# Tangential Acceleration Along an Oddly Shaped Road

1. Nov 4, 2007

### 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.

2. Nov 4, 2007

### Staff: Mentor

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

3. Nov 4, 2007

### JeYo

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

4. Nov 4, 2007

### Staff: Mentor

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

5. Nov 4, 2007

### JeYo

Apparently I was incorrect in my reasoning. I realise 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.

6. Nov 4, 2007

### Staff: Mentor

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.

7. Nov 4, 2007

### JeYo

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

8. Nov 4, 2007

### Staff: Mentor

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

9. Nov 4, 2007

### JeYo

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

10. Nov 4, 2007

### Staff: Mentor

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

11. Nov 4, 2007

### 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?

12. Nov 4, 2007

### Staff: Mentor

No. Realize that the cars are on curved tracks and that they are centripetally accelerating.
Once you correctly find the friction, you'll use Newton II to find the tangential acceleration.

13. Nov 4, 2007

### 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.

14. Nov 4, 2007

### Staff: Mentor

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

Last edited: Nov 4, 2007
15. Nov 4, 2007

### JeYo

But A is accelerating in the opposite direction, because of the braking. And I realise 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.

16. Nov 4, 2007

### Staff: Mentor

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.

17. Nov 4, 2007

### 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...?

18. Nov 4, 2007

### Staff: Mentor

w = m_A*g
(a_y = 0)
Good. The horizontal force on car A is $F = \mu N = \mu mg$. Applying Newton II gives you the horizontal acceleration, $a_x = \mu g$.

19. Nov 4, 2007

### 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?

20. Nov 4, 2007

### Staff: Mentor

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