## Engineering Mechanics: Statics & Dynamics (14th Edition)

$F=303lb$
The required force can be determined as follows: $m=\frac{\gamma}{g}\times Q=\frac{62.4}{32.2}\times 2=3.876slug/s$ Similarly $v_{Ay}=\frac{2}{0.02778}=72ft/s$ and the velocity of the water in the horizontal direction is $v_{Bx}=\frac{2}{0.08333}=24ft/s$ We apply the impulse and momentum principle in the vertical direction $F_y=\frac{dm}{dt}(v_{Ay}-v_{By})$ $\implies F_y=3.876(72-0)=279.072lb$ Now we apply the impulse and momentum principle in the horizontal direction $\Sigma F_x=\frac{dm}{dt}(v_{Ax}-v_{Bx})$ We plug in the known values to obtain: $24-F_x=3.8760(0-24)$ $\implies F_x=117.01lb$ The required resultant force can be calculated as $F=\sqrt{F_x^2+F_y^2}$ We plug in the known values to obtain: $F=\sqrt{(117.01)^2+(279.02)^2}$ This simplifies to: $F=303lb$