When you take your 1300-kg car out for a spin, you go around a corner of radius 57.8 m with a speed of 15.2 m/s. The coefficient of static friction between the car and the road is 0.82. Assuming your car doesn't skid, what is the force exerted on it by static friction?
Please show work, thank you!
That would equal the centripetal force,
M V^2/R,
if the car does not skid. It will not depend upon the static friction coefficient
To find the force exerted on the car by static friction, we need to consider the centripetal force acting on the car as it goes around the corner. The centripetal force is provided by the static friction force between the tires and the road.
The centripetal force can be calculated using the formula:
F = (m * v^2) / r
Where:
F is the centripetal force
m is the mass of the car (1300 kg)
v is the velocity of the car (15.2 m/s)
r is the radius of the curve (57.8 m)
Plugging in the values, we get:
F = (1300 kg * (15.2 m/s)^2) / 57.8 m
Simplifying the equation:
F = (1300 kg * 231.04 m^2/s^2) / 57.8 m
F = 52640 kg⋅m/s^2
Now, since the car does not skid, the maximum force of static friction can be calculated using the formula:
F_friction = μ_s * F_normal
Where:
F_friction is the force of static friction
μ_s is the coefficient of static friction (0.82)
F_normal is the normal force acting on the car, which is equal to the weight of the car (m * g), where g is the acceleration due to gravity (approximately 9.8 m/s^2)
F_friction = 0.82 * (1300 kg * 9.8 m/s^2)
Simplifying the equation:
F_friction = 1011.56 kg⋅m/s^2
Therefore, the force exerted on the car by static friction is approximately 1011.56 kg⋅m/s^2.