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Question Number 31474 by NECx last updated on 09/Mar/18
$${A}\:{vacuum}\:{chamber}\:{has}\:{a}\:{door} \\ $$$${with}\:{surface}\:{dimensions} \\ $$$$\mathrm{20}{cm}×\mathrm{20}{cm}.{How}\:{many}\:\mathrm{70}{kg}\:{men}, \\ $$$${each}\:{exerting}\:{a}\:{pull}\:{equal}\:{to}\:{his} \\ $$$${weight},{will}\:{it}\:{take}\:{to}\:{pull}\:{the} \\ $$$${door}\:{open}\:{when}\:{the}\:{chamber}\:{is} \\ $$$${evacuated}\:{to}\:{a}\:{pressure}\:{of}\:\mathrm{0}.\mathrm{1}\:{atm}? \\ $$
Commented by NECx last updated on 09/Mar/18
$${please}\:{show}\:{workings}.^{} \\ $$
Answered by ajfour last updated on 09/Mar/18
$${F}=\mathrm{0}.\mathrm{9}×\mathrm{1}×\mathrm{10}^{\mathrm{5}} ×\frac{\mathrm{1}}{\mathrm{25}}\:=\:{N}×\mathrm{700} \\ $$$$\Rightarrow\:{N}=\frac{\mathrm{9}×\mathrm{10}^{\mathrm{4}} }{\mathrm{25}×\mathrm{700}}=\frac{\mathrm{9}×\mathrm{100}}{\mathrm{25}×\mathrm{7}}\:=\:\frac{\mathrm{36}}{\mathrm{7}} \\ $$$$\Rightarrow\:{At}\:{least}\:\mathrm{6}\:{men}\:. \\ $$
Commented by NECx last updated on 10/Mar/18
$${how}\:{come}\:\mathrm{0}.\mathrm{9},\:{and}\:\mathrm{1}/\mathrm{25} \\ $$
Commented by ajfour last updated on 10/Mar/18
$${Net}\:{pressure}\:{from}\:{outside} \\ $$$$=\:\mathrm{1}{atm}−\mathrm{0}.\mathrm{1}{atm}\:=\:\mathrm{0}.\mathrm{9}\:{atm} \\ $$$${Area}=\:\mathrm{20}{cm}×\mathrm{20}{cm}\:=\:\frac{\mathrm{1}}{\mathrm{25}}{m}^{\mathrm{2}} \:. \\ $$