At the start of a race the rear drive wheels B of the 1640lb

At the start of a race, the rear drive wheels B of the 1640-lb car slip on the track. The coefficient of kinetic friction is ?k = 0.74, and the mass center of the car is at G. The front wheels are free to roll. Neglect the mass of all the wheels.

Part A: Determine the normal reaction the track exerts on the front pair of wheels A.

Part B: Determine the normal reaction the track exerts on the rear pair of wheels B.

Part C: Determine the car\'s acceleration.

Solution

>> Let, Reactions at A and B are : Ra and Rb

>> Now, as Weight of the car,W = mg = 1640*32.2 = 52808 lb-f

>> As, Car has no motion in vertical direction.

So, Net forces in vertical direction = 0

=> Ra + Rb = 52808 .....(1)......

>> Also, car will not rotate.

So, Ma = 0

=> 6*52808 = 10.75*Rb

=> Rb = 31684.8 lbf

From (1), Ra = 21123.2 lbf

>> Now, as Kinetic Fricetoin Coefficient, k = 0.74

and, friction force, f = k*R

As, Net Forward Force = fb - fa = k( Rb - Ra) = 0.74*(31684.8 - 21123.2) = 7815.584 lb-f

Let, acceleration of vehicle = a

=> a = 7815.584/1640= 4.766 ft/s² ......ANSWER........