Pezmachine9000idn
Answered

Get expert advice and community support for all your questions on IDNLearn.com. Discover in-depth answers from knowledgeable professionals, providing you with the information you need.

The radius of curvature of a highway exit is r = 93.5 m. The surface of the exit road is horizontal, not banked. (See figure.)
If the coefficient of static friction between the tires of the car and the surface of the road is μ[tex]_{s}[/tex] = 0.402, then what is the maximum speed at which the car can safely exit the highway without sliding?

The Radius Of Curvature Of A Highway Exit Is R 935 M The Surface Of The Exit Road Is Horizontal Not Banked See Figure If The Coefficient Of Static Friction Betw class=

Sagot :

Static friction keeps the car from skidding off the road and points toward the center of the curve. By Newton's second law, the car experiences

• net vertical force

F [normal] - F [weight] = 0

• net horizontal force

F [friction] = ma = mv²/r

where v is the tangential speed of the car.

It follows that

F [normal] = F [weight] = mg

and when static friction is maximized at the car's maximum speed,

F [friction] = µ F[normal] = 0.402 mg

Solve for v :

0.402 mg = mv²/r   ⇒   v = √(0.402 g (93.5 m)) ≈ 19.2 m/s

Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for visiting IDNLearn.com. We’re here to provide dependable answers, so visit us again soon.