An airtight box has a removable (massless) lid of area 1·10-2 m2. A lid is placed on it while it is on top of
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An airtight box has a removable (massless) lid of area 1·10-2 m2. A lid is placed on it while it is on top of a mountain (where Patm=0.86·105 Pa). It is then taken to sea level, where Patm = 1.013·105 Pa. How much force will be required to remove the lid?
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You can assume that the pressure inside the box is 0.86x10^5 Pa. So the force pushing up aginst the lid from inside is:
F=PA = (0.86x10^5) * (1x10^-2) = 860 N.
The pressure pushing down on the lid when at sea level is 1.013x10^5 Pa. So the force pushing down is:
F=PA = (1.013x10^5) * (1x10^-2) = 1013 N
So the force required to remove the lid is (1013-860) = 153 N.
Calculate the pressure differential at the two elevations (perhaps in pounds per square inch) and multiply that times the area (in square inches) of the cover. You can do similar calculations in metric if you reduce terms to physical quantities.
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