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Let the radius of the circular segment be 105 m, the mass of the car 2011 kg, and the coefficient of the static friction between the road and the tire 0.7. Find the magnitude of the normal force N which the road exerts on the car at the optimal speed (the speed at which the frictional force is zero) of 85 km/h. The acceleration due to gravity is 9.8 m/s 2 . Answer in units of N.

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Answer:

Step-by-step explanation:

Let N be the reaction force making angle α with the vertical.

So N Cos α = mg ( this makes friction zero )

N Sin α = mv²/r ( centripetal force is provided by sin component of N)

Squiring and adding


N=\sqrt{(mg)^2((mv^2))/(r)}

N = m( g² + v⁴/r² )^1/2

Putting the values

N = 2011 X ( 9.8² + 23.61⁴/105² )^1/2 [ v = 23.61 m/s given , radius r = 105 m ]

N = 22382 N

User Goran Rakic
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