Max Velocity for Banked Road Problem

[jrmed13]

Okay, so I am doing a problem involving a car driving on a banked, frictionless, circular track (theta=31degrees) and i am supposed to find the maximum velocity that the car can drive. I know that to find the velocity, i have to find the centripetal acceleration by saying that (mv^2)/r = nsin(theta). Then, I have to solve for n by saying that ncos(theta)=mg. However, I am confused... why can't n=mgcos(theta)? My understanding is that two forces are equal in magnitude if the object doesn't move in either direction. The car doesn't move into the road or out of the road... or does it?? please help! I have a test on monday.

 

[learningphysics]

It is true that the car doesn't move into the road or out of the road. However, the component of acceleration into the road is not 0.

In other words... suppose you take the $$\Sigma{F} = ma$$ equation perpendicular to the road. N - mgcos(theta) = ma... here a is not 0! this occurs because of the centripetal acceleration.

 

[m2003]

Hello, thank you for sharing your problem with me. I can provide some clarification on your question.

Firstly, your understanding is correct that in order for an object to remain in equilibrium, the forces acting on it must be equal in magnitude and opposite in direction. In this case, the two forces acting on the car are its weight (mg) and the normal force (n) from the banked road.

However, in this specific scenario, the normal force is not acting directly opposite to the weight. This is because the road is banked at an angle (theta) and the normal force is acting perpendicular to the road's surface. Therefore, you cannot simply use the equation n=mgcos(theta) to find the normal force.

Instead, you need to consider the components of the weight force and the normal force that are acting in the direction of the car's motion. These components can be found using trigonometry, and the equation you mentioned, (mv^2)/r = nsin(theta), takes into account these components.

To summarize, the normal force is not equal to mgcos(theta) because it is not acting directly opposite to the weight. I hope this helps you understand the problem better and good luck on your test!