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Car on banked turn - net force

  1. Sep 24, 2005 #1

    Suppose there is no friction between the car and the road -> if car is on a banked turn , the normal force ( which is always perpendicular to the road's surface ) is no longer vertical . The normal force now has a horizontal component , and this component can act as the centripetal force on the car

    But doesn't gravity force have a component that is of equal size , but in opposite direction than normal force ? In which case net force pulling on car would be zero and thus couldn't act as centripetal force ?
    So how can net force equal horizontal component of normal force ?

    Also , what must be angles between the components of normal force and how are they related to the banking angle between the road and the horizontal ?

    thank you
  2. jcsd
  3. Sep 24, 2005 #2

    Andrew Mason

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    Gravity is equal to the normal force if the road is horizontal. If you bank the road, the component of gravity that is perpendicular to the road surface is reduced but gravity now has a component that is parallel to the surface (ie pulling it sideways). The horizontal centripetal force has a component that is perpendicular to the road and a component that is parallel to the road. If there is no parallel or perpendicular motion, the parallel forces must sum to 0 and the perpendicular forces must sum to 0.

    Resolve all forces into components parallel to and perpendicular to the surface of the road. The perpendicular components are: [itex]\vec{N}[/itex] being the Normal force (pointing upward), [itex]mgsin\theta[/itex] and [itex]cos\theta mv^2/R[/itex] where [itex]\theta[/itex] is the angle of the road surface to the vertical. The parallel components are [itex]mgcos\theta[/itex] and [itex]sin\theta mv^2/R[/itex]

    Since there is no motion in either the parallel or perpendicular direction. So:

    (1)[itex]N - mgsin\theta - cos\theta mv^2/R = 0[/itex] and

    (2)[itex]mgcos\theta - sin\theta mv^2/R = 0[/itex]

  4. Sep 25, 2005 #3


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    There is a "Normal Force" by a surface because
    (after compressing a bit, if it doesn't break,)
    it reacts to keep the object from sinking into it.

    in a jump, while you straighten your legs,
    to accelerate your center-of-mass -
    you think the "Normal Force" is equal mg ?
    If so, how would anything get vertical speed?
    Standing up needs N>mg ; tossing a pencil ...

    In the case of a banked road, the Surface Force
    is "convinced" to become stronger by curving the road
    as well as banking it. The road then must push inward
    so the car doesn't go in a straight line thru the road.

    It's traditional to start with level (horizontal) turns ...
    AndrewM presumed yours is level automatically.
    Here it is clearest to use axes horizontal & vertical.

    The vertical Forces are gravity (mg) and enough
    surface Force to so the car doesn't sink in (Ncos theta).
    The horizontal Force is only Nsin(theta).

    The driver wants Nsin(theta) to provide mv^2/R .
    But with no friction, the road must push perp to its Area.

    (A.M. has an extra "N")
  5. Sep 25, 2005 #4
    Uh , could you pretend I'm mildly retarded and please break it into little steps ?! I'm sooo confused I don't even know exactly what to ask ... but I will try

    So somehow forces are alligned in such way that net force becomes centripetal force ?!

    You mean forces perpendicular to road - wouldn't that be a normal force and a component of gravity force ?

    But if normal force is perpendicular to surface area and at the same time it is greater than component of the opposite gravity force also perpendicular to surface area , then resulting vector should point upwards and perpendicular to surface ?! In that case car would start moving to the side and also horizontally

    Theta being angle of the road surface to the vertical ?

    I got it just the other way around . For component of gravity force parallel to the road I got sin(angle)*m*g and component perpendicular to road I got cos(angle)*m*g .

    In any case , notice that components of both centripetal force and gravity force parallel to the road have the same direction so I don't see how they could cancel each other out

    Dont's the parallel components of gravity force and centripetal force have opposite directions ?
  6. Sep 25, 2005 #5
    nevermind my initial questions . I forgot that since N is greater in magnitude its vertical component also has component perpendicular to the roaqd , but it's not of the same size as normal force

    My question now would instead be what must an angle between banking road and horizontal be in order for all this to work ( assuming there is no friction on the road ) ?
  7. Sep 25, 2005 #6

    Doc Al

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    There are only two forces acting on the car: the normal force and the weight. Set up the equations of motion, noting that the car accelerates centripetally:
    (1) Vertical forces must add to zero
    (2) Horizontal forces must provide the centripetal force

    Combine these equations to solve for the angle of banking for a given speed and curve radius.
  8. Sep 25, 2005 #7


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    Since you call yourself "beginner", it will be a quite educative experience to try the following approach after you've followed Doc Al's approach and solved the problem in that manner:
    [itex]\vec{F}=m\vec{a}[/itex] is valid however you choose to decompose your vectors.
    Thus, instead of decomposing forces and accelerations in the vertical and horizontal directions, try to decompose them in the directions normal to and tangential to the banked turn.
    Your answers will, of course, agree (at least if you do it correctly).
  9. Sep 25, 2005 #8
    got it

    I hope you can help me with just one more question

    It seems that on banked turn (no friction) the resulting force doesn't depend on the speed of the car

    F=m*g*tan (angle)

    that would suggest if car goes into banked turn with speed less than necessary for centripetal force at that angle, car will be thrown in a direction of force (meaning it won't succed in making a turn) ?

    thank you very much for all your help
  10. Sep 25, 2005 #9


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    if a car enters a steeply-banked road way too slow,
    then it can slide down the bank ... I've seen cars
    on icy roads do this if drivers are "too cautious".

    If the driver is too aggressive, the car slips the bank
    and they end up on the outside of the curve.
    did you ever play with "slot-car" racers?

    It is certainly BEST in statics and dynamics questions
    to begin with a Free-Body-Diagram, then
    add the Force vectors tail-to-tip so that the
    resultant points in the direction of the acceleration.
    |- - - ->/
    | . . . /
    | . . /
    |. . /
    |. /
    That way, you can more wisely choose axis directions
    (trying to do as few vector components as possible)
    Last edited: Sep 25, 2005
  11. Sep 26, 2005 #10
    Though I'm a bit confused why on banked turn (no friction) the resulting force doesn't depend on the speed of the car . I mean , common sense would suggest otherwise , wouldn't it ?
  12. Sep 26, 2005 #11

    Andrew Mason

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    I am not sure why you think that it is independent of speed. The centripetal force that is required to make the turn has to be supplied by the road and gravity:

    (1) [tex]Nsin\theta = mv^2/r[/tex]

    The vertical component of the normal force is supplied by the road to counter gravity:

    (2)[tex]Ncos\theta = mg[/tex]

    Substituting 2 into 1:

    [tex]mg\frac{sin\theta}{cos\theta} = mv^2/r[/tex]

    [tex]v = \sqrt{rgtan\theta}[/tex]

    Any speed higher than this, the car slips up the bank. Any speed lower than this, the car slides down the bank.

  13. Sep 26, 2005 #12

    Doc Al

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    Since the net force is a centripetal force, it certainly depends on the speed. ([itex]F_{net} = m v^2/r[/itex]) The greater the speed, the greater the required centripetal force. (And, as lightgrav and Andrew Mason explain, the banking angle is tuned to a particular speed. Change the speed and you'll need a new angle.)
  14. Sep 26, 2005 #13
    What I was trying to say is that if at particular angle of a banked curve car has certain mass m , and radius is R then net force (centripetal force ) will have the same magnitude (say 10 N) , no matter how fast car is going .
    So whether car has speed of 100 km/hour or 200 km/hour , centripetal force will be the same in both cases ( 10 N ) . Only difference is that at say 100 km/hour centripetal force will have just the right magnitude to keep the car on the road , while at speed 200 km/hour car will go off

    So in short , I was implying that the magnitude of centripetal force doesn't depend on car's velocity , but on its mass , radius and angle of a banked curve . But in order for car to be able to drive trought that particular curve car has to have just the right speed

    or am I wrong ?
  15. Sep 27, 2005 #14
    Not to be pushy , but could you at least tell me if , again , I'm totally off with my theory made in my las post ?
  16. Sep 27, 2005 #15

    Doc Al

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    I would say that you your statements were not totally correct. The bank angle is set so that a car can traverse the curve with zero friction and without sliding at a particular speed. If you go faster, the net force will change: the centripetal component will increase and, in addition, there will be a component of force accelerating you up the incline. (But I'll have to give this some more thought.)
  17. Sep 27, 2005 #16

    Andrew Mason

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    The centripetal force is supplied by the horizontal component of the normal force. Centripetal force is always proportional to speed. I think you are overlooking the fact that the normal force supplied by the road IS speed dependent. This can be best seen if the bank angle is 90 degrees (vertical road surface). In such case, the normal force is equal to the centripetal force as there is no vertical component. As speed increases, the normal force / centripetal force increases (as the square of the speed).

    With an angle less than 90 degrees, the normal force has an upward vertical component as well so there is only one speed at which that upward component equals the force of gravity.

  18. Sep 27, 2005 #17
    But if you look at the formula for centripetal force at particular angle of banked road

    F = m*g*tan(angle)

    you will noticed magnitude of force depends only on size of an angle and mass of an object

    If force was dependant on speed at certain angles then I assume it would be possible to drive trough the banked road with any speed since centripetal force would increase accordingly ?
  19. Sep 27, 2005 #18

    Doc Al

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    That's only true for the speed at which the banking was designed for. (The speed that satisifies the relationship [itex]\tan \theta = v^2/(rg)[/itex].)

    But if you go too fast you would not be able to negotiate the turn due to sliding off the road.
  20. Sep 27, 2005 #19
    I understand this from the formula's point of view , but somehow I can't get it into my head rationally . I mean

    m*g*tan(angle) = m*v^2/r

    Left sides equal right side exactly when ....

    I don't see anything on the right side of equation that would predict the magnitude of horizontal component of normal force when car has certain speed

    prob am not making much sense

    exactly what kind of force does slide the car off the road if speed is too great
  21. Sep 27, 2005 #20

    Andrew Mason

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    But this is not the formula for centripetal force. Where did you get this?

    The centripetal force is given by [itex]f = mv^2/r[/itex]. It is supplied by the horizontal component of the normal force. The normal force is not supplied by gravity alone. The car pushes on the road (trying to go in a straight line) and the road pushes back. If the car is making the turn and not going off the road, the normal force will 'self adjust' so that the horizontal component is equal to the centripetal force. If there is no friction, this occurs only at one speed for a given angle.

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