# End position of braking car on an icy incline

1. Sep 13, 2010

### imatreyu

1. The problem statement, all variables and given/known data

A car with a speed of 40.0 km/h approaches the bottom of an icy hill. The hill has an angle of inclination of 10.5 degrees. The driver applies the brakes, which makes the car skid up the hill. If the coefficient of kinetic friction between the ice and the tires if 0.153, how high, measured along the incline, is the car on the hill when it comes to rest?

2. Relevant equations

F=ma. . .and. . .eventually kinematics?

3. The attempt at a solution
Wy= Wsin(theta)--> mgsin(theta)
Wx= Wcos(theta)--> mgcos(theta)

Y: N-mgsin(theta) = 0 (there is no y acceleration?)
N= mg sin(theta)

X: N(mu) - mgcos(theta) = ma
mgsin(theta)mu-mgcos(theta)=ma
--> m cancels out-->
gsin(theta)mu-gcos(theta)=a

Where can I go from here? What I have I done wrong? :( Thank you in advance!

2. Sep 13, 2010

### rl.bhat

Since car is going up, the friction force and the component of weight along the inclined plane act in the same direction.

3. Sep 13, 2010

### imatreyu

But doesn't the friction force make the car skid up the hill (which would make it opposite from the weight x component)?

4. Sep 13, 2010

### rl.bhat

The car moves up due to its initial KE. Frictional force acts in opposite direction to the motion of the car. The car is moving up. So Fr acts in the downward direction and brings the car to rest.

5. Sep 13, 2010

### imatreyu

So, once I correct the direction regarding friction, should the rest of the process lead me to the solution?

(Supposedly 19 m)

Because I did correct the direction of the friction, and the acceleration I am getting is suspiciously near 9.8 m/s2. . I think there must be a mistake elsewhere. .?

6. Sep 13, 2010

### rl.bhat

I am getting correct answer. Show your calculations.

7. Sep 13, 2010

### imatreyu

Y: N-mgsin(theta) = 0 (there is no y acceleration?)
N= mg sin(theta)
---->
X: N(mu) + mgcos(theta) = ma
mgsin(theta)mu+mgcos(theta)=ma
--> m cancels out-->
gsin(theta)mu+gcos(theta)=a

so:

9.8sin(10.5)(.153) + 9.8cos(10.5) = a
a= 9.90914. . .m/s^2

vi= 40.0 km/h --->11.1111m/s
vf= 0 m/s

vf^2 = vi^2 + 2ad
0^2= (11.111) + 2 (9.90914) d
d= - 0.56064

:( Something is wrong. . .

Still getting wrong answer.

Last edited: Sep 13, 2010
8. Sep 13, 2010

### rl.bhat

N(mu) + mgcos(theta) = ma

Normal reaction is mg*cosθ

And vf^2 = vi^2 - 2ad

9. Sep 13, 2010

### housemartin

i think in you got x components of forces wrong. Component of "gravity (mg)" along the hill is not mgcos(teta). Check your diagram. And one more important thing, before writing Newtons second law, always indicate your coordinate axis ;] this way you can keep track of positive and negative acceleration etc.

10. Sep 13, 2010

### imatreyu

rl.bhat,

mu (-9.8) cos10.5 - 9.8sin10.5 =a
a= 03.08675439

--> 0= 11.111^2 + 2 a d

solve for d= 19.99. . .

I think this is about right?

housemartin, The coordinate axis is aligned so that the positive x is going up the direction of the incline. Is the component of gravity parallel to the hill actually supposed to be mgsin(theta)?

Thank you to all. . .

Last edited: Sep 13, 2010
11. Sep 13, 2010

### housemartin

so if positive x is uphill component of gravity points in which direction? +x or -x? And in which direction does friction force points?

12. Sep 13, 2010

### imatreyu

Gravity is pointing negative. Friction forces the car positive.

. . .I don't even understand how "The driver applies the brakes, which makes the car skid up the hill. " is possible. .

13. Sep 13, 2010

### housemartin

Friction force always opposes motion - it makes things go slower - velocity decrease, and that mean what? it's not in the positive direction ;]
If driver didn't applied brakes, car would slow down only because of gravity, which component uphill is negative. With brakes, another force "helps" gravity to slow the car faster.

14. Sep 13, 2010

### imatreyu

Aha! Thank you, rl.bhat and housemartin. =)