A wooden log of 0.6 m diameter and 5m in length is floating in river water. Find the depth of the wooden log in water when the specific gravity of the log is 0.7.

SOLUTION:

Given,

Diameter of wooden log, `D=0.6 m`

Length of log,`L=5 m`

Specific gravity of log,`S=0.7`

Here,

Weight of wooden log `=V*rhog`

`=A*L*rhog`

`=pi/4**0.6^2**5**0.7**1000**9810`

`=9707.99 N`

Now, for equilibrium,

Weight of wooden log = weight of water displaced

=Volume  of water displaced ** weight density

or, volume of water displaced `=9707.99/(9.81**1000)`

`=0.9896 m^3`

Let h be the depth of immersion.

Now, 

Volume of log inside water = volume of water displaced 

or.... Show More

SOLUTION:

Given,

Diameter of wooden log, `D=0.6 m`

Length of log,`L=5 m`

Specific gravity of log,`S=0.7`

Here,

Weight of wooden log `=V*rhog`

`=A*L*rhog`

`=pi/4**0.6^2**5**0.7**1000**9810`

`=9707.99 N`

Now, for equilibrium,

Weight of wooden log = weight of water displaced

=Volume  of water displaced ** weight density

or, volume of water displaced `=9707.99/(9.81**1000)`

`=0.9896 m^3`

Let h be the depth of immersion.

Now, 

Volume of log inside water = volume of water displaced 

or, area of ACDA ** Length =0.9896

or, area of sector OADC + area of `triangle AOC` = `0.9896/5`

or,`pi**0.3^2**(180+2theta)/360+½*rsintheta*(rcostheta+rcostheta)=0.9896/5`

Solving this, we get,

`theta=18.64^0`

Again from figure,

`h=r+rsintheta`

`=0.3959 m`