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
Answers
The questions means that if we apply the brakes then then Friction becomes greater than the component of the Weight acting along the incline thus causing a retardation so that block can stop easily.
So, Condition for that will be,
μmgcosθ - mgsinθ = ma [θ is the angle of the Incline.]
⇒ μgcosθ - gsinθ = a
⇒ 40/3cosθ - 10Sinθ = a
For a,
Initial velocity = 36 km/hr = 10 m/s.
Distance = 5 m.
⇒ v² - u² = 2aS
⇒ - 100 = 2 × a × 5
⇒ a = -10 m/s².
∴ Retardation = 10 m/s².
⇒ 40/3cosθ - 10Sinθ = a
∴ 40/3cosθ - 10Sinθ = 10
⇒ 40cosθ - 30Sinθ = 30
∴ 4cosθ - 3Sinθ = 3
Diving each term by 5,
4/5 cosθ - 3/5 Sin θ = 3/5
Now, Let us assume one right angled triangles, in which sides are 3, 4 and 5.
∴ 4/5 = Sinα and 3/5 will be Cos53°.
⇒ Cosα cosθ - Sinα Sin θ = Cos53°
⇒ Cos (α + θ) = Cos 53°
∴ α + θ = 53°
We know that α was 37°. Because, Cosα = 4/5 will give, α = 37°
⇒ θ = 53° - 37°
∴ θ = 16°.
Hope it helps .
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