Chapter 13.3, Problem 30PP

To determine

To Calculate:The weight of bricks that could be stacked on a slab of foam.

Answer to Problem 30PP

  $\text{8}\text{.8}{\text{×10}}^{\text{2}}\text{N}$

Explanation of Solution

Given data:

Volume of slab of foam, $V=\text{1}\text{.0}\text{m}\text{×}\text{1}\text{.0}\text{m}\text{×}\text{0}\text{.10}\text{m}$

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 plastic foam, ${\rho }_{foam}\text{=0}\text{.10 }{\rho }_{water}\text{=1}{\text{.0×10}}^{\text{2}}{\text{kg/m}}^{\text{3}}$

Formula used:

The upward force or buoyant force can be calculated by using formula, $F_{buoyant}\text{ =}\rho Vg$ .

Where, $\rho$ is the density of

Calculation:

The foam would displace $V=\left(\text{1}\text{.0}\text{m}\right)\left(\text{1}\text{.0}\text{m}\right)\left(\text{0}\text{.10}\text{m}\right)\text{=}\text{0}\text{.10}{\text{m}}^{\text{3}}$ of water.

The weight of foam is $F_{g,foam}={\rho }_{foam}Vg$

  $=\left(1.0\times {10}^2{\text{kg/m}}^3\right)\left(0.10{\text{m}}^3\right)\left(9.80{\text{m/s}}^2\right)$

  $=\text{98}\text{N}$

The buoyant force is $F_{buoyant}\text{ =}{\rho }_{water}Vg$

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

  $\text{=}\text{980}\text{N}$

The weight of brick that a person can stack is

  $F_{g,brick}=F_{buoyant}-F_g{}_{foam}$

  $\text{=}\text{980}\text{N}\text{-}\text{98}\text{N}$

  $=\text{8}\text{.8}{\text{×10}}^{\text{2}}\text{N}$

Conclusion:

The weight of bricks is $\text{8}\text{.8}{\text{×10}}^{\text{2}}\text{N}$ .