File:C:\Users\19205\Documents\ME 361\EES\HW1 Part 2.EES
9/7/2022 6:38:41 PM Page 1
EES Ver. 11.416: #100: For use only by Students and Faculty, College of Engineering University of Wisconsin - Madison
{ HW1 Part 2
ME 361
Name: Ethan Stelzer }
{ knowns }
D = 2 [ft] * convert(ft,m)
r = D/2
th = 0.375 [in] * convert(in,m)
rho_s = 7996 [kg/m^3]
L = 10 [ft] * convert(ft,m)
L_w = 4 [ft] * convert(ft,m)
rho_a = 1 [kg/m^3]
rho_w = 1000 [kg/m^3]
{diameter of cylinder}
{radius of cylinder}
{thickness of the wall of cylinder and end caps}
{density of steel}
{length of cylinder}
{depth of cylinder that is submerged in water}
{density of air}
{density of water}
{ Q: a) What is the weight of steel associated with clyinder and end caps (N)? }
vol_cylinder = pi# * r^2 * L
vol_air = pi# * ((r-th))^2 * (L-2*th)
vol_steel = vol_cylinder - vol_air
{volume of the air inside cylinder}
{volume of the steel in the cylinder}
{total volume of the cylinder}
m_s = vol_steel * rho_s
g = 9.81 [m/s^2]
w_s = m_s * g
{mass of steel}
{gravity of Earth}
{weight of steel in cylinder}
{ Answer: 4702 N }
{ Q: b) What is the weight of air inside the cylinder (N)? }
m_a = vol_air * rho_a
w_a = m_a * g
{mass of air}
{weight of air in cylinder}
{ Answer: 8.139 N }
{ Q: c) Determine the buoyancy force (N)? }
vol_submerged = pi# * r^2 * L_w
{volume of water displaced by the cylinder}
f_b = vol_submerged * rho_w * g
{buoyancy force on the submerged cylinder}
{ Answer: 3491 N }
{ Q: d) What is the force associated with the line from the crane (N)? }
f_line = w_s + w_a - f_b
{force on the line from the crane}
{ Answer: 1220 N in the upwards direction }
{ Q: e) How far must the cylinder be submerged before the force on the line from the crane goes to zero and cylinder floats (ft)? }
f_line_new = 0
{force on line from crane is now zero}
f_b_new = w_s + w_a
{new buoyancy force allowing the cylinder to float}
pi# * r^2 * L_w_new = f_b_new / (rho_w * g)
{ Answer: 5.398 ft (1.645 m ) }
{equation to find depth of submerged cylinder}
File:C:\Users\19205\Documents\ME 361\EES\HW1 Part 2.EES
9/7/2022 6:38:41 PM Page 2
EES Ver. 11.416: #100: For use only by Students and Faculty, College of Engineering University of Wisconsin - Madison
m
D = 2 [ft] · 0.3048 ·
ft
r =
D
2
th = 0.375 [in] · 0.0254 ·
m
in
r s = 7996 [kg/m3]
m
L = 10 [ft] · 0.3048 ·
ft
m
L w = 4 [ft] · 0.3048 ·
ft
r a = 1 [kg/m3]
r w = 1000 [kg/m3]
vol cylinder
= 3.142 · r
2
· L
vol air = 3.142 · r – th
2
· L – 2 · th
vol steel = vol cylinder – vol air
m s = vol steel · r s
g = 9.81 [m/s2]
ws = m s · g
m a = vol air · r a
wa = m a · g
vol submerged
= 3.142 · r
2
· Lw
f b = vol submerged · r w · g
f line = w s + w a – f b
f line,new = 0
f b,new = w s + w a
3.142 · r
2
· L w,new =
f b,new
rw · g
File:C:\Users\19205\Documents\ME 361\EES\HW1 Part 2.EES
9/7/2022 6:38:41 PM Page 3
EES Ver. 11.416: #100: For use only by Students and Faculty, College of Engineering University of Wisconsin - Madison
SOLUTION
Unit Settings: SI K Pa J mass rad
D = 0.6096 [m]
fb,new = 4711 [N]
fline,new = 0 [N]
L = 3.048 [m]
Lw,new = 1.645 [m] {5.398 [ft]}
ms = 479.4 [kg]
3
ra = 1 [kg/m ]
3
rw = 1000 [kg/m ]
volair = 0.8297 [m3]
volsteel = 0.05995 [m3]
wa = 8.139 [N]
No unit problems were detected.
fb = 3491 [N]
fline = 1220 [N]
g = 9.81 [m/s2]
Lw = 1.219 [m]
ma = 0.8297 [kg]
r = 0.3048 [m]
3
rs = 7996 [kg/m ]
th = 0.009525 [m]
volcylinder = 0.8896 [m3]
volsubmerged = 0.3558 [m3]
ws = 4702 [N]