I'm stuck with this problem that I have almost solved but the last bit seems impenetrable. Please take a look at the problem on the image i provided.

Plan: I started by calculating the momentum flow on the two sides by integrating over the two sides. After i was done calculating that I found the difference of the two to get the net momentum flow and this should be equal to the force on the tank.

Problem: I am to find the acceleration of the tank.The net momentum flow came out to be gh²rho(L2 - L1) which should be equal to the force on the tank but I can't find the volume of the tank and so can't find the mass. The answer to the problem is given as

a= {gh²rho(L2-L1)}/ 6m

Where a is the acceleration. The mass apparently will be 6m but I can't see how. So can anyone help me to find the total mass.

enter image description here

  • $\begingroup$ Irrespective of your force result ( which I don't agree with), mass of each wall or base is m. There is no roof and two walls are taken away. So you are left with two walls and a base leading to total mass of 3m. $\endgroup$ – npojo Apr 13 at 21:13
  • $\begingroup$ The water in the tank must also contribute something to the mass, and why don't you agree with the force result? Could you tell me then what is that you think should be the force on the tank? $\endgroup$ – Lucifer Apr 14 at 0:16
  • $\begingroup$ I used Bernoulli energy conservation equal to compute the momentum of fluid from face, since the speed varies with depth I appropriately integrated over the surface and similar procedure was followed for the second fave as well $\endgroup$ – Lucifer Apr 14 at 0:19
  • $\begingroup$ Momentum is of course conserved but the interplay between water movement and tank movement makes this solution path too complex. Alternatively, you can calculate total force on all walls in the axis of interest and immediately derive acceleration with mass being $3m$. $\endgroup$ – npojo Apr 14 at 7:15
  • $\begingroup$ How do you even do that.. would be immensely helpful if you elaborated a bit $\endgroup$ – Lucifer Apr 14 at 10:20

The expression for hydrostatic force acting on a surface has been derived here. https://i.stack.imgur.com/aNDip.jpg These are the forces acting on the tank, and they are balanced. (https://i.stack.imgur.com/iXReV.jpg) On removing the 2 surfaces, F1 and F2 are removed. There is no change in eql along Y axis. But along x axis, only F4(x) and F3(x) act. That is the net force on the tank, now of mass 3m. Hence the acceleration.


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