A wooden plank of length $$1$$ m and uniform cross section is hinged at one end to the bottom of a tank as shown in the figure. The tank is filled with water up to a height of $$0.5$$ m. The specific gravity of the plank is $$0.5$$. If the angle θ by the inclination of that the plank makes with the vertical in the equilibrium position (exclude$$ the case θ $$=$$ $$0). Find the value of $$1cos2$$⁡θ.

Question

A wooden plank of length $$1$$ m and uniform cross section is hinged at one end to the bottom of a tank as shown in the figure. The tank is filled with water up to a height of $$0.5$$ m. The specific gravity of the plank is $$0.5$$. If the angle θ by the inclination of that the plank makes with the vertical in the equilibrium position (exclude$$ the case θ $$=$$ $$0). Find the value of $$1cos2$$⁡θ.

Infinity Learn · Accepted Answer

Let y be the length of the plank inside water.$$y=0.5cos$$⁡θLet A be the $$cross-sectional$$ area of the plank. Then the buoyant force on it is $$Fb=V$$ρω$$g=(Ay)$$ρwgSince the plank is in rotational equilibrium, so ∑τ→$$0=0$$ or mg×$$l2sin$$⁡θ−Fb×$$y2sin$$⁡θ$$=0$$ or mgl−Fb×$$y=0$$ or $$(Al$$×$$0.5)gl$$−$$(Ay$$ρ$$wyg=0$$ or $$0.5l2=y2$$ or $$0.5$$×$$(1)2=0.5cos$$⁡θ$$2$$ or $$cos2$$⁡θ$$=12$$

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