You\'re late for class and entering the freeway via an unbanked curve to the rig

Question

You're late for class and entering the freeway via an unbanked curve to the right with radiu; 40 m. Assume the mass of your car is 2000 kg. The coefficient of friction between the tire rubber and the roadway is mu_s = 0.90. What is the highest speed you can drive through this curve without slipping? Now that you are on the freeway, you have to drive over a banked, curved bridge with radius 70 m. What is the maximum speed you can drive on this curve without slipping, if the curve is banked at 15 degree ? The coefficient of friction between the tire rubber and the roadway is a much lower mu_s-wet = 0.60 when the roadway is wet. What are the maximum speeds for parts a and b if the roadway is wet? Similar to the problem worked in class, Sam (80 kg) takes off up a 60 m high, 10degree slope on his jet-powered skis. The coefficients of kinetic and static friction between the skis and the ramp are 0.30 and 0.50, respectively. The skis have a thrust of 250 N and remain tilted at 10degree throughout the problem. When Sam is in the air, he experiences a drag force due to the wind of 70 N. What is the net force acting on Sam as he travels up the ramp? What is Sam's acceleration which he is on the ramp? How far does Sam land from the base of the cliff?

Explanation / Answer

1. radius of the track R = 40 m

Mass of car M = 2000 kg

Coefficient of friction = 0.90

(a.) Centripetal force on curve = MV2 / R = 2000*V2 / 40 = 50 V2

       Limiting friction force = 0.90*Mg = 0.90 * 2000*9.8 = 17640 N

For maximum speed both the force will be balancing each other

i.e        50 V2 = 17640

or              V = 18.78 m/s

(b.) Freeway

Centripetal force on curve = MV2 /R

Force due to banking = MgSin15

For limiting case both the forces will balance each other

i.e             ( MV2 /R   )Cos15 = MgSin15

                     V2 = 9.8*70*0.259

           or        V = 13.32 m/s

(c.) Centripetal force on curve = MV2 / R = 2000*V2 / 40 = 50 V2

       Limiting friction force = 0.60*Mg = 0.60 * 2000*9.8 = 11760 N

Now              50 V2 = 11760

            or           V = 18.78 m/s

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