| 1B-8 : | Force Required to Lift an Underwater Gate | 5 pts | |||||
| A horizontal 2 m diameter man-hole is located in the bottom of a water tank as shown here. Determine the extra upward force, F, that a man or machine must exert on the man-hole cover to just barely lift it. | |||||||
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| Read : | The key to this problem is to recognize that the TOTAL force required to just barely lift the man-hole cover is equal to the force exerted on the top surface of the man-hole 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 man-hole cover. The force a man or machine would need to exert in order to lift the man-hole cover is less because atmospheric pressure is also acting on the outer or bottom surface of the man-hole cover. | ||||||
| Given : | D | 2 | m | ||||
| h | 5 | m | |||||
| g | 9.8066 | m/s2 | |||||
| gc | 1 | kg-m/N-s2 | |||||
| Assumptions : | Patm | 100 | kPa | ||||
| rH2O | 1000 | kg/m3 | |||||
| Solution : | The total force required to just barely lift the man-hole cover is: | ||||||
| Eqn 1 | |||||||
| The gate is circular, so : | Eqn 2 | ||||||
| Plug values into Eqn 2 : | Agate | 3.142 | m2 | ||||
| Next, we can use the Barometer Equation to determine the pressure at the bottom of the tank. | |||||||
| Eqn 3 | |||||||
| Pbottom | 149 | kPa | |||||
| Finally, we substitute values into Eqn 1 to answer the question : |
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| F | 468 | kN | |||||
| This is the total force required to lift the man-hole 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 man-hole cover, we must subtract the upward force attributable to atmospheric pressure below the cover. | |||||||
| Eqn 4 | F | 314 | kN | ||||
| The additional force that must be exerted to lift the man-hole cover is the difference between the total force required and the force exerted by the atmosphere. | |||||||
| Eqn 5 | F | 154 | kN | ||||
| 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. | |||||||
