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Car on a banked roadway

  1. Oct 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider a wet banked roadway, where there is a coefficient of static friction of 0.30 and a coefficient of kinetic friction of 0.25 between the tires and the roadway. The radius of the curve is R=50m

    A: If the banking angle is β=25∘, what is the maximum speed the automobile can have before sliding up the banking? (Answer is in the low 20s)

    B: What is the minimum speed the automobile can have before sliding down the banking?
    (Answer is between 5-10 m/s)

    2. Relevant equations
    Newton's first law:
    ƩF = 0 if a body is in equilibrium (at rest/constant velocity). This is per-axis.

    Newton's second law (if accelerating)
    ƩF = m*a

    Magnitude of friction force:
    f_fric (static or kinetic) = u*n where u is the coefficient of friction (either u_s or u_n) and n is the normal force.

    Forces in uniform circular motion:

    [itex]a_rad = \frac{v^2}{R} = \frac{4pi^2R}{T^2}[/itex]


    3. The attempt at a solution

    I'm reviewing this problem for an exam. I remember it was a beast. I'm having trouble figuring out how to do a free body diagram efficiently. I have dysgraphia and dyslexia - I've attached my work but I'm worried it might not be legible.

    So essentially we're looking at a car on a wedge but our acceleration formula is different. The forces in the X direction include the centripetal force as well as the coefficient of static friction (the car is not moving in the X direction) times the amount of weight pressed against the wedge in the X direction.

    The sum of the forces in the Y direction should be the coefficient of static friction (no movement on the Y axis) times the amount of weight still heading downwards.

    For my free body diagram, I drew a right hand triangle with the hypotenuse facing right and the 90 degree angle at the origin. The car is a dot on this triangle. The angle β=25° is represented by the other corner of the triangle along the X axis.

    Starting at the bottom and going counterclockwise.

    The force mg is going straight down.

    The force n is perpendicular to the hypotenuse of the triangle.

    In part A the force of friction opposes the direction of movement, which is 'up' the banking angle. This makes the free body diagram of part A have the force of friction directed down the slope of the triangle.

    Using this free body diagram, I am able to see that the Y component of weight is u_s * sin(25) * m. The x component should be u_s*cos(25)*m. Intuition says the Y component > the x component.

    Checking this:
    .30*sin(25)*1 = .1268
    .30*cos(25)*1 = .2719

    I've broken it here. However, my free body diagram shows that this should not be the case.

    So I'm going to correct this - but I want to know why I have it wrong!

    So the sum of the forces in the X direction is:

    [itex]ƩF_x = m\frac{V^2}{R} + m u_s sin(25) = 0[/itex] (Equal to zero since the system is in equilibrium)

    This is equal to
    [itex]ƩF_x = m\frac{V^2}{50} + m .3 sin(25) = 0[/itex]
    [itex]ƩF_x = \frac{V^2}{50} + .3 sin(25) = 0[/itex] (Mass factors out)

    The sum of the forces in the Y direction is:
    [itex]ƩF_y = m g + .3 m cos(25) = 0[/itex]
    [itex]ƩF_y = g + .3 cos(25) = 0[/itex] (Mass factors out)

    Equating these and solving for V, I get:

    [itex]ƩF = \frac{V^2}{50} + .3 sin(25) = g + .3 cos(25) [/itex]
    Solving this for V gets me very, very close to the answer for part A, but after rounding it is higher than the answer is. I tried using 9.8 and 9.81 for my values of g, but it isn't quite correct.

    I'm going to continue because as I was typing this I realized I was very nearly getting it!

    For part B, we're trying to avoid sliding down the embankment. Looking at part A's sums of the forces, for X we have mass times the centripetal acceleration plus the force resulting from gravity working in our favor. In part B, gravity is working against us. So...

    For part A:
    [itex]ƩF_x = m \frac{V^2}{50} + m .3 sin(25) = 0[/itex]
    [itex]ƩF_y = m g + .3 m cos(25) = 0[/itex]

    For part B, gravity is assisting the motion we do not want and is therefore negative/working against our positive forces (if someone could clarify this concept I would really appreciate it!)

    [itex]ƩF_x = m \frac{V^2}{50} - m .3 sin(25) = 0[/itex]
    [itex]ƩF_y = m g - .3 m cos(25) = 0[/itex]

    Equating, dividing through by m:

    [itex]ƩF = \frac{V^2}{50} - .3 sin(25)g = g - .3 m cos(25)[/itex]

    Solving for V we get...the same answer as above? Hmmm. Well, .3*sin(25) and .3*cos(25) are very small numbers. So changing the signs doesn't really do much to gravity, which is huge in comparison. I am missing a multiplier.

    ...

    Or I entered it into my calculator (useful when dealing with dysgraphia+dyslexia, perilous none-the-less) wrong and neglected gravity. Woe unto me.
    Part A corrected:
    [itex]ƩF = \frac{V^2}{50} + .3 sin(25) g = g + .3 cos(25) g [/itex]
    Solving for V gives us the wrong value. Flipping trigonometric functions:
    [itex]ƩF = \frac{V^2}{50} + .3 cos(25) g = g + .3 sin(25)*g [/itex]
    Gets VERY CLOSE to the answer!
    ...switching to the coefficient of kinetic friction on the left side of the equation, the 'x' side, gets us to the answer.
    [itex]ƩF = \frac{V^2}{50} + .25 cos(25) g = g + .3 sin(25) g [/itex]

    Alright. Let's look at part B again. We're apparently still moving on the X axis.
    Part B:
    [itex]ƩF = \frac{V^2}{50} - .25 sin(25)g = g - .3 m cos(25) g[/itex]
    Neither solving that nor this:
    [itex]ƩF = \frac{V^2}{50} - .25 cos(25)g = g - .3 m sin(25) g[/itex]
    for V yields the correct answer.

    I've meandered enough and come VERY CLOSE to the answer, I think. What am I missing in part B?

    Thanks!
     

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    Last edited: Oct 28, 2013
  2. jcsd
  3. Oct 28, 2013 #2

    haruspex

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    One of the angles you have marked as 25 degrees isn't.
     
  4. Oct 28, 2013 #3

    haruspex

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    The centripetal force is not an applied force to be 'included' with other supplied forces. It is that resultant force which is required if the object is to achieve the arc of motion. In the ƩF=ma form, it is part of the ma.
    That's not a good way to think of it. Weight acts vertically, so its component in the X direction would be 0. Think of it in terms of the normal force.
    The Y component of weight is mg. u_s * sin(25) * m is not even a force.
     
  5. Oct 29, 2013 #4
    Ah!

    I had a longer post reworking the problem the wrong way! I started digging through Google and found this site: http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/carbank.html

    Their diagram makes things make more sense. However, I'm still not crystal clear on this and have spent an additional 45 minutes on it. I plan on asking my instructor if I can before the exam.

    I'm still confused by the trig, but here's my understanding of the problem now:

    The forces in the X direction F_net are at equilibrium with m*v^2/r.

    There are two forces acting at equilibrium with m*v^2/r. First is the normal force, which is opposite mg. Then there is the friction force opposing the direction of motion.


    I don't understand the free body diagram. Specifically why the normal force has the value theta coming off of the head. I might need to see this drawn out so that I can get my trigonometric functions right.


    On the Y axis, there are three forces acting. The normal force opposing mg. The friction force, and the weight. The weight and friction forces are both negative - they act at an angle on the negative Y axis if the normal force is considered in the positive Y axis (Normal force is in quadrant 2. Weight is straight down the Y axis. Friction is pointing into quadrant 3).


    For minimum speed I would wager we flip a few signs. Specifically, on the X axis the friction force has reversed directions and is now negative. On the Y axis, the friction force has flipped signs as well because the direction of motion has been rotated 180 degrees.


    I think I almost have this! :)
     
  6. Oct 29, 2013 #5

    haruspex

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    No, Fnet is not 'in equilibrium' with mv2/r. It is mv2/r.
    In uniform circular motion, the v2/r is the a in ƩF = ma.
    Not sure what you mean by the normal force being opposite mg. The normal force and the frictional force act together to balance mg.
    It is showing that N is at theta to the vertical.
    That's all basically correct.l
     
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