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The wooden plank of length 1 m and unifo...

The wooden plank of length 1 m and uniform cross section is hinged at one end to the bottom of a tank as shown in figure. The tank is filled with water up to a height of 0.5 m. The specific gravity of the plank is 0.5. Find the angle `theta` that the plank makes with the vertical in the equilibrium position. (Exclude the case `theta=0`)
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The forces acting om the plank are shown in the figure. The height of water gravity of the plank is 0.5. Find the angle 0 the pank 1.0=2l. The weight of the plank acts throught he centre B of the plank. We have `OB=` The buoyant force F acts through the point A which is the middle point of the dipped part OC of the plank.
We have `OA=(OC)/(2)=(l)/(2costheta)`
Let the mass per unit length of the plank be `rho`. its weight `mg=2lrhog`.
The mass of the part OC of the plank `=((l)/(costheta))rho`.
the mass of water displaced `=(1)/(0.5)(l)/(costheta)rho(2lrho)/(costheta)`
The buoyant force F is therefore `F=(2lrhog)/(costheta)`
Now, for equilibrium the torque of mg about O should balance the torque of F about O.
So, mg `(OB)sintheta=F(OA)sintheta` or , `(2lrho)l=((2lrho)/(costheta))((l)/(2costheta))` or, `cos^(2)theta=(1)/(2)` or `costheta=(1)/(sqrt(2))`, or `theta=45^(@)`
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