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    Why is it necessary to never allow brakes to lock up in order to successfully pull out of a skid on the ice?

    © BrainMass Inc. brainmass.com December 24, 2021, 6:29 pm ad1c9bdddf
    https://brainmass.com/physics/circular-motion/circular-motion-locked-brakes-117198

    SOLUTION This solution is FREE courtesy of BrainMass!

    (Please see the attached file).

    To understand the given situation, let us first understand the motion of a vehicle on a curved track. Fig.1 shows a vehicle of mass m traveling with a speed v on a circular track of mean radius R. As we know, for any object to move on a circular track, a force is required to act on it, directed towards the centre of the circular track (centripetal force). The magnitude of this force must be at least equal to mv^2/R. From where does this force come for motion of the vehicle on the circular track? Answer is - from friction between the tires and the track as explained hereunder.

    To understand better, let us look at fig.2, which gives a view of the vehicle as seen from the back (fig.1 is a view from the top). As the vehicle is made to take a continuous turn on the circular track (by the driver), due to the inertia of direction, the vehicle has a tendency to continue to move on the straight line path. This tendency results in an inertial force acting on the vehicle in an outwards direction. As a result, a frictional force comes into play between the tires and the track acting inwards (towards the centre of the track) [friction always acts in a direction opposite to the direction in which the object tends to move]. It is this frictional force which provides the required centripetal force. Magnitude of the frictional force depends upon the value of the coefficient of friction between the tires and the track.

    Centripetal force = Frictional force = mv^2/R .........(1)

    For the vehicle to successfully move on the track, (1) must be satisfied.

    When the track is on snow, the coefficient of friction is very small and the frictional force available is negligible. Consequently, if we try to move faster in a circular path on snow, the radius R of the path has to increase in order to satisfy (1) (and vice versa) . This phenomena is known as skidding.

    Now, let us imagine a snow vehicle is skidding. What must the rider do to stop skidding? He must slow down by pressing on the brakes gradually. As v reduces, so does R in order to satisfy (1). The vehicle slows down in a spiral path of continuously reducing radius, to finally come to rest at the centre of the path.

    Now, imagine the brakes get jammed as they are pressed. The speed of the vehicle becomes zero suddenly (v=0). In order to satisfy (1) R must also become zero instantly. Thus, the vehicle moves instantly inwards, towards the centre (instead of the spiral path). Of course, it does not stop at the centre, but continues to move on due to inertia, as there is no significant opposing force to stop it (friction being negligible). Needless to say, this is a potentially dangerous situation in mountainous region as the vehicle may straight head for a ditch (or hit a rock with a force). Hence, the brakes must never get jammed.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 6:29 pm ad1c9bdddf>
    https://brainmass.com/physics/circular-motion/circular-motion-locked-brakes-117198

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