Assignment 3 Solution

hcyl 0.75:= h1 0.5:= ρ 1010:= B 0.2:= H 0.75:=

A h1 B⋅:= A 100 10 3−×=

D 0.3:= aπ D2⋅

4:= a 70.686 10 3−

×=

R ρ 9.81⋅ A⋅h12

⋅:= R 247.703 100×=

hbar2 h1⋅

3:= hbar 333.333 10 3−

×=

x H h1− hbar+( ):= x 583.333 10 3−×=

FR x( )⋅

H:= F 192.657 100

×=

h2F

ρ 9.81⋅ a⋅:= h2 275.083 10 3−

×=

Mcyl ρ a⋅ h2⋅:= Mcyl 19.639 100×=

Wcyl 9.81 Mcyl⋅:= Wcyl 192.657 100×=

ρcylMcyla hcyl⋅

:= ρcyl 370.445 100×=