In the figure below, the driver of a car on a horizontal road makesan emergency
ID: 2099448 • Letter: I
Question
In the figure below, the driver of a car on a horizontal road makesan emergency stop by applying the brakes so that all four wheelslock and skid along the road. The coefficient of kinetic frictionbetween tires and road is 0.43. Theseparation between the front and rear axles is L =3.7 m, and the center of mass of the caris located at distance d = 1.8 m behind the front axle anddistance h = 0.85 m above theroad. The car weighs 11 kN. Find the magnitude of the following.(Hint: Although the car is not in translational equilibrium, it isin rotational equilibrium.)
Explanation / Answer
Given that The coefficient of kinetic friction betweentires and road is ? = 0.43 The separation between the two axle is L = 3.7 m The distance between the front axle and the center of mass is d = 1.8 m The height above the road is h = 0.85 m The weight of the car is W = 11 kN Then the mass of the car is m = W /m = (11000N)/ (9.8m/s2) = ------------- kg -------------------------------------------------------------------------------- If the body is moving in the rough horizontal plane theaccelaration of the body is a = - ?g =----------- m/s2 The total frictional force is f = ?mg =0.43*9.8m/s2*m The verticle distance below the center of mass is h = 0.85m Then torque equlibrium leads to (4.2m)(0.85) + Fnr (1.85) - Fnf (1.8) = 0 -----------(1) since the body is in equlibrium then the Fnr + Fnf = mg --------(2) From the equations (1) and (2) we get the value of the normalforces normalforce on the each rear wheel is N1 = Fnr /2 =-------- N normal force on the each front wheel is N2 = Fnf /2 =------ N