Answer:
F=1487.52 N
μ = 0.12
Explanation:
Given that
m= 1200 kg
r= 25 m
v= 20 km/hr = 20 x 5/18 m/s
v= 5.55 m/s
To maintain the speed 5.55 m/s ,friction force should be equal to the radial force
Friction force = Radial force(F)
[tex]F=\dfrac{mv^2}{r}[/tex]
By putting the values
[tex]F=\dfrac{mv^2}{r}[/tex]
[tex]F=\dfrac{1200\times 5.55^2}{25}[/tex]
F=1487.52 N
So friction force = 1478.52 N
We also know that
Friction force = μ m g
1478.52 N = μ x 1200 x 10 ( take g =10 m/s²)
μ = 0.12
This is minimum coefficient of static friction required to keep the car in this turn.
