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.

$y = \frac{0 .5}{\cos θ}$

Let A be the cross-sectional area of the plank. Then the buoyant force on it is $F_{b} = {Vρ}_{ω} g = (A y) ρ_{w} g$

Since the plank is in rotational equilibrium, so $\sum {\vec{τ}}_{0} = 0$

$\begin{array}{l} or mg \times \frac{l}{2} \sin θ - F_{b} \times \frac{y}{2} \sin θ = 0 or mgl - F_{b} \times y = 0 \\ or (Al \times 0 .5) gl - ({Ayρ}_{w} yg = 0 or 0 .5 l^{2} = y^{2} \\ or 0 .5 \times (1)^{2} = {(\frac{0 .5}{\cos θ})}^{2} or \cos^{2} θ = \frac{1}{2} \end{array}$

Pressure in a fluid-mechanical properties of fluids

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