Question

A penguin is held under the surface of seawater by a tension of .18 pounds. His density is 1.8 slugs per cubic foot. What is his volume?

Answer 1

Task:

A penguin is held under the surface of seawater by a tension of 0.18 pounds. His density is 1.8 slugs per cubic foot. What is his volume?

Solution:

Assuming Archimedes' principle to be reformulated as follows,

$$F _ {A} = \rho g V,$$

1.8 slugs per cubic foot = $\frac{1.8 \cdot 14.593903 \, \text{kg}}{0.3048^3 \, \text{m}^3} = 927.682 \, \frac{\text{kg}}{\text{m}^3}$

0.18 pounds = $0.18 \cdot 0.45359237 \, \text{kg} = 0.082 \, \text{kg}$

As penguin is held under the surface of seawater $F_{A} = \text{tension} + \text{penguin's weight}$

$F_{A} = 0.082 \, \text{kg} \cdot 9.81 \, \frac{\text{N}}{\text{kg}} + \text{penguin's weight}$

penguin's weight = $927.682 \, \frac{\text{kg}}{\text{m}^3} \cdot V$

$1030 \, \frac{\text{kg}}{\text{m}^3} \cdot 9.81 \, \frac{\text{N}}{\text{kg}} \cdot V = 0.082 \, \text{kg} \cdot 9.81 \, \frac{\text{N}}{\text{kg}} + 927.682 \, \frac{\text{kg}}{\text{m}^3} \cdot V$

$V = 0.0000876597 \, \text{m}^3 = 0.00309567 \, \text{ft}^3$

Answer:

$V = 0.00309567 \, \text{ft}^3$