设圆柱体A的底面积和高为 sA hA,则圆柱体B的底面半径为 sB=3sA hB=3hA/2
p1=(sAhAρA+sBhBρB)÷sB
=(sAhAρA+3sA(3hA/2)ρB)÷(3sA)
=sAhA(ρA+9ρB/2)÷(3sA)
p2=sBhBρB÷sA
=(3sA)(3hA/2)ρB)÷sA
=sAhA(9ρB/2)÷sA
(上面除sA不除sB,是因为B在A上面,两者的接触面积是sA,而不是sB)
p1:p2=1:2
[sAhA(ρA+9ρB/2)÷(3sA)]:[sAhA(9ρB/2)÷sA]=1:2
(ρA+9ρB/2)÷3:9ρB/2=1:2
2ρA+9ρB=27ρB/2
2ρA=9ρB/2
ρA:ρB=9:4