The density relation of three solid cubes is ρ 1 Find the relationship between these three forces

The density relation of three solid cubes is ρ 1 Find the relationship between these three forces

For the horizontal ground, the pressure is equal, that is, the gravity / bottom area is equal, g * ρ 1 * A1 ^ 3 / A1 ^ 2 = g * ρ 2 * A2 ^ 3 / A2 ^ 2 = g * ρ 3 * A3 ^ 3 / A3 ^ 2G is the acceleration of gravity, so ρ 1 * A1 = ρ 2 * A2 = ρ 3 * A3 ρ 1A3 bottom area S1 > S2 > S3, the pressure is still the same, F1 / S1 = F2 / S2 = F3 / S3, the pressure on the ground of the cube itself has disappeared

If the three solid cubes have the same pressure on the horizontal surface, and their densities are ρ 1, ρ 2, ρ 3 and ρ 1 ＞ ρ 2 ＞ ρ 3, then the three cubes have the same pressure on the horizontal surface
What is the relationship between the pressure f1f2f3 of the three cubes on the horizontal plane?

Let the edge length (side length) of a cube be L
Pressure P = f / S = mg / S = ρ VG / S = ρ (L ^ 3) g / (L ^ 2) = ρ LG
Because P is equal to L, l is inversely proportional to ρ
ρ1＞ρ2＞ρ3
so
L1

If the pressure of the two solid cubes on the horizontal ground is F1 and F2 respectively, then ()
A. F1 = f2b. F1 ＜ F2C. F1 ＞ F2D

∵ two solid cubes are respectively placed on the horizontal ground, so the pressure on the ground is p = FS = GS = MGS = ρ VGS = ρ SHGs = ρ Hg; the pressure on the ground is equal; because ρ 1 ＞ρ 2, so H2 ＞ H1, because it is a cube, so the contact area with the ground is S2 ＞ S1; so the pressure on the ground is f pressure = PS; the larger the contact area is, the greater the pressure on the ground is, so the bottom ground pressure F1 and F2 are The relationship between F 1 and F 2 is that F 1 < F 2