Question

How much extra water does a 21.5 ft, 175-lb concrete canoe displace compared to an ultra-lightweight 38-lb Kevlar canoe of the same size carrying the same load?

Answer 1

Answer:

The volume of the extra water is $2.195 ft^{3}$

Solution:

As per the question:

Mass of the canoe, $m_{c} = 175 lb + w$

Height of the canoe, h = 21.5 ft

Mass of the kevlar canoe, $m_{Kc} = 38 lb + w$

Now, we know that, bouyant force equals the weight of the fluid displaced:

Now,

$V\rho g = mg$

$V = \frac{m}{\rho}$                                  (1)

where

V = volume

$\rho = 62.41 lb/ft^{3}$ = density

m = mass

Now, for the canoe,

Using eqn (1):

$V_{c} = \frac{m_{c} + w}{\rho}$

$V_{c} = \frac{175 + w}{62.41}$

Similarly, for Kevlar canoe:

$V_{Kc} = \frac{38 + w}{62.41}$

Now, for the excess volume:

V = $V_{c} - V_{Kc}$

V = $\frac{175 + w}{62.41} - \frac{38 + w}{62.41} = 2.195 ft^{3}$