rectangular gate

[logistic_guy]

The \(\displaystyle 280\)-\(\displaystyle \text{kg}\), \(\displaystyle 6\)-\(\displaystyle \text{m}\)-wide rectangular gate shown in the figure is hinged at \(\displaystyle B\) and leans against the floor at \(\displaystyle A\) making an angle \(\displaystyle 45^{\circ}\) with the horizontal. The gate is to be opened from its lower edge by applying a normal force at its center. Determine the minimum force \(\displaystyle F\) required to open the water gate.

 

[khansaheb]

The \(\displaystyle 280\)-\(\displaystyle \text{kg}\), \(\displaystyle 6\)-\(\displaystyle \text{m}\)-wide rectangular gate shown in the figure is hinged at \(\displaystyle B\) and leans against the floor at \(\displaystyle A\) making an angle \(\displaystyle 45^{\circ}\) with the horizontal. The gate is to be opened from its lower edge by applying a normal force at its center. Determine the minimum force \(\displaystyle F\) required to open the water gate.
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[logistic_guy]

We start by calculating the torques about the hinge.

\(\displaystyle y_pF_R + \frac{l}{2}F_g\sin\theta = \frac{l}{2}F\)

\(\displaystyle F_R = (P_0 + \rho g h_{c})A = \left(P_0 + \rho g \left[h_1 + \frac{l\sin\theta}{2}\right]\right)lw\)

Or

\(\displaystyle F_R = \left(P_0 + \rho g \left[h_1 + \frac{3}{2}\right]\right)\frac{3w}{\sin\theta}\)

Since the problem did not mention anything about the atmospheric pressure, I will assume that it is zero in this problem.

Plug in numbers.

\(\displaystyle F_R = \left(0 + 1000(9.8) \left[0.5 + \frac{3}{2}\right]\right)\frac{3(6)}{\sin 45^{\circ}} = 498935 \ \text{N}\)

 

[logistic_guy]

We continue.

The full distance from the surface of the water to the center of pressure in the direction of the slope of the rectangular gate is:

\(\displaystyle y_f = y_c + \frac{I_{xx}}{(y_c + P_0/\rho g \sin\theta)A}\)


\(\displaystyle = \frac{l}{2} + l_1 + \frac{\frac{wl^3}{12}}{(\frac{l}{2} + l_1 + 0/\rho g \sin\theta)lw}\)


\(\displaystyle = \frac{h_2}{2\sin\theta} + \frac{h_1}{\sin\theta} + \frac{\frac{wh_2^3}{12\sin^3\theta}}{\left(\frac{h_2}{2\sin\theta} + \frac{h_1}{\sin\theta}\right)\frac{h_2w}{\sin\theta}}\)

Plug in numbers.

\(\displaystyle y_f= \frac{3}{2\sin 45^{\circ}} + \frac{0.5}{\sin 45^{\circ}} + \frac{\frac{6(3)^3}{12\sin^3 45^{\circ}}}{\left(\frac{3}{2\sin 45^{\circ}} + \frac{0.5}{\sin 45^{\circ}}\right)\frac{3(6)}{\sin 45^{\circ}}} = \frac{19}{4\sqrt{2}} \ \text{m} = 3.3587572 \ \text{m}\)

 

[logistic_guy]

we continue.

\(\displaystyle y_p = y_f - l_1 = yf - \frac{h_1}{\sin\theta} =3.3587572 - \frac{0.5}{\sin 45^{\circ}} = 2.6516504 \ \text{m}\)

 

[logistic_guy]

Now we have all the pieces of the puzzle.

\(\displaystyle y_pF_R + \frac{l}{2}F_g\sin\theta = \frac{l}{2}F\)


\(\displaystyle F = \frac{2y_pF_R}{l} + F_g\sin\theta\)


\(\displaystyle F = \frac{2y_pF_R\sin\theta}{3} + mg\sin\theta\)

Plug in numbers.

\(\displaystyle F = \frac{2(2.6516504)(498935)\sin 45^{\circ}}{3} + 280(9.8)\sin 45^{\circ} = 625609.044 \ \text{N} = \textcolor{blue}{625.609044 \ \text{kN}}\)