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A car stopping on a slippery surface on an uphill

  1. Jan 8, 2016 #1
    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.
     
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  3. Jan 8, 2016 #2
    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.
     
  4. Jan 8, 2016 #3

    PeroK

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    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.
     
  5. Jan 8, 2016 #4
    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.
     
  6. Jan 8, 2016 #5

    PeroK

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    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.
     
  7. Jan 8, 2016 #6
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant 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?

    3. The attempt at a solution I need to make the equations fit together, first
     
    Last edited by a moderator: Jan 8, 2016
  8. Jan 8, 2016 #7

    berkeman

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    Please do not cross-post -- it is against the PF rules. Instead, click on "Report" on your post, and ask the Mentors to move the thread. I will merge your two threads now.
     
  9. Jan 8, 2016 #8

    haruspex

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    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.
     
  10. Jan 8, 2016 #9
    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.
     
  11. Jan 8, 2016 #10
    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.
     
  12. Jan 8, 2016 #11

    haruspex

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    What given information have you not used?
     
  13. Jan 8, 2016 #12
    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.
     
  14. Jan 8, 2016 #13

    haruspex

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    There's another distance mentioned.
     
  15. Jan 8, 2016 #14
    19.5 m to stop on level ground if starting at 25 mph. There must be a way to get mu from that.
     
  16. Jan 8, 2016 #15

    haruspex

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    Yes indeed.
     
  17. Jan 10, 2016 #16
    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
     
    Last edited: Jan 10, 2016
  18. Jan 10, 2016 #17

    haruspex

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    Check your signs.
    Anyway, it would be more direct to use a SUVAT equation that does not involve time.
     
    Last edited: Jan 10, 2016
  19. Jan 10, 2016 #18
    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.
     
  20. Jan 10, 2016 #19
    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...
     
  21. Jan 10, 2016 #20
    I am curious what equation you would use to find mu without finding time.
     
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