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The car rests on an incline with the (now) two wheels in contact with the rough surface at A and B. We draw a perpendicular through the center of gravity, G at some point above the road, and let it cut the road (slope) at C, which may be somewhere between A and B or elsewhere.

At A and B there will be a reaction, perpendicular to the slope, due to the weight of the car in that direction, which will cause a force through friction perpendicular to it, ie: along the road, in a direction which will stop the car from moving. In fact this frictional force must match the weight of the car resolved in that direction (action and reaction being equal and opposite).

Let these two forces be R and S at A and B.

Now the force will be that of a braked friction or a rolling friction dependant on whether the wheel is braked or not. We have been told to ignore rolling friction, so the weight of the car in the direction of the slope must match the braked friction at A or B, ie: µR or µS.

Since the car is in equilibrium we can take moments about the point C (which eliminates the weight of the car) and we must have - for balances forces - R.AC = S.CB otherwise the car would turn off the slope. Since the friction µR or µS matches the resolved weight of W*sin(theta) where W is the weight and theta the angle of inclination, the greatest possible value for theta for a given µ corresponds to the greater value of R or S, which means that the wheels at A or B should be locked according as AC or CB is the smaller. This is the required condition.