Answer
$F_x=19.5lb$, $F_y=1.9lb$
Work Step by Step
We know that:
$\frac{dm}{dt}=\rho\times v\times A$
$\frac{dm_A}{dt}=\frac{62.4}{32.2}\times 12\pi\times (\frac{0.25}{12})^2=0.03171slug/s$
Similarly
$\frac{dm_B}{dt}=\frac{62.4}{32.2}\times 25\pi \times (\frac{0.25}{12})^2=0.06606slug/s$
and $\frac{dm_c}{dt}=\frac{dm_A}{dt}+\frac{dm_B}{dt}=0.03171+0.06606=0.097777slug/s$
Now the balancing force in the $x$ direction is given as
$\Sigma F_x=\frac{dm_A}{dt}v_{Ax}+\frac{dm_B}{dt}v_{Bx}+\frac{dm_C}{dt}v_{Cx}$
We plug in the above known values to obtain:
$F_x=19.5lb$
and the balancing force in the $y$ direction is
$\Sigma F_y=\frac{dm_A}{dt}v_{Ay}+\frac{dm_B}{dt}v_{By}+\frac{dm_C}{dt}v_{Cy}$
We plug in the known values to obtain:
$F_y=1.9lb$
