Answer to Question #130105 in Mechanics | Relativity for Ariyo Emmanuel

Question

A long solid cylinder of radius 0.8 m hinged
at point A is used as an automatic gate, as
shown in Fig. beside. When the water level
reaches 5 m, the gate opens by turning about
the hinge at point A. Determine (a) the
hydrostatic force acting on the cylinder and
its line of action when the gate opens and (b)
the weight of the cylinder per m length of the
cylinder.

Answer 1

solution

given data in question

radius of cylinder(r)=0.8 m

water level ($h$ )=5 m

density of water ($\rho$ )=1000 Kg/m^3

figure can be shown as below

a)horizontal component of hydro static force

$F_H=PA=\rho gh'A$

where $h'=(s+r/2)$ ; s is height from water level to hinge point A

$=1000\times9.8\times (4.2+0.8/2)\times0.8\times 1$

$=361\times 10^2 N$

Vertical component of hydro static force

$F_V=\rho ghA$

$=1000\times9.8\times 5\times0.8\times 1$

$=392\times10^2 N$

Weight of fluid block per meter length

$W=mg=\rho gV$

(Where $V=(r^2-\pi r^2/4)$ )

$=1000\times 9.8\times(0.8)^2\times(1-\pi/4)\times1$

$=13\times10^2 N$

Net vertical force is become

$F_V'=F_V-W=392\times10^2-13\times10^2$

$=379\times10^2 N$

Magnitude of hydro static force is given

$F_R=\sqrt{(F_V')^2+(F_H)^2}$

$=\sqrt{(361\times10^2)^2+(379\times10^2)^2}$

$=523\times10^2 N$

Direction of hydro-static force is

$\tan\theta=\frac{F_V'}{F_H}$

$\theta=\arctan{(\frac{379\times10^2}{361\times10^2})}$

$\theta=46.4^0$

hydro static force is acting at at an angle $46.4^0$ with horizontal axis.

b)when the gates are about to open then normal reaction force on the cylinder is zero and cylinder will be in rotational equilibrium taking moment about point A.

$F_R\space r\sin\theta-W_c\space r=0$

$W_c=F_R\sin\theta$

$=523\times10^2\sin(46.4^0)$

$=379\times10^2N$

therefore weight of cylinder is 37900 N.