IIT JEE , AIEEE Forum - Physics , Mechanics

varun.tinkle

a turn of radius 20m is banked for vehicles ging at a speed of 10ms. if the coefficient of static friction between the tyres and the road is 0.4 , what r the possible speeds that the vehicle neithe skids up or slips down

plss expln in detail

computer001

root(60) and root(140) ???
il post soln if its rite
EDIT:
i assumed g=10

EDITED:pls refer my later proof i made a mistake here

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edited

computer001

in 1 case friction in other case no friction..
10 m/s in absence of friction
EDIT:ive written this here as a reply to prev post.

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edited again
sorry.. dint read qsn properly

varun.tinkle

just verify my working
in 1st case the speed is 2 much so friction comes into play
then
Nsin@+un=mv2/r
Ncos@=mg
in second case friction is helping centrepital force
Nsin@=mv2/r+un
ncos@=mg
just verify wether in both cases ncos@=mg

computer001

to find @ the angle of banking:
the eqn for eqbm of vehicle in absense of friction: (v^2/r)cos@=mgsin@
tan@=1/2
now friction comes in to the pic
friction can act upward or downwards so 2 cases
2 move neithe up or down,eqbm eqn give:
(mv^2/r)cos@ +(or - ) (

mgcos@+(mv^2/r)sin@)=mgsin@
we get v^2/r=7 or 3

hence v=root(60) or root(140)-edited after the correction im gettimg same as varun
the value of N is N=mgcos@ +(mv^2/r)sin@

computer001

pl post the ans...only if i know if im rite i can look into ur logic dont u agree

varun.tinkle

so where does mg cos@ come into picture

computer001

n=mgcos@
not ncos@=mg
hey pllsss post ans...iwant to know if im rite..thts least u cud do :\

varun.tinkle

no in h.c.v it says ncos@=mg

computer001

edited

varun.tinkle

ans is between 14.km/h and 54km/hr

eistien

no da gokul remember the banking sum we proved v^2=gr we got that from here only but there we didnt have any friction

but if we have friction V^2=Rg(u-tan(@)/1+utan(@);

use that formula no???

and Ncos(@) = Mg-Fsin(@)

Nsin@+Fcos@=Mv^2/R

check it out!!!!

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