Question

A rectangular barge floats in freshwater. When a 400 kg block is loaded on the 5 m long by 2 m wide barge,
the barge sinks a few centimeters deeper. How much deeper does the barge lower?

Answer 1

Answer on Question #49184, Physics, Mechanics | Kinematics | Dynamics

A rectangular barge floats in freshwater. When a $400\mathrm{kg}$ block is loaded on the $5\mathrm{m}$ long by $2\mathrm{m}$ wide barge, the barge sinks a few centimeters deeper. How much deeper does the barge lower?

Solution.

Before:

Due to $1^{\mathrm{st}}$ Newton's law:

$$M g = F _ {A 1}$$

After:

Due to $1^{\mathrm{st}}$ Newton's law:

$$M g + m g = F _ {A 2}$$

By definition:

$$F _ {A} = \rho_ {\text {w a t e r}} g V _ {\text {u n d e r \_ w a t e r}}$$

So:

$$\left\{ \begin{array}{c} M g = \rho_{w} g S h \\ M g + m g = \rho_{w} g S (h + \Delta h) \end{array} \right.$$

So:

$$M g + m g = \rho_{w} g S (h + \Delta h) = \rho_{w} g S h + \rho_{w} g S \Delta h = M g + \rho_{w} g S \Delta h$$ $$m g = \rho_{w} g S \Delta h$$ $$\Delta h = \frac{m}{\rho_{w} S} = \frac{m}{\rho_{w} \cdot a b}$$

Where $a$ is the width and $b$ is the length of barge.

Numerically:

$$\Delta h = \frac{400 \, kg}{1000 \, \frac{kg}{m^{3}} \cdot 5m \cdot 2m} = 0.04m = 4cm$$

Answer: $\Delta h = 4cm$