Chapter 19, Problem 53A

Phil can support 100 N of iron $\left(\rho =78,000kg/m^3\right)$ in water. How many newtons can he support in air?

To determine

Weight supported by person P in the air.

Answer to Problem 53A

The weight supported by person P in air is 87 N.

Explanation of Solution

Given:

Weight supported by person P in water is 100 N.

Density of iron, ${\rho }_{\text{iron}}=7800{\text{kg/m}}^3$

Formula Used:

The buoyancy force is given as,

  $F_{\text{buoyancy}}=\rho Vg$

Where, $\rho$ is density, $V$ is volume of liquid displaced and $g$ is gravitational acceleration.

Calculation:

When the iron is supported in water, the buoyancy force on iron is

  ${\rho }_{\text{w}}{V}_{\text{iron}}g={\rho }_{\text{w}}\left(\frac{m}{{\rho }_{\text{iron}}}\right)g=mg\left(\frac{{\rho }_{\text{w}}}{{\rho }_{\text{iron}}}\right)$

Taking density of water ${\rho }_{\text{w}}=1000{\text{kg/m}}^3$

Now substituting the value in above equation

  $\begin{array}{l}{F}_{\text{buoyancy}}=\left(100\text{N}\right)\left(\frac{1000{\text{kg/m}}^3}{7800{\text{kg/m}}^3}\right)\\ \Rightarrow F_{\text{buoyancy}}=13\text{N}\end{array}$

There the weight that person P can support in the air is the difference of weight in water and buoyancy,

  $\text{100}\text{N}-13\text{N}=\text{87}\text{N}$

Conclusion:

Therefore, the weight supported by person P in air is 87 N.