A 3000-lb car is parked on a 30° slope, facing uphill. The center of mass of the car is halfway between the frontand rear wheels and is 2 ft above the ground. The wheels are 8 ft apart. Find the normal force exerted by the roadon the front wheels and on the rear wheels.0 = 30°l = 2 ft2 = 4 ftmgHints :The car is in stationary equilibrium, so :1) the total force on the car must be 0 (i.e., along the incline and perpendicular to it),2) the total torque about any point must be 0. (choose as point of rotation the center of mass)Remember friction is: f = µ NThree equations , three unknowns. Solve for N1 and N2.

Given data m=3000 lgθ=30° Here m is the mass and theta is the angle. The distance between the wheels…

A 3000-lb car is parked on a 30° slope, facing uphill. The center of mass of the car is halfway between the front and rear wheels and is 2 ft above the ground. The wheels are 8 ft apart. Find the normal force exerted by the road on the front wheels and on the rear wheels. I le 0 = 30° 4 = 2 ft ½ = 4 ft mg

A 3000-lb car is parked on a 30° slope, facing uphill. The center of mass of the car is halfway between the front and rear wheels and is 2 ft above the ground. The wheels are 8 ft apart. Find the normal force exerted by the road on the front wheels and on the rear wheels. I le 0 = 30° 4 = 2 ft ½ = 4 ft mg