The wooden beam shown in figure is 200 mm ** 200 mm ** 5 m long. It is hinged at A and remains in equilibrium at `ta` with horizontal. Find inclination `ta`. The specific gravity of wood can be taken as 0.70.

SOLUTION:

Let, the specific weight of water `=w`

Thus, the specific weight of wood becomes `0.6w`

Let `x m` length of the beam is immersed under water as shown in figure.

Now, the weight of the beam acting at point G i.e, c.g. of the beam is,

`=V*rho*g`

`=0.2**0.2**5**0.6w`

`=0.12w`N

Similarly the upthrust on the wooden beam is,

`=V_(in)*rho_w*g`

`=0.2**0.2**x.w`

`=0.04 wx` N

This upthrust acts at point E i.e, at the point of centre of buoyancy.

`A....Show More

SOLUTION:

Let, the specific weight of water `=w`

Thus, the specific weight of wood becomes `0.6w`

Let `x m` length of the beam is immersed under water as shown in figure.

Now, the weight of the beam acting at point G i.e, c.g. of the beam is,

`=V*rho*g`

`=0.2**0.2**5**0.6w`

`=0.12w`N

Similarly the upthrust on the wooden beam is,

`=V_(in)*rho_w*g`

`=0.2**0.2**x.w`

`=0.04 wx` N

This upthrust acts at point E i.e, at the point of centre of buoyancy.

`AG=5/2=2.5`

`AE=5-x/2`

Now, taking moments about the hinge A, we get,

`0.12w**2.5 costheta=0. 04 wx (5-x/2)costheta`

or,`0.3=0.20x-0.02x^2`

Solving this, we get,

`x=1.838 m`

Also,

`sintheta=1/(5-1.838)`

or,`theta=18^0 26'`