Supplemental Problems - Solutions
SUP 1
a) ρ = m/V = 12 / (0.5x0.2x0.2) = 600 kg / m3
b) The block will float based on its density. For floating, block weight = buoyant force.
mobjg = ph20 Vdisp g mobj = ph20 Asqaure(hsubmerged) 12 = 1000(0.2x0.2)hsub hsub = 0.3 m
c) The extra weight added should equal the extra buoyant force created by submerging the remaining 0.2 m of height.
Fb(extra) = ph20 Vdisp g = (1000)(0.2x0.2x0.2)(9.8) = 78.4 N 78.4 / 9.8 = 8kg of extra mass.
SUP 2
a) Using fluid continuity. A1v1 = A2v2 (4)(10) = (2) (v2) v2 = 20 m/s
b) Bernoulli. ρgy1 terms cancel out since the pipe stays on the same level.
P1 + ρgy1 + ½ ρv12 = P2 + ρgy2 + ½ ρv22
2x105 + 0 + ½ (1000)(10)2 = P2 + 0 + ½ (1000)(20)2
P2= 50000 Pa.
Since P1 was the gauge pressure and did not include Po, P2 will also come out as the gauge pressure.
SUP 3
a) Bernoulli. ρgy1 terms cancel out since the height difference is negligible.
P1 + ρgy1 + ½ ρv12 = P2 + ρgy2 + ½ ρv22 … rearrange equation so we can find P2 – P1 which is the ΔP
P2 – P1 = ½ ρv12 – ½ ρv22 ΔP = ½ (1.2) (502 – 402) = 540 Pa
b) i) ΔP = Flift / A
540 = Flift / (9x2 wings)
Flift = 9720 N
ii) Fnet = 0
Flift =mg
9720 = m (9.8)
m = 992 kg.
SUP 4
This problem involves floating objects, so weight of object = buoyant force mobj g = ρfluid Vdisp
In general … mobj g= ρobjVobj
Giving … ρobjVobj g = ρfluid Vdisp g … ρobjVobj = ρfluid Vdisp
Water
ρobjVobj = ρfluid Vdisp
ρwV = (1000)(2/3V)
ρw =666.67 kg/m3
Oil
ρobjVobj = ρfluid Vdisp
(666.67)V = ρoil (9/10V)
ρoil = 740.74 kg/m3