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

$50.0lb$
We can determine the required force as follows: We know that: $\frac{dm}{dt}=\rho Q$ $\implies \frac{dm}{dt}=(1040)(0.04)=41.6=Kg/s$ The velocity of the jet can be calculated as $v_a=\frac{Q}{A}$ $\implies v_a=\frac{0.04}{(0.051)^2)(3.14)}=4.90m/s$ Now, $F=\frac{dm}{dt}[(v_b)_y-(-v_a)_y]$ We plug in the known values to obtain: $F_y=41.6(0+4.9)$ $F=203.64N=50.0lb$