Question

When an open-faced boat has a mass of 5750 kg, including its cargo and passengers, it floats with the water just up to the top of its gunwales (sides) on a freshwater lake. (a) What is the volume of this boat? (b) The captain decides that it is too dangerous to float with his boat on the verge of sinking, so he decides to throw some cargo overboard so that 20% of the boat’s volume will be above water. How much mass should he throw out?

Short answer

a) The volume of the boat is $5.75{\mathrm{m}}^3$.

b) The mass of cargo thrown out by the captain must be 1150 kg .

Step-by-step solution

Identification of given data

The given data can be listed below,

• The mass of the open-faced boat is, $m_b=5750\mathrm{kg}$.
• The cargo thrown overboard is $20\%$.

Significance of buoyant force

A buoyant object will be buoyant due to the top and bottom forces. This is because height is directly proportional to the base and the top.

Buoyant force results from multiplying the difference between pressure and the area of the object, which corresponds to the net force on that object.

(a) Determination of the volume of the boat

When a boat floats, so the buoyant force on it equals the weight of the object given by,

$B=m_{b}g$

Also, the buoyant force can be expressed as,

$B={\rho }_{w}gV$

Here, ${\rho }_w$is the density of water whose value is $\left(1.00\times {10}^3\mathrm{kg}/{\mathrm{m}}^3\right)$, V is the volume of the boat, and g is the acceleration due to gravity.

Equate the equations mentioned above; the volume of the boat can be calculated as,

$$\begin{gathered} {\rho }_{w}gV=m_{b}g \\ V=\frac{m_b}{{\rho }_w} \end{gathered}$$

Substitute values in the above equation.

$$\begin{gathered} V=\frac{5750\mathrm{kg}}{1.00\times {10}^3\mathrm{kg}/{\mathrm{m}}^3} \\ =5.75{\mathrm{m}}^3 \end{gathered}$$

Thus, the volume of the boat is $5.75{\mathrm{m}}^3$.

(b) Determination of the mass of cargo thrown by the captain

After cargo is thrown in water, 20% of the boat’s volume will be above water, so 80% of the boat is submerged.

The volume of the submerged boat is expressed as,

$V_{sub}=0.80V$

Here, V is the volume of the boat.

The mass of the submerged boat can be expressed as,

$m={\rho }_w{V}_{sub}$

Here, ${\rho }_w$is the density of water and $V_{sub}$is the volume of the boat submerged in water.

Substitute values in the above equation.

$$\begin{gathered} m={\rho }_w\left(0.80V\right) \\ =\left(1.00\times {10}^3\mathrm{kg}/{\mathrm{m}}^3\right)\left(0.80\right)\left(5.75{\mathrm{m}}^3\right) \\ =4600\mathrm{kg} \end{gathered}$$

The mass of cargo thrown out is calculated as,

$m_{cargo}=m_b-m_{sub}$

Substitute values in the above equation.

$$\begin{gathered} m_{cargo}=5750\mathrm{kg}-4600\mathrm{kg} \\ =1150\mathrm{kg} \end{gathered}$$

Thus, the mass of cargo thrown out by the captain must be $1150\mathrm{kg}$.