Answer
(a) $9.58\times 10^{-3}m^3$
(b) $10.1Kg$
Work Step by Step
We know that
$F_b+N-mg=0$
$\implies \rho_w Vg+N-mg=0$
This can be rearranged as:
$V=\frac{mg-N}{\rho_wg}$
We plug in the known values to obtain:
$V=\frac{685-407}{(1000)(9.81)}=0.01916m^3$
Now $V_{leg}=\frac{V}{2}=\frac{0.01916}{2}=9.58\times 10^{-3}m^3$
(b) The mass of each leg can be determined as
$m_{leg}=\rho_{leg}V_{leg}=(1.05\rho_w)V_{leg}$
We plug in the known values to obtain:
$m_{leg}=1.05(1000)(0.00958m^3)=10.1Kg$
