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Some mechanics question

  1. Mar 31, 2010 #1
    well my first qs is
    lets say i have a pulley and rope system
    and two blocks each of mass m hanging from either end of the rope which goes over a pulley
    the pulley has a mass of 1 kg, frictionless and rope is massless and frictionless
    now if i were to analyze the force on the pulley it would be equal to 2mg
    now what i want to know is how is the system applying this 2mg force on the pulley
    cuz for all the points on the rope that touch the pulley have tension tangential to the pulley
    physically i can see that it would be 2mg but how do u explain it through physics

    second lets say i have two black placed one over the other
    and the ground and the surface of the blacks all have friction
    now lets say i apply some force on the block(the force isnt high to exceed the limiting friction from either the ground or the block above) touching the ground
    so how to judge why and which of the two, ground and the upper block,
    will apply the friction force first or is it that both will apply the friction simultaneously

    when a car goes on a horizontal circular turn with velocity v , the friction provides the centripetal force. now will a friction also act in the direction opposite to v? if not why?

    and lastly
    why on a circular turn do the tyre ski outwards
    well i mean how come there is relative motion radially which gives rise the friction force in radial direction
    and lets say the friction is not high enough to prevent the skidding
    then what would be the force or velocity with which the car will skid outwards(i.e. if it is possible to calculate)

    i wud be thankful if u can answer even one of the qs
    cya
     
  2. jcsd
  3. Mar 31, 2010 #2

    rock.freak667

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    Homework Helper


    I don't understand what you want to know. Unless the rope is stretching, the tension is equal to the weight mg. That is how the forces acting are mg and mg on opposite sides of the pulley.

    Well if the tires are not slipping, there will be a tangential acceleration which is affected by friction.

    The motion of the car will be to move outwards, the friction forces will act opposite in this direction. Since the car does not move outwards, it can only mean that the frictional force is providing enough force to keep the car in the circular motion.


    The formula would be the same mv2/r. It is just that there might not be enough 'r' ( or well road) for the car to maintain the circular turn.
     
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