Question

A 6900-pound vehicle is driven at a speed of 39 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? (Round your answer to one decimal place.)

Answer 1

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Answer 2

The road surface must exert 12482.57N frictional force on the tires.

To find the answer, we need to know about the expressions velocity at the turning of road and frictional force.

What will be the velocity expression when a vehicle moves at a turn without skidding?

• To avoid the skidding at a turning, the vehicle's frictional force must be equal with the centripetal force at that turning point.
• The expression of the velocity= √(μ×radius of curvature ×g)
• From the above expression, expression of coefficient of friction (μ) = V²/(R×g)

What will be the coefficient of friction if the speed is 39miles/hour and radius is 250 feet?

• First, we have to convert the speed and radius into S.I. units. 39miles/hour = 17.433 m/s and 250feet= 76.2m
• Coefficient of friction (μ)= 17.433²/(76.2×9.8)= 0.41

What is the expression of frictional force?

• The frictional force is given as μmg.
• If m= 6900pounds = 3129.787 Kg, and =0.41
• Frictional force= 0.41×3129.787×9.8= 12482.57N.

Thus, we can conclude that the frictional force is 12482.57N.

Learn more about frictional force here:

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