Thank you any help would be awesome. Given: rho water = 1000 kg.m-3 and g = 9.8m

Question

Thank you any help would be awesome.

Given: rho water = 1000 kg.m-3 and g = 9.8m.s-2 Oil with density rho 1 floats on water with rho 2. Tank is placed on a frictionless horizontal plane. Assume incompressible liquids and mu = 0. For the values of rho 1, h1, h2, h3, d1, d2, and anchoring force FA listed below, Determine FA required to hold the tank stationary. Determine h1 such that FA = 0. Determine d2 such that FA = 0. A different liquid with density rho 1 replaces the oil. Determine rho 1 such that FA = 0. rho 1(kg.m-3 h1 (m) h2 (m) h3 (m) d1 (cm) d2 (cm) FA (N) (a) 900 1.0 0.3 0.5 5 2 ? (b) 900 ? 0.3 0.5 5 2 0 (c) 900 1.0 0.3 0.5 ? 2 0 (d) ? 0.14 0.3 0.5 5 2 0

Explanation / Answer

a) Force required to hold the tank = Net rate of Momentum in rightward direction F = (AV)1V1 - (AV)2V2 = 1(/4*d12)*V12 - 2(/4*d22)*V22

But, V1 = (2gh1) and V2 = (2g(h1+h2+h3)) and 2 = 1000 kg/m3

Hence, F = 1(/4*d12)*(2gh1) - 1000*(/4*d22)*2g(h1+h2+h3)

a) F = 900*(/4*0.052)*(2*9.81*1) - 1000*(/4*0.022)*2*9.81*(1+0.3+0.5)

F = 23.56 N

b) 0 = 900*(/4*0.052)*(2*9.81*h1) - 1000*(/4*0.022)*2*9.81*(h1+0.3+0.5)

h1 = 0.17 m

c) 0 = 900*(/4*d12)*(2*9.81*1) - 1000*(/4*0.022)*2*9.81*(1+0.3+0.5)

d1 = 0.028 m = 2.8 cm

d) 0 = 1*(/4*0.052)*(2*9.81*0.14) - 1000*(/4*0.022)*2*9.81*(0.14+0.3+0.5)

1 = 1074.3 kg/m3

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