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# A turn of radius 20m is banked for vehicles going at a speed of 36km/hr.If the co-efficient of static friction between the road and the tyre is 0.4,what are the possible speeds of the vehicle so that it niether slips down nor skids up? (hc verma pgno:115)19th problem

anurag v.
38 Points
10 years ago

animesh shrivastav
14 Points
10 years ago

N=mv2/r sinθ + mgcos?

mgsinθ ~ mv2/r cosθ = μN

(Difference symbol because if centrifugal greater then friction stops skid outward

where as if component of weight greater then friction stops slip inwards)

Solve above 2 equation 2 get:

V2/gr=(tan?θ-μ)/(1+μtanθ) or (tanθ+μ)/(1-μtanθ)

Put value of g,r,μ,tanθ=362/rg

v=4.082m/s or 15m/s

multiply by 18/5 to get kph

v=14.7kph or 54 kph