/r$.

The presentation is somewhat suggestive: $\sum F = 0$ would mean uniform linear motion, not a circular trajectory. Here $\sum F \ne 0$. In fact, $\sum F= {mv^2\over r}$.

CWatters
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Gold Member
Second thing I don't know is the direction of the centripetal force in such situation: isn't it from the centre of gravity of the vehicle towards the centre of the circular motion? Ie. isn't it PARALLEL to the surface of the banked road? Because here, he has it parallel to the horizon, instead of parallel to the banked road!
The drawing is correct. Note how the radius is specified.

haruspex