Tangential Acceleration Along an Oddly Shaped Road

[JeYo]

Three cars are driving at 29.0m/s along the road shown in the figure. Car B is at the bottom of the hill and car C is at the top. Suppose each car suddenly brakes hard and starts to skid. What are the tangential accelerations of Cars A B and C.

Here should be the picture: http://i216.photobucket.com/albums/cc200/zucchinigrandma/knight_Figure_07_43.jpg


I am not really sure how to even approach this problem, for I have never attempted a problem of this type and have never been very clear on how to find tangential acceleration.

 

[Doc Al]

How does the friction force compare for each car? Hint: Consider how the normal force differs for each car.

 

[JeYo]

I would assume that the frciton would be like f_k = $$\mu$$_k * cos$$\theta$$ * n? But what would the angle be?

 

[Doc Al]

Kinetic friction is $f_k = \mu_k N$. (I don't understand where you got that $\cos\theta$ factor.)

 

[JeYo]

Apparently I was incorrect in my reasoning. I realize that the normal force for car A is mg, but I honestly have no idea of what the normal forces for cars B and C would be, if they are different, or how to find them.

 

[Doc Al]

You find the normal force by applying Newton's 2nd law. Realize that the cars B and C are centripetally accelerating. Start by indentifying the forces acting on each car.

 

[JeYo]

The forces acting on the car should only be gravity, normal, and the braking, which I am not sure how to represent, right?

 

[Doc Al]

Right. The three forces are gravity, normal force, and friction. Now apply Newton II to the vertical forces.

 

[JeYo]

But how does friction account into the vertical forces when the force of friction is only along the horizontal?

 

[Doc Al]

It doesn't. But to calculate friction you need the normal force, which is a vertical force.

 

[JeYo]

Ooh, well, since there is no acceleration in the y direction, i would assume that n = w = mg, yes? and then the friction is equal to $$\mu$$_k * mg. But how does that enable me to find the acceleration?

 

[Doc Al]

Ooh, well, since there is no acceleration in the y direction, i would assume that n = w = mg, yes?

No. Realize that the cars are on curved tracks and that they are centripetally accelerating.

and then the friction is equal to $$\mu$$_k * mg. But how does that enable me to find the acceleration?

Once you correctly find the friction, you'll use Newton II to find the tangential acceleration.

 

[JeYo]

I do not understand...I mean what I thought would not be right for car A either? I mean, I don't want you to give me the answer or anything, but I just do not understand.

 

[Doc Al]

Car A is on a flat road--no vertical acceleration there. Cars B and C are on curved sections. Big difference.

 

[JeYo]

But A is accelerating in the opposite direction, because of the braking. And I realize that there is a difference in the motions of car A and cars B and C; however, I do not understand how to use that difference in a way that helps me solve the problem.

 

[Doc Al]

I meant to say no vertical acceleration of Car A, thus its normal force is simply equal to mg. But cars B & C have centripetal acceleration, which is vertical. Calculate the normal force by applying Newton II to the vertical forces.

 

[JeYo]

Alright, take a look at this:

$$\Sigma$$(F_A)_y = n - w = m_A * a_y = 0N
$$\Sigma$$(F_A)_x = -f_k = m_A * a_x

So, f_k = m_A * a_x, $$\mu$$_k * g = a_x

Right...?

 

[Doc Al]

Alright, take a look at this:

$$\Sigma$$(F_A)_y = n - w = m_A * a_y = 0N

w = m_A*g
(a_y = 0)

$$\Sigma$$(F_A)_x = -f_k = m_A * a_x

So, f_k = m_A * a_x, $$\mu$$_k * g = a_x

Good. The horizontal force on car A is $F = \mu N = \mu mg$. Applying Newton II gives you the horizontal acceleration, $a_x = \mu g$.

 

[JeYo]

Sweet...now, how exactly would I begin to classify the forces on cars B and C as in either the x- or y-direction?

 

[Doc Al]

The three forces are gravity, normal force, and friction. Which way does each act?

 

[JeYo]

Well, gravity goes straight down. My first instinct is to say that normal goes perpindicular, but perpindicular to what? So, maybe it is oposition to gravity, but I doubt it. And I am also not sure about friction, I know it goes in the opposite direction of the front of the car, but whether or not it has a vertical component, I am not sure.

 

[Doc Al]

Well, gravity goes straight down.

Right.

My first instinct is to say that normal goes perpindicular, but perpindicular to what?

The surface. And the plane of the surface is defined by the tangent to the surface at the point in question. (There's a reason why cars B and C are at the exact bottom and exact top of the hills.)

So, maybe it is oposition to gravity, but I doubt it. And I am also not sure about friction, I know it goes in the opposite direction of the front of the car, but whether or not it has a vertical component, I am not sure.

Friction is parallel (tangential) to the surface.

 

[JeYo]

So the forces on cars B and C are pointing in the same directions as the forces on car A?

 

[Doc Al]

So the forces on cars B and C are pointing in the same directions as the forces on car A?

Absolutely.

 

[JeYo]

But the accelerations could not be the same for all three cars, could they?

 

[Doc Al]

But the accelerations could not be the same for all three cars, could they?

No. As I suggested before, first figure out the normal force (as that will determine the friction and thus the tangential acceleration). Do that by analyzing the vertical forces and acceleration.

 

[JeYo]

But if all of the cars have the same mass, and the mass cancels out when you do the y-direction of the sum of the forces, will it not come out to be the same acceleration?

 

[Doc Al]

Try it and see. What does Newton's 2nd law tell you about vertical forces?

 

[JeYo]

It tells me that the normal force - the force of gravity = m_car * a_y = 0, yes?

 

[Shipman515]

I have this same problem on my homework, except with 21 m/s.

How can you use these formulas if you have no mass or coefficient of friction?