Question

The friction coefficient between a road and the Tyre of a vehicle is 4/3. Find the maximum incline the road may have so that once hard brakes are applied and the wheel starts skidding, the vehicle going down at a speed of 36 km/hr is stopped within 5 m.Concept of Physics - 1 , HC VERMA , Chapter 6:Friction "

Answer 1

Solution :
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Given : 
s=5m
μ=4/3
g=10m/s2
u=36km/hr=36x5/18=10 m/s
V=0 m/s
a=v²-u²/2s
a=0-10²/2x 5
a= 10 m/s2
From the free body diagram :
R-mgcosθ=0

R=mgCosθ
Again, ma+mg Sinθ-μR=0----------------1
ma+mg Sinθ-μmg Cosθ=0
a+g sinθ-μgcosθ=0
10+10x sinθ-4/3x 10 Cosθ=0
30 + 30 sinθ-40 Cosθ=0
3+3sinθ-4Cosθ=0
4cosθ-3Sinθ=3

SQUARING WE GET :
=4√(1-sin²θ)=9 +9 sin²θ+18 sin θ
sinθ=18 +√18²-4(25)*(-7) / 2 x 25
Sinθ=-18+32/50=14/50 =0.28[ Taking only positive sign]
θ=sin⁻¹[0.28)=16⁰

Hence maximum angle is 16⁰

Answer 2

Answer:

Given :  

s=5m

μ=4/3

g=10m/s2

u=36km/hr=36x5/18=10 m/s

V=0 m/s

a=v²-u²/2s

a=0-10²/2x 5

a= 10 m/s2

From the free body diagram :

R-mgcosθ=0

R=mgCosθ

Again, ma+mg Sinθ-μR=0----------------1

ma+mg Sinθ-μmg Cosθ=0

a+g sinθ-μgcosθ=0

10+10x sinθ-4/3x 10 Cosθ=0

30 + 30 sinθ-40 Cosθ=0

3+3sinθ-4Cosθ=0

4cosθ-3Sinθ=3

SQUARING WE GET :

=4√(1-sin²θ)=9 +9 sin²θ+18 sin θ

sinθ=18 +√18²-4(25)*(-7) / 2 x 25

Sinθ=-18+32/50=14/50 =0.28[ Taking only positive sign]

θ=sin⁻¹[0.28)=16⁰

Hence maximum angle is 16⁰