# Homework Help: Wet vs. dry friction

1. Sep 21, 2014

### Wapapow10

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

A driver makes an emergency stop and inadvertently locks up the brakes of the car, which skids to a stop on dry concrete. Consider the effect of rain on this scenario. If the coefficients of kinetic friction for rubber on dry and wet concrete are
μK(dry) = 0.81 and μK(wet) = 0.53,
how much farther would the car skid (expressed in percentage of the dry-weather skid) if the concrete were instead wet?

2. Relevant equations

3. The attempt at a solution
F=uk*g
F=.81*9.8
F=7.93

F=.53*9.8
F=5.19

5.19/7.93=65.44%

Got this wrong and I'm not sure what I'm missing

2. Sep 21, 2014

### guitarphysics

The force isn't directly proportional to the displacement, so you can't just take the ratio of the forces. Figure out first how far each one will go (use kinematics, and remember that F=ma), then take that ratio. Keep in mind that you should take the ratio of wet/dry displacement, so you'll get a number like 1.53, and that means that in wet concrete the car travels 153% of what it travels in dry concrete, or that it travels 53% more.

3. Sep 21, 2014

### Wapapow10

Thats what I'm lost on. I don't know the mass or acc of the car. I don't know where to start we didnt get a chance to go over this in class and the work is due tomorrow.

4. Sep 21, 2014

### guitarphysics

You don't need to know the mass of the car. All the unknown quantities (mass, initial velocity I think are the only two) will cancel out when you take the ratio, because you can obviously assume it's the same car and everything :D

5. Sep 21, 2014

### Wapapow10

I guess I missed something in class. I can't get it to work out I changed it over to F=ukg that was no help. tired just f=ma as well

6. Sep 21, 2014

### guitarphysics

Ah, allright- I'll get you part of the way there ;).
From kinematics, we know that when there's constant acceleration (let's assume it's constant), the following holds: $$v^2=v_0^2+2aD$$
Now, the situation we have describes zero velocity at the end, so $$D=-\frac{v_0^2}{2a}$$
This is the general case (for any acceleration). Now, let's get specific: for the wet concrete, we have some acceleration $a_w$ for the car, and for dry concrete we have an acceleration $a_d$. What we want to find is the ratio $D_w/D_d$. Keep in mind that $$D_w=-\frac{v_0^2}{2a_w}$$ and that $$D_d=-\frac{v_0^2}{2a_d}$$
Now to find the respective accelerations, just use dynamics; I think you were ok on that part.

Afterwards, when you take the ratio of the distances, the $v_0^2$ terms (as well as the mass) will cancel and everything will be ok.

7. Sep 21, 2014

### KL7AJ

In a real world scenario, where hydroplaning occurs, the coefficient of friction is basically zero. :)