A 1200kg car rounds a curve of radius of 50m with a speed of 80km/hr. What is the coefficient of friction if the car is just about to skid off the road?
I got:
ac= 9.88m/s2
Fc= 11851N
I also know Ff=uFN but have no idea how to figure this out. Thx.
Ac = v^2/R
F = m Ac = m v^2/R
but
F = mu m g
so
mu m g = m v^2/R
mu = (v^2/R)/g
ratio of centripetal acceleration to gravitational acceleration :)
by the way on earth g is more like 9.81 m/s^2, not 9.88
To find the coefficient of friction (μ) when the car is just about to skid off the road, we can start by calculating the centripetal force (Fc) acting on the car as it rounds the curve.
The centripetal force is given by the equation Fc = ma_c, where m is the mass of the car and ac is the centripetal acceleration.
Given:
- Mass of the car (m) = 1200 kg
- Radius of the curve (r) = 50 m
- Speed of the car (v) = 80 km/hr = 22.22 m/s
The centripetal acceleration (ac) can be calculated using the formula ac = v^2 / r:
ac = (22.22 m/s)^2 / 50 m
ac ≈ 9.88 m/s^2
The centripetal force (Fc) can now be calculated using the formula Fc = ma_c:
Fc = (1200 kg)(9.88 m/s^2)
Fc = 11851.2 N
Now, let's find the frictional force (Ff) using Ff = μFN, where FN is the normal force acting on the car. The normal force is equal to the weight of the car (mg).
Given:
- Mass of the car (m) = 1200 kg
- Acceleration due to gravity (g) = 9.8 m/s^2
FN = mg = (1200 kg)(9.8 m/s^2)
FN = 11760 N
Substituting these values into the equation, we get:
Ff = μFN
11851.2 N = μ(11760 N)
Now, we can solve for the coefficient of friction (μ):
μ = Ff / FN = 11851.2 N / 11760 N
μ ≈ 1.01
Therefore, the coefficient of friction (μ) when the car is just about to skid off the road is approximately 1.01.