Chapter 13.3, Problem 27PP

To determine

To Calculate: The net force on a $\text{0}\text{.20}{\text{m}}^{\text{3}}$ block of bricks under water.

Answer to Problem 27PP

The net force is $\text{1}{\text{.6×10}}^{\text{3}}\text{N}$ .

Explanation of Solution

Given data:

Volume of block, $V\text{=0}\text{.20}{\text{m}}^{\text{3}}$

Acceleration due to gravity, $g=\text{9}\text{.8}{\text{m/s}}^{\text{2}}$

Density of water, ${\rho }_{water}\text{=1}{\text{.00×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}$

Density of brick, ${\rho }_{brick}=1.8{\rho }_{water}=\text{1}{\text{.8×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}$

Formula used:

The apparent weight can be expressed by the equation

  $F_{apparent}=F_g -F_{buoyant}$

Where, $F_g$ is the force of gravity and $F_{buoyant}$ is the buoyant force.

Calculation:

  $F_{apparent}=F_g -F_{buoyant}$

  $={\rho }_{brick}Vg-{\rho }_{water}Vg$

  $=\left({\rho }_{brick}-{\rho }_{water}\right)Vg$

By putting all the values in above formula,

  $\text{=}\left(\text{1}{\text{.8×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}\text{-}\text{1}{\text{.00×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}\right)\left(\text{0}\text{.20}{\text{m}}^{\text{3}}\right)\left(\text{9}\text{.80}{\text{m/s}}^{\text{2}}\right)$

  $\text{=1}{\text{.6×10}}^{\text{3}}\text{N}$

Conclusion:

The net force on a $\text{0}\text{.20}{\text{m}}^{\text{3}}$ block of bricks under wateris $\text{1}{\text{.6×10}}^{\text{3}}\text{N}$ .