A car with a load equal to 1/4 the total mass of the empty car has its center of gravity halfway between the front and rear axles. The drive wheels (rear) start slipping when the car is driven up a 20 degree incline. How far back must the load be shifted for the car to get up a 25 degree slope. (The distance between the axles is 10 ft.) I am really stuck on this one. I write the force equation as mg sin 20 - u mg cos 20 = 0 And it seems to me that the additional frictional force that must be supplied by moving the load would then be given by sin 25 - sin 20. But I have no idea how to turn this value into a distance. I have tried using torques around the center, like this: mgd/4 = (sin 25 - sin 20)mg But this is so small it can't be right. Any suggestions that would help shake me loose would be appreciated.