# Normal reaction in banked road circular motion problem

• physea
In summary, the equation of forces for the axis parallel to the banked road depends on the direction of the car's slipping. If the car is in perfect equilibrium, the equation would be R1+R2=mgcosθ, with the normal force and mgcosθ causing the centripetal acceleration. However, if the car is not in equilibrium, the equation would be R1+R2=mgcosθ+mV^2/r or R1+R2=mgcosθ-mV^2/r, depending on the direction of slipping. The variables R1 and R2 represent the normal reactions of the ground to the vehicle's wheels, mg is the vehicle weight, V is its speed, and θ is
physea
What is the equation of forces please for the axis parallel to the banked road? Is it R1+R2-mgcosθ=mV^2/r or R1+R2-mgcosθ=-mV^2/r? Can you advise me please?

physea said:
What is the equation of forces please for the axis parallel to the banked road? Is it R1+R2-mgcosθ=mV^2/r or R1+R2-mgcosθ=-mV^2/r? Can you advise me please?
The friction will be directly opposite mgsinθ if the car is going slow enough, but it will be the other way if the object is going fast enough. Therefore, the equation depends on which way the car is slipping. If it is in perfect equilibrium, then mgsinθ and friction cancel out and you're left with only components perpendicular to the road (such as the normal force and mgcosθ) causing the centripetal acceleration.

physea said:
What is the equation of forces please for the axis parallel to the banked road? Is it R1+R2-mgcosθ=mV^2/r or R1+R2-mgcosθ=-mV^2/r? Can you advise me please?
If you want an answer to that then it is probably better if you actually define all the variables in those equations, along with a suitable diagram. Also, you will get a better response if you can manage to be a bit more polite with your replies; the customer is not always right here because there are no (paying) customers. "That's all".

R1 and R2 are the normal reactions of the ground to the vehicle's wheels
mg is the vehicle weight
V it's speed
θ the angle of the banked road
I think now it's clear

physea said:
R1 and R2 are the normal reactions of the ground to the vehicle's wheels
mg is the vehicle weight
V it's speed
θ the angle of the banked road
Is the angle measured from the horizontal or from the vertical? Is positive for R1, R2 and the centripetal acceleration taken in the sense of an inward net force? I assume yes. Is positive for mg cos(θ) taken in terms of down=positive or up=positive. That would obviously have a key influence on the answer you seek.

Since you have adopted coordinates in which the centripetal acceleration is neither aligned with the x-axis or with the y-axis and have restricted your attention to the x axis, one would expect only part of the centripetal force (##\frac{mv^2}{r}##) to be relevant. Yet there is no cos(θ) or sin(θ) factor on the term to account for this. Similarly, there is no cos(θ) or sin(θ) factor on the terms for R1 or R2. So I have to question whether either equation can be correct.

Still, a simple argument can be made. If we select a track with a degenerate bank angle so that cos(θ) = 0 then the equations reduce to

##R_1 + R_2 = \frac{mv^2}{r}## and
##R_1 + R_2 = -\frac{mv^2}{r}##

Assuming a convention that positive is inward/up then ##R_1## and ##R_2## must be positive. ##\frac{mv^2}{r}## will neccessarily be positive. So the second equation is obviously wrong. Though, as above, that does not mean that the first equation must be right.

## What is normal reaction in a banked road circular motion problem?

The normal reaction in a banked road circular motion problem refers to the force exerted by the road surface on a vehicle as it moves in a circular path. This force is perpendicular to the surface of the road and is responsible for keeping the vehicle from sliding off the road.

## How does the angle of banking affect the normal reaction in a banked road circular motion problem?

The angle of banking has a significant impact on the magnitude of the normal reaction in a banked road circular motion problem. A steeper banked angle will result in a higher normal reaction force, while a shallower angle will result in a lower normal reaction force.

## What role does friction play in the normal reaction in a banked road circular motion problem?

Friction plays a crucial role in the normal reaction in a banked road circular motion problem. It acts in the direction opposite to the motion of the vehicle and helps to balance the forces acting on the vehicle, thereby keeping it on the road.

## What happens to the normal reaction when the speed of the vehicle changes in a banked road circular motion problem?

The normal reaction will change as the speed of the vehicle changes in a banked road circular motion problem. As the speed increases, the normal reaction force decreases, and as the speed decreases, the normal reaction force increases. This is due to the changes in centripetal force and centrifugal force as the vehicle accelerates or decelerates.

## Can the normal reaction force be greater than the weight of the vehicle in a banked road circular motion problem?

Yes, the normal reaction force can be greater than the weight of the vehicle in a banked road circular motion problem. This is possible when the banked angle is steep enough and the speed of the vehicle is high enough to generate a large enough centripetal force to counteract the weight of the vehicle.

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