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Banked turn

  1. Mar 10, 2005 #1
    I need some help, I am a relative noob at physics. I need this to get credit on test corrections to boost my test grade. Here's the problem,

    A 1300 kg car goes around a turn with radius 50 m, at a speed of 75 km/h. The coefficient of static friction is .45, (a) what is the minimum angle of the bank needed to keep the car in the turn? (b) If the car changes tires, which increase the coefficient of static friction to .65, what will be the maximum velocity with which the car will be able to take the turn, assuming the same banked angle?

    This is probably easy for you guys. Any help would be appreciated, Thanks.
     
    Last edited: Mar 10, 2005
  2. jcsd
  3. Mar 10, 2005 #2

    arildno

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    1) The car moves in a horizontal circle; what is its (centripetal) acceleration?
    2) List the forces acting upon the car, and their directions.

    Relate 1) and 2) by Newton's 2.law of motion.
     
  4. Mar 10, 2005 #3
    1) It's centripetal acceleration is 20.83^2/50= 8.68 m/s^2
    2) The forces acting on it are gravity, the normal force (perpendicular to the incline), and friction (down the incline)

    so Fnx+Fsx=ma
    Fn*(Sin theta)+.45*Fn*Cos (theta)= 11284.72

    This is where I get stuck, I was taught to use another equation to solve for Fn, but I just can't get Fn. the equation I used was Fny+Fg+Fsy=0, so Fn*(Cos theta)+.45*Fn*(Sin theta)=-fg (=12740). After this I don't know what to do.
     
  5. Mar 10, 2005 #4

    arildno

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    1) Try to write this in VECTOR form
    2) Do NOT introduce numerical values; they just confuse you.
    Pay attention to the structure of the equation instead.
     
  6. Mar 10, 2005 #5
    I just tried to solve for (a) and I got 41.53 degrees, do you know if this is right?

    Thanks for all of the help, I have to go to school now. Thanks again.
     
    Last edited: Mar 10, 2005
  7. Mar 10, 2005 #6

    arildno

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    I haven't bothered to check.
    Here's one way of solving this properly:
    1) Centripetal acceleration: [tex]\vec{a}=-\frac{V^{2}}{R}\vec{i}_{r}[/tex]
    where V is the constant speed, R the radius, and [tex]\vec{i}_{r}[/tex] the outward radial vector in the horizontal [tex]\vec{i},\vec{j}[/tex]-plane.
    2) Forces:
    a)Weight: [tex]-Mg\vec{k}[/tex]
    (M being the mass, g the acceleration of gravity)
    b) Normal force: [tex]N\vec{n}[/tex]
    c) Frictional force: [tex]-F\vec{t}[/tex]
    where [tex]\vec{n}[/tex] is normal to the banked curve, and [tex]\vec{t}[/tex] the appropriate tangent vector. (F and N being magnitudes of forces)

    By geometry, we have:
    [tex]\vec{t}=\cos\theta\vec{i}_{r}+\sin\theta\vec{k}[/tex]
    [tex]\vec{n}=-\sin\theta\vec{i}_{r}+\cos\theta\vec{k}[/tex]

    Hence, Newton's 2.law of motion states:
    [tex]N\vec{n}-F\vec{t}-Mg\vec{k}=-M\frac{V^{2}}{R}\vec{i}_{r}[/tex]
    Taking the dot product of this equation with [tex]\vec{n}[/tex] yields:
    [tex]N-Mg\cos\theta=M\frac{V^{2}}{R}\sin\theta[/tex]
    Taking the dot product with [tex]\vec{t}[/tex] yields:
    [tex]-F-Mg\sin\theta=-M\frac{V^{2}}{R}\cos\theta[/tex]
    Or, F and N must satisfy for a purely horizontal motion:
    [tex]N=M(g\cos\theta+\frac{V^{2}}{R}\sin\theta)[/tex]
    [tex]F=M(\frac{V^{2}}{R}\cos\theta-g\sin\theta)[/tex]

    In addition, if we have achieved the critical value of static friction, [tex]F=\mu{N}[/tex]
    we can solve for the (minimum) angle:
    [tex]tan(\theta)=\frac{\frac{V^{2}}{R}-\mu{g}}{\mu\frac{V^{2}}{R}+g}[/tex]
    If [tex]\frac{V^{2}}{R}\leq\mu{g}[/tex] then the minimum angle is 0.

    Now, you can plug&chug numerical values into this to your heart's content.
    (Use a consistent set of units, though.)
     
    Last edited: Mar 10, 2005
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