A steel plate rests on the horizontal bottom of a water tank that is 3 m deep. What upward force, in N,
must be applied to the steel plate to just barely lift it straight upward?
The plate is 20 cm in diameter. 
Assume the weight of the plate is negligible. 


Read: 
The key to this
problem is to recognize that the TOTAL force required to just barely lift the
manhole cover is equal to the force exerted on the top surface of the
manhole cover by both the atmosphere and the water. This force is equal to the absolute
pressure at the bottom of the tank times the area of the manhole cover. The force a man or machine would need to
exert in order to lift the manhole cover is less because atmospheric
pressure is also acting on the outer or bottom surface of the manhole cover. 










Given: 
D 
0.2 
m 


g_{c} 
1 
kgm/Ns^{2} 


h 
3 
m 
















Find: 
F_{up} 
??? 
kN 
















Assumptions: 
1 Assume: 

g 
9.8066 
m/s^{2} 







P_{atm} 
100 
kPa 







r_{H2O} 
1000 
kg/m^{3} 













Equations
/ Data / Solve: 


















The total
force required to just barely lift the manhole cover is: 



Eqn 1 














The gate is circular, so
: 





Eqn 2 

Plug values into Eqn 2 : 




A_{gate} 
0.03142 
m^{2} 











Next, we can use the
Barometer Equation to determine the pressure at the bottom of the tank. 















Eqn 3 

















P_{bottom} 
129.4 
kPa 











Finally, we substitute
values into Eqn 1 to answer the question : 
F 
4066 
N 











This is the total
force required to lift the manhole cover.
But atmospheric pressure is also acting on the outside or bottom
surface of the manhole cover as well.
So, since we want to determine how much additional force must be applied in order to lift the manhole cover, we
must subtract the upward force attributable to atmospheric pressure below the
cover. 


















Eqn 4 

















F_{atm} 
3142 
N 











The additional force
that must be exerted to lift the manhole cover is the difference between the
total force required and the force exerted by the atmosphere. 












Eqn 5 



F_{up} 
924.3 
N 











If the door had a
hinge, the problem would be a bit more challenging because it would involve
the computation of torques acting around the axis of the hinge. Torques and lever arms are not part of this
thermodynamics course. 










Answers: 
F_{up} 
924 
N 















