Question

A ship of draft 10 m is loading in salt water density 1015 kg/m3. Calculate the cheng in draught on entering salt water of density 1025 kg/m3.

Answer 1

Answer on Question 49750, Physics, Mechanics | Kinematics | Dynamics

Question:

A ship of draft $10\mathrm{m}$ is loading in salt water density $1015\mathrm{kg / m3}$ . Calculate the change in draft on entering salt water of density $1025\mathrm{kg / m3}$ .

Solution:

Let's consider a ship that moves from salt water of density $1015\frac{kg}{m^3}$ to salt water of density $1025\frac{kg}{m^3}$ . The ship must displace the same mass of water in each case. The mass of water displaced in case of salt water with density $1015\frac{kg}{m^3}$ will be:

$$m _ {w. o l d} = \rho_ {w. o l d} V _ {\text {s h i p u n d e r w a t e r l i n e}} = \rho_ {w. o l d} (w / l) b d _ {o l d},$$

where $\rho_{w.old} = 1015\frac{kg}{m^3}$ is the old density of salt water, $w / l$ is the length of the ship at waterline, $b$ is the beam of the ship, the ship's width at the widest point as measured at the ship's nominal waterline, $d_{old} = 10m$ is the old draft.

The mass of water displaced in case of salt water with density $1025\frac{kg}{m^3}$ will be:

$$m _ {w. n e w} = \rho_ {w. n e w} V _ {\text {s h i p u n d e r w a t e r l i n e}} = \rho_ {w. n e w} (w / l) b d _ {n e w}.$$

So, we equate both expressions:

$$\rho_{w.old} \left(w / l\right) b d_{old} = \rho_{w.new} \left(w / l\right) b d_{new}.$$

From this formula we obtain the new draft:

$$d_{new} = \frac{\rho_{w.old} \cdot d_{old}}{\rho_{w.new}} = \frac{1015 \frac{kg}{m^3} \cdot 10m}{1025 \frac{kg}{m^3}} = 9.9m.$$

**Answer:**

The draft in salt water of density $1025 \frac{kg}{m^3}$ will be 9.9 meters.

https://www.AssignmentExpert.com