a car is to make a turn withg a radius of 60m at a speed of 27m/s at what angle should the road banked for the

Answer 1

Let the banking angle be '$'.
The reaction force will be perpendicular to the banking surface.
Let the reaction force be 'R'.
Let 'm' be the mass of the car.
velocity of car (v)=27 mt/s
radius(r)=60 mt.

The centrifugal force will be in horizontal direction. And this force will be balanced by a component of the reaction force 'R'.
R sin$= mv^2/r ---------------------------(1)

The weight of the car will be vertically downward. This weight will be balanced by Rcos$.
Rcos$=mg ----------------(2)

Now dividing eqn(1) by eqn(2)

Rsin$/Rcos$=mv^2/r /mg
=>tan$=v^2/rg = 27 x 27/60 x 9.8 = 729/588 = 1.239

$= tan(inverse)(1.239)

solve above tan(inverse) function. and find the value of the banking angle

Answer 2

Tan theta=v^2/ rg

so tan theta=27*27/10*60

Answer 3

a = v²/r = 27²/60 = 12.15m/s²
A = arctan(12.15/9.80662) = 51°