A vacuum chamber has a door with surface dimensions 20cm×20cm.How many 70kg 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 0.1 atm?


Question and Answers Forum

All Questions Topic List
Heat and Theromdynamics Questions
Previous in All Question Next in All Question
Previous in Heat and Theromdynamics Next in Heat and Theromdynamics
Question Number 31474 by NECx last updated on 09/Mar/18

$A v a c u u m c h a m b e r h a s a d o o r$ $w i t h s u r f a c e d i m e n s i o n s$ $20 c m \times 20 c m . H o w m a n y 70 k g m e n,$ $e a c h e x e r t i n g a p u l l e q u a l t o h i s$ $w e i g h t, w i l l i t t a k e t o p u l l t h e$ $d o o r o p e n w h e n t h e c h a m b e r i s$ $e v a c u a t e d t o a p r e s s u r e o f 0.1 a t m ?$
Commented by NECx last updated on 09/Mar/18

$p l e a s e s h o w w o r k i n g s .^{}$
	Answered by ajfour last updated on 09/Mar/18

	$F = 0.9 \times 1 \times 10^{5} \times \frac{1}{25} = N \times 700$ $\Rightarrow N = \frac{9 \times 10^{4}}{25 \times 700} = \frac{9 \times 100}{25 \times 7} = \frac{36}{7}$ $\Rightarrow A t l e a s t 6 m e n .$
	Commented by NECx last updated on 10/Mar/18

	$h o w c o m e 0.9, a n d 1 / 25$
	Commented by ajfour last updated on 10/Mar/18

	$N e t p r e s s u r e f r o m o u t s i d e$ $= 1 a t m - 0.1 a t m = 0.9 a t m$ $A r e a = 20 c m \times 20 c m = \frac{1}{25} m^{2} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum