Chapter 10, Problem 80GP

To determine

To Find: The buoyant force on a 3 kg granite rock.

Answer to Problem 80GP

The buoyant force on a 3 kg granite rock is $99.96\text{ N}\text{.}$ The rock will not float as buoyant force on rock is more than the buoyant force on water.

Explanation of Solution

Given info:

Acceleration, $a=2.4g$

Mass of rock, $m=3\text{ kg}$

Specific gravity of rock, $S{G}_{\text{rock}}=2.7$

Formula used:

Write the expression for force.

  $F_{{\text{b}}_{\text{water}}}-m_{\text{water}}\times g=m_{\text{water}}\times a$

Write the expression for density.

  $\text{Density }\rho \text{ = }\frac{\text{Mass (}m\text{)}}{\text{Volume (}V\text{)}}$

Write the expression for mass.

  $m=\rho \times V$

Calculation:

Substitute the given values, the buoyant force exerted by wateris,

  $\begin{array}{l}{F}_{{\text{b}}_{\text{water}}}-m_{\text{water}}\times g=m_{\text{water}}\times a\\ F_{{\text{b}}_{\text{water}}}=m_w(a+g)\\ m=\rho \times VandV=\frac{m}{\rho }\\ F_{b_w}={\rho }_w\times V_{\text{rock}}\times (a+g)\\ F_{b_w}={\rho }_w\times \frac{m_{\text{rock}}}{{\rho }_{\text{rock}}}\times (a+g)\\ F_{b_w}=\frac{m_{\text{rock}}}{S{G}_{\text{rock}}}\times (a+g)\\ F_{b_w}=\frac{3}{2.7}\times (2.4g+g)=\frac{3}{2.7}\times (3.4\times 9.8)\\ F_{b_w}=37.022\text{ N}\end{array}$

Buoyant force exerted by rock is calculated by following.

  $\begin{array}{l}{F}_{{\text{b}}_{\text{rock}}}-m_{\text{rock}}\times g=m_{\text{rock}}\times a\\ F_{{\text{b}}_{\text{rock}}}=m_r(a+g)\\ F_{{\text{b}}_{\text{rock}}}=3(2.4g+g)\\ F_{{\text{b}}_{\text{rock}}}=3\times 3.4g\\ F_{{\text{b}}_{\text{rock}}}=99.96\text{ N}\end{array}$

So, the buoyant force of rock is more than the buoyant force of water. So, force of water does not support the weight of rock. That's why rock will sink.

Conclusion:

Thus, the buoyant force on a 3 kg granite rock is $99.96\text{ N}\text{.}$ The rock will not float as buoyant force on rock is more than the buoyant force on water.