Question

derive an expression for banking of road with the help of diagram?

Answer 1

fully derive an expression baking of road

Answer 2

Banking Of Roads

Angle of Banking

Consider a vehicle of mass ‘m’ with moving speed ‘v’ on the banked road with radius ‘r’. Let ϴ be the angle of banking, with frictional force f acting between the road and the tyres of the vehicle.

Total upwards force = Total downward force

NcosΘ=mg+fsinΘ

Where,

NcosΘ : one of the components of normal reaction along the verticle axis

mg: weight of the vehicle acting vertically downward

fsinΘ : one of the components of frictional force along the verticle axis

therefore, mg=NcosΘ−fsinΘ (eq.1)

mv2r=NsinΘ+fcosΘ (eq.2)

Where,

NsinΘ : one of the components of normal reaction along the horizontal axis

fcosΘ : one of the components of frictional force along the horizontal axis

mv2rmg=NsinΘ+fcosΘNcosΘ−fsinΘ (after diving eq.1 and eq.2)

therefore, v2rg=NsinΘ+fcosΘNcosΘ−fsinΘ

Frictional force f=μsN v2rg=NsinΘ+μsNcosΘNcosΘ−μsNsinΘ v2rg=N(sinΘ+μscosΘ)N(cosΘ−μssinΘ) v2rg=(sinΘ+μscosΘ)(cosΘ−μssinΘ) v2rg=(tanΘ+μs)(1−μstanΘ)

therefore, v=rg(tanΘ+μs)(1−μstanΘ)−−−−−−−−√ vmax=rgtanΘ−−−−−−√ tanΘ=v2rg Θ=tan−1v2rg

Above is the expression for the angle of banking.