A car weighing 15,000N rounds a curve of 60m at an angle of 30 degrees. Find the friction force acting on the tires when the car is travelling at 100kph. The coefficient of friction between tires and the road is 0.90

Answer 1

First, convert the speed of the car from kph to m/s:

100 kph * (1000m/1 km) * (1 hr/3600 s) = 27.78 m/s

Next, we will find the critical speed at which the car would slip using the formula:

v_c = sqrt(mu * R * g)

Where v_c is the critical speed, mu is the coefficient of friction, R is the radius of the curve, and g is the acceleration due to gravity (approx. 9.8 m/s^2).

v_c = sqrt(0.90 * 60m * 9.8 m/s^2)
v_c = sqrt(529.2 m^2/s^2)
v_c = 23.0 m/s

Since the car's speed (27.78 m/s) is greater than the critical speed (23.0 m/s), we need to find the required friction force to keep the car from slipping.

First, let's find the horizontal component of the car's weight which can contribute to the required friction force, W_horizontal:

W_horizontal = 15000N * sin(30 degrees)
W_horizontal = 15000N * 0.5
W_horizontal = 7500N

Now, let's find the required friction force:

Specifically, we will use the formula for centripetal force:

F_centr = m * v^2 / R

Where F_centr is the required centripetal force, m is the car's mass (which can be found by dividing the weight by acceleration due to gravity; m = 15000N/9.8m/s^2 = 1531kg), v is the car's speed, and R is the radius of the curve.

F_centr = 1531kg * (27.78m/s)^2 / 60m
F_centr = 1531kg * 772.1m^2/s^2 / 60m
F_centr = 19949 N

Next, we will find the friction force:

F_friction = F_centr - W_horizontal
F_friction = 19949 N - 7500 N
F_friction = 12449 N

When the car is traveling at 100 kph around the curve of 60 meters, the friction force acting on the tires is approximately 12,449 N.