What causes a car to skid outwards on a circular turn?

[clavin]

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 let's say i have two black placed one over the other
and the ground and the surface of the blacks all have friction
now let's say i apply some force on the block(the force isn't 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 let's 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

 

[rock.freak667]

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

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.

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?

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

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 let's 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)

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.