Chapter 13.1, Problem 3PP

To determine

To Calculate:The pressure that the brick exerts on the ground.

Answer to Problem 3PP

  $23\text{ kPa}$

Explanation of Solution

Given data:

Dimensions of brick are:

  $\begin{array}{l}h=5.0\text{ cm}=0.05\text{ m}\\ w=10.0\text{ cm}=0.1\text{ m}\\ l=20.0\text{ cm}=0.2\text{ m}\end{array}$

Lead has a density = $\rho =11.8{\text{ g/cm}}^{\text{3}}$

Formula used:

The volume of a brick: $V=lwh$

Mass of the brick:

  $m_{brick}=\rho V=\rho lwh$

A force (per unit area) that acts on any object is defined by pressure. Pressure on any object is calculated by using formula

  $P=\frac{F}{A}$

Where, F = force, A = Area

  $\begin{array}{l}F=m_{brick}g\\ A=hw\end{array}$

Calculation:

  $\begin{array}{c}F=m_{brick}g\\ =\rho \times l\times w\times h\times g\\ =(11.\text{8}{\text{g/cm}}^3)({10}^{-3}\text{ kg/g})\times \left(\text{5}\text{.0 cm}\right)\left(\text{10}\text{.0 cm}\right)\text{×}\left(\text{20 cm}\right)\text{(9}{\text{.80m/s}}^{\text{2}})\\ \approx 116\text{ N}\end{array}$

Calculate the area

  $\begin{array}{c}A=h\times w\\ =\left(\text{0}\text{.050 m}\right)\left(\text{0}\text{.100 m}\right)\\ =0.0050{\text{ m}}^{\text{2}}\end{array}$

Substitute in the formula to find the pressure:

  $\begin{array}{c}P=\frac{F}{A}\\ \text{=}\frac{\left(\text{116 N}\right)}{\left(\text{0}{\text{.0050 m}}^{\text{2}}\right)}\\ \approx \text{23 kPa}\end{array}$

Conclusion:

The pressure that the brick exerts on the ground is $23\text{ kPa}$ .