A wooden block of volume `0.03 m^3` weighing 210 N is attached to one end of a 3.5 m long wooden rod are under hinged to a wall. The block and part of the rod are under water in the position shown in the figure. Find the inclination `theta` of the rod with the horizontal. The rod weighs 15 N and has a uniform sectional area of 2000 `mm^2`.

SOLUTION:

Here,

Weight of the wooden block =210 N

Weight of the rod = 15 N

Thus, the upthrust on block is,

`=V*rho*g`

`=0.03**1000**9.81`

`=294.3`

Similarly,the upthrust on rod is,

`=V_(i n)*rho*g`

`=2000**10^(-6)**1000**9.81**(3.5-0.4/(sintheta))`

`=19.62 (3.5-0.4/(sintheta))`

Now, taking moment about A, we get,

 `294.3**3.5 costheta+19.62 (3.5-0.4/(s.... Show More

SOLUTION:

Here,

Weight of the wooden block =210 N

Weight of the rod = 15 N

Thus, the upthrust on block is,

`=V*rho*g`

`=0.03**1000**9.81`

`=294.3`

Similarly,the upthrust on rod is,

`=V_(i n)*rho*g`

`=2000**10^(-6)**1000**9.81**(3.5-0.4/(sintheta))`

`=19.62 (3.5-0.4/(sintheta))`

Now, taking moment about A, we get,

 `294.3**3.5 costheta+19.62 (3.5-0.4/(sintheta))**1/2 (3.5+0.4/(sintheta))*costheta=210**3.5costheta+15**3.5/2costheta`

or,`1030.05+9.81 (12.25-0.16/(sin^2theta))=735+26.25`

Solving this,we get,

`theta=3^0 28'`