# Integration and forces

• Jan 21st 2009, 08:20 PM
razorfever
Integration and forces
the question tells us that the buoyant force is equal to the weight of the fluid the object displaces

po = object density
pf = fluid density
g = acceleration due to gravity
A(y) = cross-sectional area of object
h = height of object underwater
L = total height of object
F = buoyant force
W = weight

and then

F = pf*g * integral from -h to 0 of A(y) dy

and

W = po*g * integral from -h to (L-h) of A(y) dy

show that the percentage of the volume of the object above the surface of the liquid is

100 * (pf - po) / pf
• Jan 22nd 2009, 01:56 AM
Volumes and integration
Hello razorfever
Quote:

Originally Posted by razorfever
the question tells us that the buoyant force is equal to the weight of the fluid the object displaces

po = object density
pf = fluid density
g = acceleration due to gravity
A(y) = cross-sectional area of object
h = height of object underwater
L = total height of object
F = buoyant force
W = weight

and then

F = pf*g * integral from -h to 0 of A(y) dy

and

W = po*g * integral from -h to (L-h) of A(y) dy

show that the percentage of the volume of the object above the surface of the liquid is

100 * (pf - po) / pf

First, note that, since the body is floating in equilibrium:

$F = W$ (1)

Then let the volume of the body above water be $V_a$, and the total volume $V$.

Then $V = \frac{W}{\rho_og}$

and $V_a = \int_0^{L-h}A(y)dy$

$= \int_{-h}^{L-h}A(y)dy -\int_{-h}^0A(y)dy$

$= \frac{W}{\rho_og} - \frac{F}{\rho_fg}$

$= \frac{W}{g}\left(\frac{\rho_f-\rho_o}{\rho_o\rho_fg}\right)$, from (1)

Can you complete it now?