As shown in the attacked image, a tank has an inclined wall at an angle of 450 to the horizontal. On this wall, there is a 1m square door that is hinged at A and has a simple latch at B. The distance between A and the free surface of sea water measured along the inclined wall is 2 m. (Specific gravity of sea water sg = 1.1). Calculate the force F on the door due to the oil

enter image description here MY WORKING OUT could anyone pls help check if i am right :) P=F/A SO F=PA than F=densitygravitysin(angle)Areaheight now in order to find the height cause it's inclined i did sin 90/2=sin45/x cross multiply to get x-1.414 back to the formula 1100*9.81*sin45*1m^2*1.414 and i got 10.789kN as my force


1 Answer 1


Because hydrostatic pressure depends on depth, the problem requires integration.

Pressure on door.

The pressure as a function of $y$ according to Pascal's Law is:

$$p(y)=p_0+\rho g y\sin \theta$$ Where $p_0$ is the atmospheric pressure and $\theta=45\:\mathrm{degrees}$. $y\sin \theta$ is the depth.

On an infinitesimal piece of door of length $dy$, at position $y$ and with width $W$, the pressure exerts a force $dF$: $$dF=p(y)Wdy$$ Or: $$dF=W(p_0+\rho g y\sin \theta)dy$$ To find the total force $F$ we integrate: $$F=\int_{y_1}^{y_2}W(p_0+\rho g y\sin \theta)dy$$ $$F=Wp_0(y_2-y_1)+\frac12 W\rho g \sin \theta (y_2^2-y_1^2)$$ Where: $\rho=1100\:\mathrm{kg/m^3}$




$\sin 45=\frac{\sqrt{2}}{2}$



Note that the term $Wp_0(y_2-y_1)$ is often discarded because atmospheric pressure acts also on the dry side of the door. By discarding the atmospheric term we're calculating the net force on the door.

  • $\begingroup$ Yes, it's the same way total force on a dam (or part of it) is calculated. Thanks for the upvote! $\endgroup$
    – Gert
    Commented Apr 30, 2016 at 21:24
  • $\begingroup$ hello there, so now if i were to calculate the positiong of centre of pressure on the door, i found Ixc using ab^3/12 to be 0.0833m than i used hp equation Ic/AHc sin^2(angle)+Hc putting the values 0.0833/1*(0.707^2)+1 i get 1.0416 metres am i right please ? $\endgroup$
    – Mark
    Commented Apr 30, 2016 at 22:29
  • $\begingroup$ I think it should be (ignoring the atmosph. component):$\frac23 \frac{y_2^3-y_1^3}{y_2^2-y_1^2}=2.53$. So the resultant force $F$ acts at a little over the half of the door's length. $\endgroup$
    – Gert
    Commented Apr 30, 2016 at 22:53
  • $\begingroup$ 2.53 meters when the door is 1 metre ? could you have a look at it here please physics.stackexchange.com/questions/253194/… $\endgroup$
    – Mark
    Commented Apr 30, 2016 at 23:02
  • $\begingroup$ 2.53 m from the top of the tank, so 0.53 m from the hinge. $\endgroup$
    – Gert
    Commented Apr 30, 2016 at 23:09

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