In the figure below, the driver of a car on a horizontal road makes an emergency
ID: 2099785 • Letter: I
Question
In the figure below, the driver of a car on a horizontal road makes an emergency stop by applying the brakes so that all four wheels lock and skid along the road. The coefficient of kinetic friction between tires and road is 0.41. The separation between the front and rear axles is L = 3.7 m, and the center of mass of the car is located at distance d = 1.8 m behind the front axle and distance h = 0.70 m above the road. The car weighs 11 kN. Find the magnitude of the following. (Hint: Although the car is not in translational equilibrium, it is in rotational equilibrium.)
Explanation / Answer
a )Given that weight of the car is W = mg = 11kN = 11000N Thenmass of the car is m = W/g = 11000/9.8 = 1122.448kg We knowthat breaking force F = ma = -?kmg ?k=0.41 Thenbraking acceleration of the car is a = -?kg =0.41 *9.8 =-4.018m/s2 b)Total frictional force on the car is f =?kmg = 0.41*11000 N = 4510N Now forthe car to be in equilibrium then net force acting on the car to bezero and also net torque acting on the car to be zero. Letnormal force on the rear wheel of the car be Fnr ' normal force onthe front wheel of the car be Fnf ' h= 0.70 m L = 3.7m d= 1.8 m Then Fnr + Fnf = mg -------1 Fnr + Fnf = 11000 N -------2 Then f* h +Fnr*(L-d) -Fnf*d = 0-----------------3 4510N * 0.70 m + Fnr*(3.7m -1.8m) -Fnf*1.8m = 0-----------4Solvingequations 2 and 4 we get Fnr =4498.108N Thistotal normal force on the rear wheel is divided into two partsbecause the car consists of 2 rear wheels Thennormal force on each rear wheel is Fnr ' = Fnr/2 = 2249.054N c ) NowFnf =6501.892 N Now normalforce on each front wheel is Fnf ' = Fnf/2= 3250.946N d)breaking force on each rear wheel is frear = ?kFnr ' = 0.41 *2249.054= 922.11214N e )breaking force on each front wheel is ffront= ?kFnf ' = 0.41 *3250.946= 1332.88N b)Total frictional force on the car is f =?kmg = 0.41*11000 N = 4510N Now forthe car to be in equilibrium then net force acting on the car to bezero and also net torque acting on the car to be zero. Letnormal force on the rear wheel of the car be Fnr ' normal force onthe front wheel of the car be Fnf ' h= 0.70 m L = 3.7m d= 1.8 m Then Fnr + Fnf = mg -------1 Fnr + Fnf = 11000 N -------2 Then f* h +Fnr*(L-d) -Fnf*d = 0-----------------3 4510N * 0.70 m + Fnr*(3.7m -1.8m) -Fnf*1.8m = 0-----------4
Solvingequations 2 and 4 we get Fnr =4498.108N Thistotal normal force on the rear wheel is divided into two partsbecause the car consists of 2 rear wheels Thennormal force on each rear wheel is Fnr ' = Fnr/2 = 2249.054N c ) NowFnf =6501.892 N Now normalforce on each front wheel is Fnf ' = Fnf/2= 3250.946N d)breaking force on each rear wheel is frear = ?kFnr ' = 0.41 *2249.054= 922.11214N e )breaking force on each front wheel is ffront= ?kFnf ' = 0.41 *3250.946= 1332.88N