1) The force of static friction between a car\'s tires and the road is what prov

Question

1) The force of static friction between a car's tires and the road is what provides the centripetal force necessary to maintain a turn.
A level curve has a radius of 43 meters.
Under normal (dry) conditions, the maximum speed which at which a car can safely make the turn is 67 km/h.
What is the coefficient of static friction between the tires and the road?

When the road becomes wet, the coefficient of static friction decreases to 0.40.
What is the maximum speed which the car could safely take the curve when wet?

2) Calculate the force of gravitational attraction between the Sun and a 89 kg person on standing on the Earth.

Explanation / Answer

1)

u = v^2/rg = (67/3.6)^2/(43*9.8) = 0.822 <------------answer(coefficient of friction)

max speed = sqrt(urg) = sqrt(0.4*43*9.8) = 12.98 m/s = 46.73 km/h

2)

force of gravity = 6.67*10^-11*89*1.99*10^30/(1.496*10^11)^2 = 0.528 N

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