Help with Friction and Stopping Car

[Husker70]

Homework Statement

A car is traveling at 50.0 mi/h on a horizontal highway. (a) If the coefficient of
static friction between road and tires on a rainy day is .100, what is the minimum
distance in which the car will stop? (b) What is the stopping distance when
the surface is dry and coefficient of static friction is .600?


2. Homework Equations [/]

The Attempt at a Solution

I'm not sure how to get started as I don't know the normal force, and I'm not
sure what to solve for. If someone can help get the ball rolling I should be able
to finish. Any help would be appreciated.

Kevin

 

[tiny-tim]

A car is traveling at 50.0 mi/h on a horizontal highway. (a) If the coefficient of
static friction between road and tires on a rainy day is .100, what is the minimum
distance in which the car will stop? (b) What is the stopping distance when
the surface is dry and coefficient of static friction is .600?
…
I'm not sure how to get started as I don't know the normal force, and I'm not
sure what to solve for.

Hi Husker70!

Call the mass of the car m (it will cancel out later ).

Then the normal force (on a horizontal surface) is just mg.

Now use F = ma to find the acceleration … which will be constant, so you can then use the usual uniform accleration kinetic equations.

 

[Husker70]

Thanks for the start and I figured that the mass shouldn't matter but in order to
slove for the acceleration, I'm not sure how to find the force.

Thanks,
Kevin

 

[tiny-tim]

… in order to
slove for the acceleration, I'm not sure how to find the force.

Hi Kevin!

Since it's a horizontal road, the only force slowing it down is the friction force …

so multiply the normal force by µ.

 

[Husker70]

Does this seem right?
50 mi/h = 80.4672 km/h = 22.352 m/s

a= .100n
ma = .100(g)
a = .98

solving for time
v=vo + at
22.352m/s=0+.98(t)
t=22.81s

solving for distance
d=do+vot+at^2/2
d= 0+0+.98(22.815)^2/2
d= 255m

Thanks for everyone's help
Kevin

 

[tiny-tim]

Hi Kevin!

Right answer, long-winded method …

Does this seem right?
50 mi/h = 80.4672 km/h = 22.352 m/s

a= .100n
ma = .100(g)
a = .98

a = .98 is right, but the other lines should be:

ma= .100n
a = .100(g)

solving for time
v=vo + at
22.352m/s=0+.98(t)
t=22.81s

solving for distance
d=do+vot+at^2/2
d= 0+0+.98(22.815)^2/2
d= 255m

hmm … you have v_f and v_i and you need s but not t …

so it would have been a lot quicker just to use v_f² = v_i² + 2as

 

[Joyci116]

Hello,

I am working on the same problem, but I do not comprehend the work that was done to solve this problem.
Problem statement and variables given:
Q: A car is traveling 50.0 mi/h on a horizontal highway. (a.) If the coefficient of static friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop? (b.) What is the stopping distance when the surface is dry and mu(s)=0.600? (mu(s) I think means static friction)
Attempt:
What I did was draw a free-body diagram where the weight force(earth is acting upon the car) is pointing down, normal force(road acting upon the car) is pointing up, and there is a friction force going to the left. I did convert 50.0mi/hr to 22.35m/s, but after that I am lost.

Thank you,

Joyce

 

[tiny-tim]

Welcome to PF!

Hello, Joyce! Welcome to PF!

(have a mu: µ )

What I did was draw a free-body diagram where the weight force(earth is acting upon the car) is pointing down, normal force(road acting upon the car) is pointing up, and there is a friction force going to the left. I did convert 50.0mi/hr to 22.35m/s, but after that I am lost.

i] First, find the friction force.

ii] Then find the acceleration.

What do you get? ​

 

[Joyci116]

Hello Tiny-Tim,

Would the friction force be: F=-w+N (friction equals negative weight force plus normal force)? I think I might be confused on how to obtain the friction force.

Thank you,

Joyce Kuang

 

[tiny-tim]

Hello Joyce!

Would the friction force be: F=-w+N (friction equals negative weight force plus normal force)?

That is a valid equation, but it's the equation for the net vertical force …

in this case the vertical acceleration is zero,

so (from good ol' https://www.physicsforums.com/library.php?do=view_item&itemid=26") mass times 0 = net force = -W + N.

so W = N …

in other words, when something stays on a flat surface (so that its vertical acceleration is zero), then the normal force equals the weight.

(you can get a similar equation for the normal force on a slope, by doing 0 = net perpendicular force)

To get the https://www.physicsforums.com/library.php?do=view_item&itemid=39" force, you don't use F = ma at all, you use F = µN,

ie friction = µ times normal force

This works either for kinetic friction (µ_k), or for maximum static friction (µ_s) …

in this case, the static friction is maximum, since the question says so (because it asks for the minimum stopping distance). ​

 

[Joyci116]

Hello,

What I have so far is:
Summation of F=m(0)=net force=-w+N
In which you said w=N
Then you said to use F=\mu N
I know that \mu is 0.100
so, F=(.100)N
What is N? I know that it is equal to the weight force, but it does not give me a value for the weight or normal force. How can I obtain that value?

And after finding the normal force, then I know the friction force, right?
After finding the friction force why do I have to find the acceleration? Is it because we need acceleration to find the minimum distance?

Thank you,

Joyce

 

[tiny-tim]

Hello Joyce!

… I know that \mu is 0.100
so, F=(.100)N
What is N? I know that it is equal to the weight force, but it does not give me a value for the weight or normal force. How can I obtain that value?

They don't give you the mass of the car, so call it "m" (it will cancel out later, when you find "a") …

then W = mg.

After finding the friction force why do I have to find the acceleration? Is it because we need acceleration to find the minimum distance?

Yes, once you know "a", you can use one of the standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations.

 

[Joyci116]

I'm so sorry, but I am still a little confused.
The normal force (N in Newtons) is equal to the weight force.
You said to use "m" to represent the mass, but what do I do after that?
What do I do with my F=\muN equation? How does that relate to the mass?
Maybe that is where my trouble lies. I do not see a connection between the mass and the equation, or I do not understand.

Thank you,

Joyce

 

[Aggression200]

Hey, Joyce.

You know that F=\muN. You also know that N = W = mg.

Now, gravity is simply an acceleration in the y-direction, correct?

So, now you know that N = ma_{y}


You already knew that F=\muN. Now F, in this situation, is friction force. So, it is a force in the x-direction. So, given that, you could say that...

ma_{x}=\muma_{y}

Since the mass is the same for both sides, the masses cancel out (since you would divide both sides by m). That leaves you with...

a_{x}=\mua_{y}

From that you can solve for a_{x} and then use kinematics to solve for \Deltax.


Enjoy,
Alex

 

[Joyci116]

Aggression200,

Thanks, I got it.
What I did was:
F=\mu_sN
N=mg
N-ma_y (g is a_y)
ma_x=\mu_s(ma_y)

a=(.100)(-9.8)= -0.98
V_i=22.35 m/s
V_f=0 m/s
x=?

So take the equation V_f²=V_i² +2ax
Solved for x, which gives you
(V_f²-V_i²)/(2a)
x=255m?

Thank you very much,

Joyce