# Normal reaction in banked road circular motion

1. May 11, 2016

### physea

1. The problem statement, all variables and given/known data

2. Relevant equations
ΣF=0

3. 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!

2. May 11, 2016

### physea

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!

3. May 11, 2016

### BvU

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).
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, $\sum F= {mv^2\over r}$.

4. May 11, 2016

### CWatters

The drawing is correct. Note how the radius is specified.

5. May 11, 2016

### haruspex

Centripetal force is a resultant force, not an applied force. It is the component of the resultant that is normal to the direction of travel. Where the speed is constant, that makes it the entire resultant: ΣFother=Fcentripetal.
Centrifugal force is a fictional force, but is an applied force, not a resultant: ΣFother+Fcentrifugal=0.
Since centripetal and centrifugal are equal and opposite, the equations are the same.
The vehicle is describing a circle in a horizontal plane. The centre of that motion is therefore in that plane, not down the bottom of the slope.