Static friction and frictional force ranking task

Question

Rank the crates on the basis of the frictional force acting on them.

Rank from smallest to largest.

1: 500kg, µ =(0.6,0.3

2: 250kg, µ=(0.2,0.1)

3: 600kg. µ=(0.8,0.5)

4: 750kg, µ=(0.6,0.5)

5: 750kg, µ=(0.4,0.3)

6: 1500kg, µ=(0.3,0.1)

Answer 1

I am not completely understanding the question. Are there 12 forces to rank? Like the mass has each of the two μ values? Or maybe we are to choose just one of the μ values provided? I will give it my best shot to answer...

The equation that applies is what I call the "fun" equation: f = μN. N = mg, but since g will be the same for every single one, we can ignore it because we are just comparing.

The first mass will have f_1a = 0.6(500)g = 300g and then also f_1b = 0.3(500)g = 150g

f_2a = 0.2(250)g = 50g and f_2b = 0.1(250)g = 25g

f_3a = 0.8(600)g = 480g and f_3b = 0.5(600) = 300g

f_4a = 0.6(750)g = 450g and f_4b = 0.5(750)g = 375g

f_5a = 0.4(750)g = 300g and f_5b = 0.3(750)g = 225g

f_6a = 0.3(1500)g = 450g and f_6b = 0.1(1500) = 150g

So if we are ranking 12 different forces, my rank would be 2b, 2a, 1b=6b, 5b, 1a=3b=5a, 4b, 4a=6a, 3a.

Maybe we were only supposed to choose one of a/b though, based on additional information. Hope this helps!