Question

A 6.00 kg aluminum metal is suspended in a rope and partially immersed in water (𝜌𝑤𝑎𝑡𝑒𝑟 = 1000 kg/m3). If the tension in the rope is 4.55 N, what volume of aluminum was immersed? 

Answer 1

$$F_b=mg-T=6.00*9.8-4.55=54.3\:\rm N$$

$$F_b=g\rho V_s$$

$$V_s=\frac{F_b}{g\rho}=\frac{54.3}{9.8*1000}=5.54*10^{-3}\;\rm m^3$$