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Circular motion on a banked curve

  1. Jan 9, 2013 #1
    how can rcosθ in a banked road be equal to mg; since r is equal to normal reaction which is equal to mgcosθ. rcos is even smaller than r.
    then how can mg=rcos when banking in curved road?
  2. jcsd
  3. Jan 9, 2013 #2


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    Welcome to PF regan1,

    mg cannot be equal to rcos(theta) simply on dimensional grounds. The former is a force, and the latter is a length. Could you post the equations that you are having trouble with more carefully and in greater detail?

    For a banked curve, if you look at the road in cross-section, it is essentially like an inclined plane, with the "downhill" direction being the direction towards the centre of the circle. In the absence of friction, the only two forces that act on the car are gravity, and the normal force from contact with the road, which can contribute to the centripetal force. This should be helpful to you as a starting point for working out the force balance equations.
  4. Jan 9, 2013 #3


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    As I read it, regan1 has used an infelicitous choice of variable names and used "r" to denote the normal reaction force that is supporting the car against gravity.

    When a physicist or mathematician sees "r cos θ" the automatic impulse is to read the "r" as radius, however that does not appear to be the intended reading in this case.

    mg > mg cos θ... yes, that's true.
    mg cos θ = r... no, that's false.
    The correct version is mg = r cos θ

    That seems to be the problem; a multiplication where division was called for.
  5. Jan 9, 2013 #4


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    If you are taking a curve on a banked road, the car is accelerating towards the center of the curve. You seen to have forgotten about that in your equations.

    This means that the reaction force r can be bigger than the weight of the car.
  6. Jan 10, 2013 #5
    Is not R the reaction force and how can Reaction force be greater than the weight of the car ?
    In my opinion the maximum value that R can have is equal to the weight of the car. But since it is a slanted surface it is always less than mg
  7. Jan 10, 2013 #6


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    By "weight of the car", I assume that you mean the mass of the car times the force of gravity. (If, instead, you mean the increased apparent weight of the car due to centrifugal force plus the force of gravity then clearly the weight of the car is exactly equal to the reaction force R).

    AlephZero has it right. The force of the road on the car must be able to both support the car against gravity and accelerate the car in its circular path -- it must be equal to the vector sum of those two components. That means that its magnitude is greater than either one of those components individually. It must be both greater than mg and greater than mv^2/radius.

    I'm not clear on why you are calling this a "reaction force".
  8. Jan 11, 2013 #7
    while banking the car in a curved road from where does the car get the force to balance it's weight and force to balance it's centrifugal force. As far as i know that force is given by the reaction force of the car's weight itself. and when you divide the reaction force in to it's component forces- one force is directed toward the center which provide the centripetal force and the next balances the weight. since the car is in equilibrium the remaining component must balance the weight of the car. so
    R cos θ=mg
    but my point if R is smaller than mg itself how can this value be equal in magnitude.

    i am not clear how you said R > mg can you please mathematically demonstrate it

    i could not get the point how road give so much force on the car greater than the car gives on the road ( max force car can give on the road is equal to mg, isn't it? ) . can you explain it further more?
    Last edited: Jan 11, 2013
  9. Jan 11, 2013 #8

    Doc Al

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    It's a 'reaction' force in the sense that the force that the road exerts on the car must be equal and opposite to the force that the car exerts on the road.

    That force would equal the weight of the car if the car were in equilibrium. But it's not. Acceleration changes things.

    Imagine yourself on an elevator. When the elevator is stationary (or moving with constant speed) the 'reaction' force of the floor on you will equal your weight. But if the elevator accelerates upward, that force will be greater than your weight.

    The same thing is going on here, but it's a bit obscured by the fact that the motion is circular.
  10. Jan 11, 2013 #9


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    No. It's not.
  11. Jan 11, 2013 #10
    if not, doesn't it violate newton's third law of motion. to be the upward pulling force of the road greater there must be other external force.
  12. Jan 11, 2013 #11

    Doc Al

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    Sounds like you are thinking that the force of the road on the car is the reaction force to the car's weight. That's not true. The car's weight is the gravitational force exerted by the earth on the car; the 'reaction' force to that is the gravitational force exerted by the car on the earth.

    The correct Newton's 3rd law pair to the force of the road on the car is the equal and opposite force that the car exerts on the road. That force depends on various things, not just the weight of the car.
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