Question

an alluminion object mass 27.0kg and density 2.70x10 kgm is attached to a string and immersed in a tank of water. Determined the tension in the string

Answer 1

The density of aluminium is

$$\rho_{Al}= 2.70\times10^3 \dfrac{kg}{m^3}= \dfrac{M_{Al}}{V_{Al}}$$

$$\Rightarrow V_{Al}=\dfrac{M_{Al}}{\rho_{Al}}$$

$=\dfrac{27.0}{2.70\times 10^2}=0.01m^3$

Applying Newton's second law for the aluminum block and noting that it is completely submerged and in equilibrium. We have

$$\sum {F=0}= T+B+mg$$ $$\Rightarrow T=mg-B\\=mg-\rho_{water}gV_{water}\\=(27.0)(9.8)-(1.00\times 10^3)(9.8)(0.01)\\=166.6N$$