Question

Frictional Force \(\quad\) A 6400 -pound vehicle is driven at a speed of 35 miles per hour on a circular interchange of radius 250 feet. To keep the vehicle from skidding off course, what frictional force must the road surface exert on the tires?

Short answer

The frictional force that the road surface must exert on the tires to prevent the vehicle from skidding is approximately 6178.4 N.

Step-by-step solution

Convert Units

The vehicle's weight is given in pounds and needs to be converted into Newtons. The weight of the vehicle can be converted from pound to Newton by multiplying by 4.44822 which gives a result of \(6400 \times 4.44822 = 28472.608 \) N. Similarly, the speed must also be converted from miles per hour to meters per second by multiplying by 0.44704 which gives \(35 \times 0.44704 = 15.6464 \) m/s. The radius is given in feet and needs to be converted into meters. One foot is equal to 0.3048 meters so \(250 \times 0.3048 = 76.2 \) m.

Apply Centripetal Force Formula

We can then plug these values into the formula for centripetal force. The frictional force also known as centripetal force is given by \(F = m v^2/r\). Here m is the weight divided by the gravitational acceleration, v is the speed and r is the radius. This gives us \(F = \frac{{28472.608}}{{9.8}} \times (15.6464)^2 / 76.2 = 6178.4203 \) N.