posted by chris on .
A cubic crate of side s = 2.5m is top-heavy: its CG is 19cm above its true center.
How steep an incline can the crate rest on without tipping over?
What would your answer be if the crate were to slide at constant speed down the plane without tipping over? [Hint: The normal force would act at the lowest corner.]
i figured out the first part using the fact that the crate will flip over once the CG is no longer above the crate, and drew a triangle of legs 1.44 m(distance from base of crate to CG), and 1.25m(distance from corner of crate to middle). at first i thought the answer would be the angle opposite the 1.44m length, but with that angle being larger than 45 degrees, i thought wouldnt a top heavy object require less angle? so i took the angle opposite the 1.25 m side and got 41 degrees, which was correct. but i have no idea where to start on part b, any hints?