Question

A crab fisherman has built a crab trap out of plastic pipe and wire mesh. The overall mass and volume of the trap are 5.59 kg and 6,213 cm^3, respectively. To catch crab, the trap must sink to the ocean floor. The fisherman has several lead weights to add to the trap to ensure it sinks. If sea water has a density of 1,021 kg/m^3 , and each lead weight has mass of 113.4 g and volume of 10.0cm^3, what is the minimum number of weights the fisherman must add so that the trap sinks to the ocean floor?

Answer 1

To develop this problem it is necessary to apply the concepts related to the force of gravity, and the force expressed in terms of density and volume, by definition we know that,

$m = \rho V$

Where,

$\rho =$Density

V = Volume

In addition we also know that by Newton's second law the weight can be defined as

$F_w = mg$

For balance to exist, the Bouyancy force is equal to the weight, therefore

$F_w = F_b$

$mg = V \rho g$

$m = V \rho$

We know that the total weight is conditioned at 5.59 Kg plus the number of objects in Kilograms, that is

$m = 5.59+n(\frac{113.4}{1000})$

We know that the Volume is the equivalent of $6,214 * 10 {- 6} m ^ 3$ plus the volume of each weight of $10 * 10{ - 6} m ^ 3$, so,

$V = \frac{6213+n*10}{10^6}$

Replacing with our values we have that

$5.59+n(\frac{113.4}{1000}) = \frac{6213+n*10}{10^6}*1021$

$559+0.1134n=6.3434+0.01021n$

$0.10319n = 0.7534$

$n = 7.301$

Therefore the minimum number of weights is 8.