give reason curved roads are banked
Answers
Answered by
1
Roads r banked so that vehicle can feel centripetal force which helps vehicle to move on mountainous roads
kokan6515:
becoz most of banked roads r found on hilly areas
Answered by
2
consider a vehicle of mass 'm' moving with a velocity 'v' and making a turn on circular path of radius 'r'. the external forces acting on the vehicle are :-
1. weight=mg
2. normal contact force= N
3. friction = f
we need to keep in mind that,
> N is vertically upwards
> f is only horizontal force acting towards the centre
> f is self adjustable
> tyres have a tendency to skid outwards and therefore, to avoid skidding friction force acts towards centre.
thus for a safe turn,
or, f=
since there is a limit to the magnitude of frictional force, it cannot exceed μN where μ is coefficient of friction.
so, for vertical equilibrium,
N= mg
so that,
f μ mg
then, for a safe turn,
m μmg
or, μ
surface of road makes an angle θ with horizontal throughout the turn. N makes an angleθ with vertical. At the correct velocity the horizontal component of N is sufficient to produce the acceleration toward the centre and the self adjustable frictional force keeps the value zero.
Applying newtons second law along the radius and first law in the vertical direction,
N sinθ=
⇒ N cosθ = mg
gives,
tanθ=
θ depends upon velocity of the vehicle as well as on the radius of the turn.
roads are banked for the average expected velocity of the vehicle.
HOPE IT HELPED YOU ENOUGH. NONE OF THIS WAS COPIED. YOU CAN EVEN CHECK ON NET TO FIND OUT. ITS MY OWN ANSWER. PLEASE MARK AS BRAINLIEST. PLEASE PLEASE PLEASE....
1. weight=mg
2. normal contact force= N
3. friction = f
we need to keep in mind that,
> N is vertically upwards
> f is only horizontal force acting towards the centre
> f is self adjustable
> tyres have a tendency to skid outwards and therefore, to avoid skidding friction force acts towards centre.
thus for a safe turn,
or, f=
since there is a limit to the magnitude of frictional force, it cannot exceed μN where μ is coefficient of friction.
so, for vertical equilibrium,
N= mg
so that,
f μ mg
then, for a safe turn,
m μmg
or, μ
surface of road makes an angle θ with horizontal throughout the turn. N makes an angleθ with vertical. At the correct velocity the horizontal component of N is sufficient to produce the acceleration toward the centre and the self adjustable frictional force keeps the value zero.
Applying newtons second law along the radius and first law in the vertical direction,
N sinθ=
⇒ N cosθ = mg
gives,
tanθ=
θ depends upon velocity of the vehicle as well as on the radius of the turn.
roads are banked for the average expected velocity of the vehicle.
HOPE IT HELPED YOU ENOUGH. NONE OF THIS WAS COPIED. YOU CAN EVEN CHECK ON NET TO FIND OUT. ITS MY OWN ANSWER. PLEASE MARK AS BRAINLIEST. PLEASE PLEASE PLEASE....
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