InterviewSolution
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InterviewSolution
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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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Answer» The force acting on the plane are shown in the ure. The height of water level is l=0.5m. The length of the plank is 1.0=2l. The weight of the lane acts through the centre B of the plank. We have OB=l. The buoyant fore 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 plane be rho`. it 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=(2/p)/(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)=((2lrho)/(costheta))(l/(2costheta))` `or, cos^2theta= 1/2` `or costheta=1/sqrt2, or theta=45^@` |
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