A cubic crate of side s = 2.5m is top-heavy: its CG is 19cm above its true center.

Question

Part A
How steep an incline can the crate rest on without tipping over?
Part B
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.]

Answer 1

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?

Answer 2

The answer for the second part is 41 degrees as well since it should be the maximum value less than the tipping point