When you take your 1000-kg car out for a spin, you go around a corner of radius 57 m with a speed of 11 m/s. The coefficient of static friction between the car and the road is 0.88.
Assuming your car doesn't skid, what is the force exerted on it by static friction?
Randy, check your 3-6-11,8:55pm post.
Heck yeah this is excalty what I needed.
To find the force exerted on the car by static friction, we can start by calculating the maximum static friction force that can act on the car.
The maximum static friction force, F(static_max), can be calculated using the formula:
F(static_max) = coefficient of static friction * normal force
The normal force (N) is the force exerted on an object perpendicular to the surface it is in contact with. In this case, since the car is on a flat road, the normal force is equal to the weight (mg) of the car, where m is the mass of the car and g is the acceleration due to gravity.
Given:
Mass of the car (m) = 1000 kg
Radius of the turn (r) = 57 m
Speed of the car (v) = 11 m/s
Coefficient of static friction (μ_static) = 0.88
Acceleration due to gravity (g) = 9.8 m/s^2
First, let's calculate the normal force (N) using the weight formula:
N = mg
N = 1000 kg * 9.8 m/s^2
N = 9800 N
Now, we can calculate the maximum static friction force (F(static_max)):
F(static_max) = μ_static * N
F(static_max) = 0.88 * 9800 N
F(static_max) = 8624 N
Therefore, the force exerted on the car by static friction is 8624 N.