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Homework Help: Circular Motion Probelm

  1. Mar 7, 2010 #1
    1. The problem statement, all variables and given/known data

    A curve of radius 65m is banked for a design speed of 105 km/h. If the coefficient of static friction is 0.34 (wet pavement), at what range of speeds can a car safely make the curve?

    I am not sure how can I get range of velocities at which the car can travel without skidding

    2. Relevant equations

    F = mA
    F = ukFN
    A = (v^2) / R

    3. The attempt at a solution

    uk x FN = mA
    uk x mg =m(v^2)/r
    v = sqrt [uk x r x g]
    v = sqrt[ 0.34 x 65 x 9.8] = 14.72 m/s -> 53 km/hr
     
  2. jcsd
  3. Mar 7, 2010 #2

    lightgrav

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    the curve is "banked" at some angle above the horizontal;
    draw your car with velocity INTO the page, so all the Force vectors ,
    and the acceleration vector, are diplayed ON the paper with angles showing.
    at the "design speed", mg + F_N = ma_c , even with no friction.

    if a car goes too slow, friction is needed ... also if a car goes too fast.
     
  4. Mar 7, 2010 #3
    This is what I found out after doing the diagram:
    *theta = angle at which the curve is banked.

    F_N = -ma_c / cos ("theta" +90)
    so I replaced into:

    mg + sin ("theta" +90) x F_N = ma_c

    mg + sin ("theta" +90) x ( (-ma_c) / (cos("theta" +90) ) = ma_c

    (-9.8) + tan ("theta"+90) x ( v^2 / r ) = v^2 / r

    Theta = 29.76181434

    Is this correct? if so, I understand this is the angle at which the car can travel at v = 29.166 m/s without friction needed, but I still don't know how to find the range of speeds.
     
  5. Mar 7, 2010 #4

    lightgrav

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    Homework Helper

    mg (down) + F_N(diagonal) = ma_c (horizontal) ... right triangle .
    F_N is the same angle from vertical , that the road is from horizontal .
    (your v^2/r is 13m/s^2 , right? ma_c this is bigger than gravity!)

    If a car is trying to creep around this curve, it will slide down the steep slope.
    what direction does friction push this car? ... how hard?
     
  6. Mar 7, 2010 #5
    Yes, my v^2/r is 13.08 m/s^2.
    If the car slides down the slope friction will act in the opposite direction the car is sliding, so friction acts up the slope.
    Force of friction = F_N x uk.
    Am I right?
     
    Last edited: Mar 7, 2010
  7. Mar 7, 2010 #6
    I tried again starting from zero, and this time I set the sum of the forces in the Y direction to zero, but now I get a negative angle.
    So:
    F_N = -ma_c / cos (90 + "theta")
    Replaced into :
    sin (90 + "theta") * F_N = mg
    and I got
    "theta" = -126.825869
     
  8. Mar 8, 2010 #7
    Hello demente182!

    I watch these forums to discourage students from cheating. Please solve the problem on your own.

    Ms. Marsh
     
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