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Question Number 30094 by Tinkutara last updated on 16/Feb/18
Commented by mrW2 last updated on 28/Feb/18
$${The}\:{thin}\:{film}\:{of}\:{water}\:{stands}\:{in}\: \\ $$$${underpressure}\:{due}\:{to}\:{surface}\:{tension}, \\ $$$${see}\:{diagram}. \\ $$$${R}\:\mathrm{cos}\:\theta=\frac{{d}}{\mathrm{2}} \\ $$$$\Rightarrow{R}=\frac{{d}}{\mathrm{2cos}\:\theta} \\ $$$$\Delta{p}=\frac{{S}}{{R}}=\frac{\mathrm{2}{S}\:\mathrm{cos}\:\theta}{{d}} \\ $$$${F}=\Delta{p}\:{A}=\frac{\mathrm{2}{AS}\:\mathrm{cos}\:\theta}{{d}} \\ $$$${with}\:\theta=\mathrm{0}° \\ $$$$\Rightarrow{F}=\frac{\mathrm{2}{AS}}{{d}} \\ $$
Commented by Tinkutara last updated on 01/Mar/18
$${But}\:\Delta{p}\:{should}\:{be}\:\frac{\mathrm{2}{S}}{{R}}? \\ $$
Commented by mrW2 last updated on 28/Feb/18
Commented by mrW2 last updated on 01/Mar/18
$${No}\:{sir}! \\ $$$$\mathrm{2}{R}×\Delta{p}=\mathrm{2}{S} \\ $$$$\Rightarrow\Delta{p}=\frac{{S}}{{R}} \\ $$
Commented by Tinkutara last updated on 01/Mar/18
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