A pickup truck has a wheelbase of 3.00 m. Ordinarily, 10,780 N rests on the front wheels and 8820 N on the rear wheels when the truck is parked on a level road. (a) A box weighing 3600 N is now placed on the tailgate, 1.00 m behind the rear axle. How much total weight now rests on the front wheels? On the rear wheels? (b) How much weight would need to be placed on the tailgate to make the front wheels come off the ground?

Solution 48P Problem (a) Step 1: Length of the wheelbase of the pickup truck L = 3.00 m Force exerted by the ground on the front wheels W = 10,780 N1 Force exerted by the ground on the rear wheels W = 8820 N 2 Weight of the box W = 333 N Position of the box behind the rear wheels x = 1.00 m Total weight of the truck W = W + W =19,6002 Step 2: To find the total weight rests on the front wheels and on the rear wheels The center of gravity of the weight has to be found before going into the problem. The effect the box is not included while calculating the center of gravity. The net force and the net torque on the truck are zero. The condition for equilibrium is satisfied. F = 0 x and F = 0 . y Step 3: Writing torque equation using the above diagram = W (0) - L (W) + W (3) = 0 R 2 cg 1 Substituting values and solving for L cg W 1*3 Lcg W 10780*3 Lcg 19600 L = 1.65 m cg The center of gravity lies 1.65 m from the rear wheels. Step 4: Consider a box of weight 3600 N is now placed 1.00 m behind the rear wheels. To find...