- #1

Guillem_dlc

- 188

- 17

- Homework Statement
- The tank in the figure contains water and a uniform pressure ##p_a## acts on its surface. We wish to determine the resultant force ##F##, and its centre of pressure ##y_c-y_G##, on a flat gate of base ##3\, \textrm{m}## and height ##1,5\, \textrm{m}##. The damper is located on a wall of the tank with an inclination of ##30## degrees with respect to the horizontal. The upper base of the damper is located at ##3\, \textrm{m}## depth. Consider the following cases: a) ##p_a=p_{\textrm{atm}}## and b) ##p_a=125\, \textrm{kPa}##.

- Relevant Equations
- ##I_{xx}=\iint y^2\, dA##

Figure:

a) CASE A ##\rightarrow p_a=101300\, \textrm{Pa}##

$$F_{\textrm{res}}?,\,\, y_c-y_{cg}=y_{cp}$$

We find ##h_{cg}\rightarrow h_{cg}=3+h##

$$h=0,75\cdot \sin (30)=0,375\, \textrm{m}\rightarrow h_{cg}=3,375\, \textrm{m}$$

$$p_{cG}=\rho_{H2O}gh_{cg}=33108,75\, \textrm{Pa}$$

We calculate ##A=1,5\cdot 3=4,5\, \textrm{m}^2##

$$\boxed{F=p_{CG}\cdot A=148989,375\, \textrm{N}}$$

$$I_{xx}=\iint y^2\, dA=\int_0^{1,5}\int_0^3 y^2\, dxdy=\int_0^{1,5}y^2\int_0^3 dxdy=3\int_0^{1,5}y^2\, dy=$$

$$=3\left[ \dfrac{y^3}{3}\right]_0^{1,5}=3,375$$

$$y_{CP}=-\dfrac{\rho_{H2O}g\sin \theta I_{xx}}{F_{CG}}=0,111$$

I don't get the ##I_{xx}## right and I don't know why? Is it calculated in this way?

$$I_{xx}=\iint y^2\, dA$$

It should give ##\dfrac{b\cdot L^3}{12}##.

a) CASE A ##\rightarrow p_a=101300\, \textrm{Pa}##

$$F_{\textrm{res}}?,\,\, y_c-y_{cg}=y_{cp}$$

We find ##h_{cg}\rightarrow h_{cg}=3+h##

$$h=0,75\cdot \sin (30)=0,375\, \textrm{m}\rightarrow h_{cg}=3,375\, \textrm{m}$$

$$p_{cG}=\rho_{H2O}gh_{cg}=33108,75\, \textrm{Pa}$$

We calculate ##A=1,5\cdot 3=4,5\, \textrm{m}^2##

$$\boxed{F=p_{CG}\cdot A=148989,375\, \textrm{N}}$$

$$I_{xx}=\iint y^2\, dA=\int_0^{1,5}\int_0^3 y^2\, dxdy=\int_0^{1,5}y^2\int_0^3 dxdy=3\int_0^{1,5}y^2\, dy=$$

$$=3\left[ \dfrac{y^3}{3}\right]_0^{1,5}=3,375$$

$$y_{CP}=-\dfrac{\rho_{H2O}g\sin \theta I_{xx}}{F_{CG}}=0,111$$

I don't get the ##I_{xx}## right and I don't know why? Is it calculated in this way?

$$I_{xx}=\iint y^2\, dA$$

It should give ##\dfrac{b\cdot L^3}{12}##.