What must be the contact area between a suction cup (completely evacuated) and a ceiling if the cup is to support the weight of an 80.0-kg student?

Question

Sienna

Mathematics

17 January, 08:23

What must be the contact area between a suction cup (completely evacuated) and a ceiling if the cup is to support the weight of an 80.0-kg student?

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Jordan Hensley · Answer 1

Contact area must be 0.00775 m2, which is the area of a circle with 10 cm in diameter.

Are you surprised that such a small cup is, in theory, enough to hold your weight? But it all comes from the equation of pressure:

P = F / S

P = Pressure, in Pascal, [P]

F = Force, in Newton [N]

S = Surface, in squared meters [S]

Pressure (atmospheric) is 101325 Pascal, and the required force (as per your question) is 80Kg = 785 N. Solving for S, you get 0.00775 m2.

You should have seen these suction cups used by glass workers, where a couple of cups are enough to lift a big piece of glass. This depends on how good the cups are and how smooth the surface of the ceiling or glass is. The idea is to have no pressure inside the cup. If you have some air inside the cup then atmospheric pressure might not be enough to hold 80Kg, as calculated.