Chapter 11, Problem 1P

To determine

The weight of the pillar in pounds.

Answer to Problem 1P

  $8.37\times {10}^3\text{lb}$

Explanation of Solution

Given:

Height of the pillar, $h=2.2\text{m}$

Radius of the pillar, $r=0.50\text{m}$

Density of concrete, $\rho =2.2\times {10}^3{\text{kg/m}}^{\text{3}}$

Formula used:

Density = Mass / Volume

Mass = Density × Volume

  $\begin{array}{l}\rho =\frac{m}{v}\\ m=\rho \times v\end{array}$

  $m=$ mass of the pillar

  $\rho =$ density of concrete

  $v=$ volume of the pillar

Calculation:

Let the mass of the pillar be $m$ .

Volume of the pillar can be denoted by $v$ .

Volume of the pillar is given by,

  $\begin{array}{l}v=\pi r^{2}h\\ $ v=\pi \times {(0.50)}^2\times 2.2$v=1.727{\text{m}}^{\text{3}}\end{array}$

Density = Mass / Volume

Mass = Density × Volume

  $\begin{array}{l}\rho =\frac{m}{v}\\ m=\rho \times v\\ m=(2.2\times {10}^3)\times 1.727\\ m=3.7994\times {10}^3\text{kg}\end{array}$

Weight of pillar = Mass × Acceleration due to gravity

  $\begin{array}{l}W=m\times g\\ W=3.7994\times {10}^3\times 9.8\\ W=37.234\times {10}^3\text{N}\end{array}$

According to the question,

  $\begin{array}{l}\text{1 N=0}\text{.2248}\text{lb}\\ W=37.234\times {10}^3\text{N}\\ W=37.234\times 0.2248\times {10}^3\\ W=8.37\times {10}^3\text{lb}\end{array}$

Conclusion:

The weight of the pillar in pounds is $8.37\times {10}^3\text{lb}$ .