IIT JEE , AIEEE Forum - Physics , Mechanics

prakhar_galaxy

A car is gong at a speed of 21.6 km/hr when it encounters a 12.8 m slope of angle 30 degree .The friction coefficient b/w road & tyre is 1/(2

3).Show dat no matter how hard the driver applies the brakes ,the car will reach the bottom with a speed greater than 36 km /hr.Take g=10m/s²

The fig is:

My doubt is:

1.Which friction will act b/w tyre & road (kinetic or static)

2. How did the brakes act in a car. That by applying brakes we stop the rotation motion of the tyre or the slipping of the tyre on the roads

Plz give me the complete & best solution with all the technical possibiliies of this ques

i wud be highly obliged for it

Thanx

apurviitjee2008

kinetic friction acts as it acts when the suraces are relatively sliding
plz give the second question a bit clearly

elessar_iitkgp

The brakes only serve to lock the wheels. Now, the main battle is between gravity and friction. The car will move down if mgsin

mgcos

That is, if tan

Which is true in the given case. (Substitute the values and check)

Now, from the FBD,
a = g(sin

-

cos

)
v(dv/dx) = g(sin

-

cos

)
Integrating the above equation,
v² = v₀² + 2g(sin

-

cos

)L
Substitute v₀ = 21.6 km/h = 6 m/s, g = 10m/s² ,

= 1/(2

3) , L = 12.8 m
v = 10 m/s = 36 km/h

So irrespective of how hard the driver applies brakes (which merely locks the wheels), the car attains a speed of 36 km/h.

PS: Note that the above equation can be obtained by energy conservation, but as you have asked from friction chapter, its possible that you haven't read upto energy conservation. So I haven't used it.

Also note that if rolling motion of wheels is present, the friction between the tyres and the incline will decrease. In that case, the velocity at the bottom of the inclined plane increases.

prakhar_galaxy

sir thanx for ur reply but i didn't understand the meaning of last line
of ur solution
secondly u didn,t tell dat which friction will act on the body kinetic or static
thirdly in which case the body will reach the bottom with greater speed

karthik2007

See, what happens is, if the car has the best brake possible, and if the driver applies them as hard as possible, the wheel will stop rotating, and the car will simply skid on the floor. This is the case taken into consideration while solving this sum. It is not possible to apply brakes harder than this. If the brakes are applied lightly, then the wheels will also rotate while going down the inclined plane, thus friction will decrease, thus the speed will be greater.

prakhar_galaxy

hey any other expert plz reply to my complete query
plzzzzz

satishrm

its very simple if u know rolling...topic...

if the wheels lock....kinetic friction acts

if the wheels dont lock.....static (why?)

after u apply brake...the wheel slows down....then it will start pure rolling.i.e velocity of lowermost point will be zero wrt floor....

but how will u maintain tht..since car is under..a?
'
static friction comes into play...it applies force of suitable magnitude...such that torque generated by it is such tht..resulting angular acc ( aa ) is such tht

a = r (aa)

a- acc of centre of mass of wheel
aa - ang acc

u have to find value of mag of st fric..which satisfies the above eqn....

but tht " expert " cooly substituted friction as mu * N
which is WRONG!!!.....how do u know maximum friction acts??/
see...............

so question is WRONG....INFORMATION INADEQUATE!!!!!!!!!!!!!!!!!!!!!!

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