When you take your 1200 kg car out for a spin, you go around a corner of radius 51 m with a speed of 12 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?
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The static friction force provides the centripetal force needed to turn on an unbanked road. With no slipping, it will be less than or equal to the maximum possible static friction, which is 0.88 Mg in this case.
Less than
To calculate the force exerted on the car by static friction, we need to use the formula:
Centripetal Force = (Mass of the car) * (Centripetal Acceleration)
The centripetal acceleration can be found using the formula:
Centripetal Acceleration = (Velocity)^2 / (Radius)
First, let's calculate the centripetal acceleration:
Centripetal Acceleration = (12 m/s)^2 / 51 m = 2.8235 m/s^2
Next, we can find the force needed to keep the car moving in a circular path using the formula:
Centripetal Force = (Mass of the car) * (Centripetal Acceleration)
Centripetal Force = (1200 kg) * (2.8235 m/s^2) = 3388.2 N
Now, we need to determine if the force of static friction is able to provide this centripetal force. The maximum force of static friction can be found using:
Maximum Force of Static Friction = (Coefficient of Static Friction) * (Normal Force)
The normal force can be calculated by multiplying the mass of the car by the acceleration due to gravity (9.8 m/s^2).
Normal Force = (Mass of the car) * (Acceleration due to gravity)
Normal Force = (1200 kg) * (9.8 m/s^2) = 11760 N
Now, we can determine the maximum force of static friction:
Maximum Force of Static Friction = (0.88) * (11760 N) = 10348.8 N
Since the maximum force of static friction (10348.8 N) is greater than the centripetal force required (3388.2 N), the force exerted on the car by static friction is:
Force exerted on the car by static friction = 3388.2 N