Question

37. A vehicle of mass 15 kg moves with a retardation of 23 m/s2 around a curve o
radius 75 m. When the speed of the vehicle is 15m/s, the total force exerted by t
vehicle on the road is (g =10m/s?)

1) 1587N 2) 1650N 3) 687 N. 4) 1566N​

Answer 1

Question statement:

A turn of radius 20 m is banked for the vehicles going at a speed of 36 km/h. If the coefficient of static friction between the road and the tyre is 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up?

Solution:

Given speed of the vehicle v =36 km/hr = 36 ×  5/18  = 10 m/s  

Turning radius r = 20 m and static friction μ  s  =0.4  

The bend angle of the road θ = tan^ −1  (  rg v^2 ) = tan^ −1  0.5  

When the car travels at maximum speed so that it slips upwards and the component μR acts downwards.

So, R1  − mgcosθ − mv^2 / r  sinθ = 0    ----(A)

and μR1 + mgsinθ −  mv^2 / r cosθ = 0 -----(B)

Solving the equation we get,

v1 = √  rg tan θ - μ / 1 + μ tanθ

v1 = √ 20 x 10 (0.1 / 1.2) = 4.082 m/s = 14.7 Km/s

Thus the possible speed will be between 14.7 Km/s and 54 km/h