# A car stopping on a slippery surface on an uphill

• Heidi Henkel
In summary, the equation for the direct deceleration due to gravity is mg(sin angle of elevation)=-.5at^2+V_0t.
Heidi Henkel
Let's say we have a car going up a 10% grade that slams on the brakes. The normal force would be mg(cos angle of elevation) in a component of Fg that is perpendicular to the road surface, and there would be a direct effect of gravity in the component of Fg that is parallel with the road surface. How can we combine this information with other stopping distance equations, so we can do things like find stopping distance if we know initial speed and the normal stopping distance on the same surface on level ground, or if we know initial speed and force of friction, or, in reverse, if we know stopping distance and want to find initial speed? I am trying to find some equations that solve for different basic aspects of the kinematics of this situation.

So we have a direct deceleration due to gravity in mg(sin angle of elevation) component and a lower force of friction than on flat ground because normal force is lower.

Heidi Henkel said:
So we have a direct deceleration due to gravity in mg(sin angle of elevation) component and a lower force of friction than on flat ground because normal force is lower.

I think you can make two possible assumptions:

a) Assume the (maximum) braking force is independent of the gradient (for small gradients like 10%).

b) Assume the (maximum) braking force is dependent on the normal force.

Neat idea. I think this depends on how slippery the road surface is. On dry pavement, probably the max braking force has to do with the brake calipers of the vehicle and aspects of the hydraulic mechanisms and so on, and is quite independent of the normal force. In a low friction situation, like braking on snow or ice, the friction is the limiter and the max braking force is dependent on normal force. I'm interested in that. The situation where the surface is slippery and the max braking force is dependent on the normal force.

Heidi Henkel said:
Neat idea. I think this depends on how slippery the road surface is. On dry pavement, probably the max braking force has to do with the brake calipers of the vehicle and aspects of the hydraulic mechanisms and so on, and is quite independent of the normal force. In a low friction situation, like braking on snow or ice, the friction is the limiter and the max braking force is dependent on normal force. I'm interested in that. The situation where the surface is slippery and the max braking force is dependent on the normal force.

If you posted this in the homework section, you might get more help.

I'll be offline now. Perhaps someone else can help.

If you assume the braking force is some factor of the normal force, then you should be able to generate some equations.

## Homework Statement

A car is going up an 8% grade on a very slippery surface, 25 mph. It has studded tires and is on packed powder. An average stopping distance for this tire and road surface combination is 19.5 meters on level ground, for this type of vehicle. The driver sees a deer and slams on the brakes, skids for 5.8 meters while braking hard, then hits the deer. How fast is the car going when it hits the deer? (the friction is the limiter of max braking force.) This is not really a homework problem, I was just told to post it here because people read these more.

## Homework Equations

Ah, that is the question. V_1^2/d_1 = V_2^2/d_2 is one of them, but then there is also the mg(cosine angle of elevation) that plays into the normal force which plays into the friction which messes up that whole thing. Ff = (mu)Fn with Fn being calculated with cosine of angle of elevation so it's the component perpendicular to the road surface. Change in kinetic energy = work done, so .5mv^2=mad=(mu)mg(cos angle of elevation), could cancel out the m and have v^2=d/2(mu)g(cos angle of elevation). But now we need to use the other component of the force of gravity acting on the car, the component parallel to the road surface, to calculate the direct effect of gravity to slow the car down. The friction is decreased by being on a hill, but the car is also being slowed directly by the force of gravity, in its other component. What is an equation for that? The only equation I can think of for that is d=-.5at^2+V_0t and I don't have any information about time, so that isn't helpful. In this case a would be g(sin angle of elevation) but it's still not helpful because it's an equation with another unknown in it. How do I combine those two dynamics into one equation? Do I need to make a third equation with time in it so I have 3 simultaneous equations?

## The Attempt at a Solution

I need to make the equations fit together, first[/B]

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Heidi Henkel said:

## Homework Statement

A car is going up an 8% grade on a very slippery surface, 25 mph. It has studded tires and is on packed powder. An average stopping distance for this tire and road surface combination is 19.5 meters on level ground, for this type of vehicle. The driver sees a deer and slams on the brakes, skids for 5.8 meters while braking hard, then hits the deer. How fast is the car going when it hits the deer? (the friction is the limiter of max braking force.) This is not really a homework problem, I was just told to post it here because people read these more.

## Homework Equations

Ah, that is the question. V_1^2/d_1 = V_2^2/d_2 is one of them, but then there is also the mg(cosine angle of elevation) that plays into the normal force which plays into the friction which messes up that whole thing. Ff = (mu)Fn with Fn being calculated with cosine of angle of elevation so it's the component perpendicular to the road surface. Change in kinetic energy = work done, so .5mv^2=mad=(mu)mg(cos angle of elevation), could cancel out the m and have v^2=d/2(mu)g(cos angle of elevation). But now we need to use the other component of the force of gravity acting on the car, the component parallel to the road surface, to calculate the direct effect of gravity to slow the car down. The friction is decreased by being on a hill, but the car is also being slowed directly by the force of gravity, in its other component. What is an equation for that? The only equation I can think of for that is d=-.5at^2+V_0t and I don't have any information about time, so that isn't helpful. In this case a would be g(sin angle of elevation) but it's still not helpful because it's an equation with another unknown in it. How do I combine those two dynamics into one equation? Do I need to make a third equation with time in it so I have 3 simultaneous equations?

## The Attempt at a Solution

I need to make the equations fit together, first[/B]
I don't understand some of your equations. What are d_1 and d_2? No, don't answer that since it won't be relevant.
In
v^2=d/2(mu)g(cos angle of elevation)​
the 2 factor seems to be in the wrong place. And that does not take into account that there is both an initial and final nonzero speed.
For the contribution of gravity to the slowing, consider PE.

What is PE? Potential energy? This is a car going uphill, so some of the kinetic energy turns into potential energy? mgh? hmm

My latest idea has been:
[things that slow down the car] = [change in kinetic energy]
so mad+[mu]mg[cos angle of elevation] = .5v_i^2-.5v_f^2. Work done by gravity + work done by friction = change in kinetic energy.
For "mad" in the first part of the equation I could substitute dg[sin angle of elevation].

d is 5.8 meters, I don't know what mu is, and otherwise everything is a number (g is the gravitational constant and angle of elevation can be easily found via trig from the grade).mgh is an interesting thing to work with relative to grade, h is easy to find...

KE_i = KE_f + PE_f (conservation of energy)
except now that makes combining the two different things contributing to the slowing, even more confusing. Because some of that KE_i does not got into either KE_f or PE_f, it gets squandered on friction, turned into heat and mechanical changes to the materials (skid marks and such).To answer your question about d_1 and d_2, the d's are the remaining stopping distances. So v_1 could be 25 mph and d_1 could be 19.5 meters and then d_2 could be (19.5-5.8)meters and then you could solve for v_2. If that was all that was happening. But it's on a hill and there's also direct gravity pulling on the car, so I don't know how to adapt it to that.

KE_i = work done by friction + KE_f + PE gained by the car going up in elevation

.5mv_i^2 = [mu]mg[cos angle of elevation] + .5mv_f^2 + mgh hahaha there's m in everything so I can dump it via distributive property
.5v_i^2 = [mu]g[cos angle of elevation] + .5v_f^2+gh

The only mystery I have left is how to calculate mu so I do not have two unknown variables.

Heidi Henkel said:
KE_i = work done by friction + KE_f + PE gained by the car going up in elevation

.5mv_i^2 = [mu]mg[cos angle of elevation] + .5mv_f^2 + mgh hahaha there's m in everything so I can dump it via distributive property
.5v_i^2 = [mu]g[cos angle of elevation] + .5v_f^2+gh

The only mystery I have left is how to calculate mu so I do not have two unknown variables.
What given information have you not used?

distance traveled between v_i and v_f
except I used the vertical component of that for h
I do not know of any equations that contain both mu and distance, but maybe one can be created by combining a bunch of different equations.

Heidi Henkel said:
distance traveled between v_i and v_f
except I used the vertical component of that for h
I do not know of any equations that contain both mu and distance, but maybe one can be created by combining a bunch of different equations.
There's another distance mentioned.

19.5 m to stop on level ground if starting at 25 mph. There must be a way to get mu from that.

Heidi Henkel said:
19.5 m to stop on level ground if starting at 25 mph. There must be a way to get mu from that.
Yes indeed.

2 equations that both contain a and t:
a = change in v / change in t
d=.5at^2+v_0t
25mph=11.176m/s^2
a = (11.176m/s - 0) / t
a=11.176/t
d=.5(11.176/t)t^2 + (11.176)t
19.5m = .5(11.176)t + (11.176)t = (1.5)(11.176)t
t=1.16s
a=11.176/1.16=9.6m/s^2

Ff=[mu]Fn
ma=[mu]mg
a=[mu]g
9.6=[mu]9.8
[mu]=0.9

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Heidi Henkel said:
d=.5(11.176/t)t^2 + (11.176)t
Anyway, it would be more direct to use a SUVAT equation that does not involve time.

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yes I am using positive accelerations all the way through, but I think it works because I am consistent about it. I could try it using negative accelerations; I am pretty sure it would come out the same.
To calculate V_f I would assign the a as negative.
I do not know how you would find acceleration without finding time. Acceleration is time dependent, and you need acceleration to use Ff=[mu]Fn. Maybe I am missing something.

d=-.5at^2+v_0(t)
19.5 = -.5 (11.176/t)t^2+11.176t
19.5 = .5(11.176)t
t=3.5s
OK so the sign for acceleration changes it

a=11.176/3.5=3.2m/s^2

a=[mu]g
3.2=[mu]9.8
mu=0.3265 OK that is more realistic, 0.9 would be above average for a dry road and I got the data from field tests

so now I want to know how fast the car is going when it hits the deer and how long a time it takes to hit the deer, from when the brakes are applied...

I am curious what equation you would use to find mu without finding time.

Heidi Henkel said:
I am curious what equation you would use to find mu without finding time.
There are five SUVAT equations. Each involves a different four of the five variables, so one omits time. Hint: it is the conservation of work equation, with mass factored out.

w=mad none of them contain that
and mass is not any part of any SUVAT equation
I am confused
maybe it is v_f^2 = v_0^2 + 2ad
Which is the SUVAT equation that makes no sense to me, seems unrelated to anything else, arbitrary or something, hard for me to use an equation that makes no sense to me
but one could plug the numbers into it and have just one unknown.
Conservation of work? Where does the 2 come from, in 2ad? ?

Heidi Henkel said:
w=mad none of them contain that
and mass is not any part of any SUVAT equation
I am confused
I said with mass factored out. Gain in PE equals loss of KE, or vice versa.

makes sense now

v_f=(11.176)^2 + 2 (3.2) (5.8) = 7.1 m/s
15.9mph

to find time, 5.8=-.5(3.2)t^2+(11.176)t
t = 0.56s
so the driver slams on the brakes, skids for 0.56 seconds, hits deer at 15.9mph

Heidi Henkel said:
v_f=(11.176)^2 + 2 (3.2) (5.8) = 7.1 m/s
15.9mph
What happened to the slope? Shouldn't you be using the equation in post #10? (But note, you left the distance out of the friction term there.)

ok back to my slope equation...
.5v_i^2 = [mu]g[cos angle of elevation] + .5v_f^2+gh
I don't see where distance fits in, though
this came from KE_0 = work done by friction + KE_f + change in PE
no distance

oh I see, in the work done by friction I just have ma basically, I don't have d. Would I use the distance parallel to the road surface, or just the horizontal component? Seems like I should use the distance parallel to the road surface, even though I am using the vertical component for h. Because it's in the friction term and so it has to do with the distance over which the friction acts, which is parallel to the road surface.
.5v_i^2 = [mu]g[cos angle of elevation]d + .5v_f^2+gh

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mu=0.3265
h=0.4633m
v_i=11.176m/s
v_f=?
d=5.8m
h=0.4633m
g=9.81
angle of elevation=4.574 degrees
cos angle of elevation=.99681517

62.451488 = .5v_f^2 + 23.062973
78.779 = v_f^2
v_f=8.876 m/s
19.856mph

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Same event on flat surface: V_f^2 = 11.176^2 + 2 (-3.2) (5.8)
v_f = 9.37m/s = 20.96 mph
so the hill slows the car down more than the loss of F_n impairs the braking.

a=2.2m/s / change in time
d=-.5at^2+v_it
5.8=-.5 (2.2/t) t^2 + 11.176t
5.8=10.076t
0.5756 seconds to hitting deer when going uphill, compared to 0.56 seconds to hitting deer on level ground

Heidi Henkel said:
78.779 = v_f^2
v_f=8.876 m/s
19.856mph
Haven't checked it in detail, but it looks right.
You can get h a little more simply. You are given that it is an 8% grade, which means the rise is 0.08 d.
Heidi Henkel said:
so the hill slows the car down more than the loss of F_n impairs the braking.
Not sure how you deduce that. The braking gives a deceleration ##\mu g##, or about 0.3 g, whereas the slope gives only 8% of g.

h is 0.08 times the horizontal, not 0.08 times the hypotenuse. So it is not 0.08d. d is the hypotenuse. You have to find the horizontal component of d.

I deduce my conclusion by calculating V_f for a slope and then for a flat, all variables the same except the slope. I wrote it on this forum, above. Did I do it wrong?

Heidi Henkel said:
h is 0.08 times the horizontal, not 0.08 times the hypotenuse. So it is not 0.08d. d is the hypotenuse. You have to find the horizontal component of d.
I believe that slope grades on roads are quoted as height/slope distance, i.e. the sine of the angle. On such a shallow grade it will make hardly any difference anyway.
Heidi Henkel said:
I deduce my conclusion by calculating V_f for a slope and then for a flat, all variables the same except the slope. I wrote it on this forum, above. Did I do it wrong?
The square root process to find velocity might mislead you here. To be certain, you need to compare the reduction in speed by braking only with the reduction in speed by coasting up the hill. But it is pretty clear that a braking coefficient of .32 is going to provide about four times the deceleration of an 8% grade.

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