Anyone ever seen a banked car problem on the MCAT/practice materials?

[axp107]

I remember there being a passage on one of the 30 min EK lectures on a banked car on an incline plane or something.

Are these types of problems too advanced for the MCAT?
I don't understand them. Is it even worth learning how to do them... or to understand banked turns?

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[mongrel]

Yah, I've seen one on some practice material (I've forgotten which/where now...) as well as in the EK book. I didn't have one on my MCAT though.

I would recommend learning how to do this. Even if you don't see one on the real thing, it's excellent practice for setting up simple force balance diagrams. There's a good chance you'll see some sort of "moving situation so it seems difficult" kind of problem-- but rest assured if it's MCAT there's got to be simple way to solve it in 45 seconds or less.

 

[axp107]

Oh ok

Can anyone explain what "bank"ing is and how it works =D
and how centripetal force and friction is involved

 

[cantabdoc]

On the last paper and pencil administration of the MCAT (August 2006), there was definitely a banked car question on my PS. The MCAT loves these types of questions because they can test multiple concepts using this type of passage.

 

[axp107]

So for a banked car, the centripetal force is equal to the friction force right? wrong? lol.. anywhere I can read about this?

 

[Nikki2002]

On the last paper and pencil administration of the MCAT (August 2006), there was definitely a banked car question on my PS. The MCAT loves these types of questions because they can test multiple concepts using this type of passage.

yes i remember it well😡

 

[Foghorn]

I remember there being a passage on one of the 30 min EK lectures on a banked car on an incline plane or something.

Are these types of problems too advanced for the MCAT?
I don't understand them. Is it even worth learning how to do them... or to understand banked turns?

The point of banking a curve at angle theta is to avoid relying on static friction to keep the car along the curved path. The principle is the same as in a conical pendulum.

For banked curves without static friction:
F net centripetal force = (mv^2)/R = N sin theta. N sin theta is the component of the normal force, N, contributing to centripetal acceleration; R = radius of curvature ; theta = banking angle.

For banked curves with static friction:
F net centripetal force = (mv^2)/R = N sin theta + f cos theta; f = frictional force = u*N; u = frictional coefficient.

Draw free body diagrams to convince yourself of these equations.

 

[axp107]

are you sure you have the sines and cosines right?

b/c if you picture the car on an inclined plane... mgsin (theta) = force alone ramp and mgcos (theta) = normal force

or maybe I'm not understanding the concept of banking..

banking is just providing an incline for the car to turn right?

 

[Foghorn]

are you sure you have the sines and cosines right?

b/c if you picture the car on an inclined plane... mgsin (theta) = force alone ramp and mgcos (theta) = normal force

or maybe I'm not understanding the concept of banking..

banking is just providing an incline for the car to turn right?

I'm 100% sure about the sines and cosines.

Some curved roads are banked because when it's wet or icy, it's more difficult to negotiate turns since frictional force is either low or absent therefore contributing little or none to the net centripetal force that allows the car to stay on it's curved path.

For a banked curved with a frictionless surface, the only force furnishing the net centripetal force is the component of the normal force, N sin theta. On this type of surface, that means a car of a given mass can negotiate the turn at a specific velocity and stay on it's curved path, without sliding up/down the incline, in the absence frictional force. For a car going at too low a speed on a banked and curved frictionless surface, it must rely on friction to keep it from sliding down the incline. For a car going too fast, it must rely on friction to keep it from sliding up along the banked curve when negotiating the turn.

 

[axp107]

I still don't understand why its different from inclined plane problems..

The Normal force here is mgcos (theta) and force along plane is mg sin (theta)

 

[themule]

In the inclined plane there is no velocity either into the page or out of the page so it is a 2 dimensional problem, up or down the plane. In the banked car you are looking at a velocity vector, into or out of the page, causing a force with enough magnitude to overcome the force pulling the car down the plane as in the inclined plane problem. Just my .02. I hate physics.

TM

 

[axp107]

For banked curves without static friction:
F net centripetal force = (mv^2)/R = N sin theta. N sin theta is the component of the normal force, N, contributing to centripetal acceleration; R = radius of curvature ; theta = banking angle.

For banked curves with static friction:
F net centripetal force = (mv^2)/R = N sin theta + f cos theta; f = frictional force = u*N; u = frictional coefficient.

Draw free body diagrams to convince yourself of these equations.

Ok I understand what you did there... except what is N equal to? mg?

 

[pandulce321]

That is a really nice picture, but the math seems to be wrong. Aren't they assuming that N equals mg when it should equal mgcos(theta)


Also what does the frictional force in the y direction account for?

 

[BrokenGlass]

I still don't understand why its different from inclined plane problems..

The Normal force here is mgcos (theta) and force along plane is mg sin (theta)

Your coordinate system is set up differently from what Foghorn is doing. Look at the way Foghorn chose X and Y axis and compare this to what you did.

 

[BrokenGlass]

I remember there being a passage on one of the 30 min EK lectures on a banked car on an incline plane or something.

Are these types of problems too advanced for the MCAT?
I don't understand them. Is it even worth learning how to do them... or to understand banked turns?

http://en.wikipedia.org/wiki/Banked_turn

 

[EECStoMed]

http://en.wikipedia.org/wiki/Banked_turn

Why doesn't static friction point the other way? Static friction is supposed to point in the opposite direction of motion-- in this case downwards which is the direction that the car wants to slide.

 

[EECStoMed]

bump, can someone answer my above post. it's freaking me out.

Edit: or does it point downward because the car wants to fly off the bank?

 

[BrokenGlass]

Why doesn't static friction point the other way? Static friction is supposed to point in the opposite direction of motion-- in this case downwards which is the direction that the car wants to slide.

Static friction doesn't always point in the direction opposite to that of motion. Imagine a book placed on a table. Put your palm on the top surface of the book and push it away from you. The static friction between your palm and the top surface of the book is in the same direction as motion. The kinetic friction between the bottom sufrace of the book and the table is in the direction opposite to that of motion. Friction acts to oppose sliding between 2 surfaces in contact. This is not always opposite to the direction of motion.

For simplicity, lets assume the turn is not banked, so that there is no component of normal force contributing to the centripetal force. In this case the centripetal force is static friction because there would be no other force that makes the car go in circles.

If the car was stationary on a banked turn, what you are saying would be true. Static friction would oppose the sliding down of the car and so static friction would be pointing up the incline. But when the car is moving, the static friction is centripetal because it opposes the car's deviation from the circular path.

 

[EECStoMed]

Wait, how is the static friction of my palm and the top of the surface of the book in the same direction as motion, it's opposing.

Static friction doesn't always point in the direction opposite to that of motion. Imagine a book placed on a table. Put your palm on the top surface of the book and push it away from you. The static friction between your palm and the top surface of the book is in the same direction as motion. The kinetic friction between the bottom sufrace of the book and the table is in the direction opposite to that of motion. Friction acts to oppose sliding between 2 surfaces in contact. This is not always opposite to the direction of motion.

For simplicity, lets assume the turn is not banked, so that there is no component of normal force contributing to the centripetal force. In this case the centripetal force is static friction because there would be no other force that makes the car go in circles.

If the car was stationary on a banked turn, what you are saying would be true. Static friction would oppose the sliding down of the car and so static friction would be pointing up the incline. But when the car is moving, the static friction is centripetal because it opposes the car's deviation from the circular path.

 

[BrokenGlass]

Wait, how is the static friction of my palm and the top of the surface of the book in the same direction as motion, it's opposing.

It's not opposing. In fact, it's the net force and it's in the same direction as motion. Static frction is what makes the book move away from you.

 

[EECStoMed]

It's not opposing. In fact, it's the net force and it's in the same direction as motion. Static frction is what makes the book move away from you.

I know that. I think we're talking w/respect to a different reference. If I put my hands on a book and push it forward. The books obviously moves away from me and let's say it's in the "+" direction. However, the static friction on the book and my hand, with respect to the book is opposite to the book's motion. I guess you were talking about the hand yes?

 

[BrokenGlass]

I know that. I think we're talking w/respect to a different reference. If I put my hands on a book and push it forward. The books obviously moves away from me and let's say it's in the "+" direction. However, the static friction on the book and my hand, with respect to the book is opposite to the book's motion. I guess you were talking about the hand yes?

It's not opposite. Think about what force makes the book move away from you. If there were no static friction, your palm would just slide over the top of the book and the book would not move. Static friction is what makes the book move away from you. Static friction is the net force. Net force is always in the direction of motion. Draw a free body diagram to convince yourself.

 

[ICUunocmeGIRL]

it says "force equations at maximum speed v, at threshold of sliding UP incline"

this relates to Forghorn's, "For a car going at too low a speed... it must rely on friction to keep it from sliding down the incline. For a car going too fast, it must rely on friction to keep it from sliding up along the banked curve when negotiating the turn." we're solving for Vmax. if we were solving for Vmin (we would be looking for the threshold of sliding down the incline), then the vector would point in the direction you were talking about and the respective signs would change in the formulae.

 

[Chargers]

Why doesn't static friction point the other way? Static friction is supposed to point in the opposite direction of motion-- in this case downwards which is the direction that the car wants to slide.

This can be easily understood by driving your car on a freeway on ramp (will not be responsible for any reckless driving). You can clearly feel the force trying to pull you (in essential the car) in the direction which will send you and the car outwards (away from the center of the circle). Hence the static friction is the only thing that's trying to keep you and the car from flying air borne.

Kids: don't try this at home or on the road.
My friend was doing 50mph on a freeway ramp and went air borne due to the lack of static friction on the wet rainy road. He's fine, but the car is in the junkyard now😀

 

[heymanooh1]

That is a really nice picture, but the math seems to be wrong. Aren't they assuming that N equals mg when it should equal mgcos(theta)


Also what does the frictional force in the y direction account for?

as opposed to your "right" 1?