Video: Determining the Friction Force Required to Provide a Centripetal Force

A car that has a mass of 900.0 kg drives along a circular unbanked curve at a constant speed of 25.000 m/s. The radius of the curve is 500.0 m. What magnitude force acts on the car to maintain its speed and direction as it follows the curve? What is the minimum possible value for the coefficient of static friction between the carโ€™s tires and the road surface?

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Video Transcript

A car that has a mass of 900.0 kilograms drives along a circular unbanked curve at a constant speed of 25.000 meters per second. The radius of the curve is 500.0 meters. What magnitude force acts on the car to maintain its speed and direction as it follows the curve? What is the minimum possible value for the coefficient of static friction between the carโ€™s tires and the road surface?

We can label the force we want to solve for capital ๐น and the minimum possible coefficient of static friction weโ€™ll name ๐œ‡ sub ๐‘ . We can start on our solution by drawing a diagram of the situation. Our car in this example, with a mass of 900.0 kilograms, moves on a circular section of road with a radius of curvature 500.0 meters. While it drives on this section, it maintains a steady speed weโ€™ve called ๐‘ฃ 25.000 meters per second.

Knowing all this, we wanna solve first for the magnitude of the force acting on the car that keeps it moving in a circular path on this unbanked road. A force that keeps an object moving in a circular path is a centripetal force. And itโ€™s equal to the mass of the object times its speed squared over the radius of the circle it moves in. In our case, weโ€™ve been given the radius ๐‘Ÿ, the speed ๐‘ฃ, and the mass of the car ๐‘š. So, weโ€™re ready to plug in and solve for ๐น.

When we enter this expression on our calculator, we find that ๐น is 1125 newtons. Thatโ€™s the center-seeking force needed to keep this car on a circular path. If we drew a picture of our car from behind while it was turning on this circular path, the direction of the force ๐น would be to the left, pointing it towards what from this perspective is the center of the circle. The physical mechanism that creates this force ๐น is the friction force on the tires of the car from the road.

Wanting to solve for the coefficient of static friction between these two materials, we can recall that the force of friction on an object is equal to the coefficient of friction it experiences multiplied by the normal force acting on it. In our case, we can write that the force of friction on the car is equal to the coefficient of static friction because the wheels donโ€™t slide on the roadโ€™s surface multiplied by the mass of the car times the acceleration due to gravity ๐‘”. Where ๐‘” weโ€™ll treat as exactly 9.8 meters per second squared.

This frictional force of the roadโ€™s surface on the tires is the physical mechanism by which the centripetal force ๐น sub ๐‘ is exerted on the car. So, we can say that the coefficient of static friction times ๐‘š๐‘” is equal to ๐‘š๐‘ฃ squared over ๐‘Ÿ. And we see that the mass of the car cancels out from this expression. If we then divide both sides of this equation by ๐‘”, we see that ๐œ‡ sub ๐‘  is equal to ๐‘ฃ squared over ๐‘Ÿ times ๐‘”.

When we plug in for the given values of ๐‘ฃ, ๐‘Ÿ, and ๐‘”, we find that, to four significant figures, ๐œ‡ sub ๐‘  is 0.1276. Thatโ€™s the minimum value that the coefficient of static friction can be between the roadโ€™s surface and the tires in order for this magnitude centripetal force to be exerted on the car.

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