Chapter 17: Problem 3

Question:        An office window is $3.43\ {\rm m}$ by $2.08\ {\rm m}$. As a result of the passage of a storm, the outside air pressure drops to $0.962\ {\rm atm}$, but inside the pressure is held at $1.00\ {\rm atm}$. What net force pushes out on the window?

Solution:         Given the conversion factor $1.00\ {\rm atm} = 1.01\times 10^5\ {\rm N/m}$, we have that

$$\vert{\bf F}\vert _{\rm out} \left( P_{\rm in} - P_{\rm out}\right) . A \ \ =\ \ 1.01\times 10... ...eft( 1 - 0.962\right)\ (2.08)\ (3.43) \ \ =\ \ 2.74\times 10^4\ {\rm N} \ \ \ .$$ = (1)