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8 years ago

*If you assume the body to be a perfect cylinder, then all pressure forces and weight of the body should balance out to produce a net translational upward force. However, in non-ideal conditions, the body is not a perfect cylinder and the force you applied is not perfectly perpendicular to the liquids surface. Hence Bouyoncy acts slightly off the COM to compensate for your imperfect force and thus produces torque about COM, causing the body to rotate untill it reaches equilibrium (parallel to the surface).*And im curious as to how you managed to get a force at an angle to the X axis in perfect conditions.

8 years ago

@ Rohan radfke

Please excuse my diagram. I dont have sufficient knowledge about designing. I have copied the above picture from google images just to explain what i meant. Ok, in the above picture which is perfectly symmetric and the cylinder is forcefully immersed under water.Ok, i am assuming the applied force(T) perpendicular to the X-Y and at an angle with the cylindrical surface denoted by length ''L'' in the picture. Now, since the cylinder is under water and is perfectly static, so my goal is to calculate the net force acting on the cylinder. All the four surfaces of the cylinder are at a different height. So, there is pressure force on all the surfaces. Let us consider the lower surface with diameter d and the upper surface with diamter d.So, the net force on the lower surface is

F(Lower surface(D) = P1A along the axial direction of the cylinder (say the unit vector along that direction is r)

f(Upper surface (D) = P2A(along -r)

Now, the net force since F>f due to depth of the surface

Net force on these two surfaces = (F- f) r

F(net) = A(P1 - P2) = A(h1*d*g - h2*d*g) (h1>h2) = V(net) along +r)

h1 depth from the point of reference to the lower surface and h2 to the upper surface.

Now, lets consider the other two surfaces (of length L)

On the top surface( L) force = P3 A1 (say the direction is along a unit vector - R)

Lower surface (L) force = P4 A1(along + R)

So, the net force is = A1(P4 - P3) P4>P3 ,

= A1 (h3*d*g - h4*d*g)- mgsina along +R (mg force acting at an angle of 90-a with -R vector)

Including my applied force which is say T acting along -R

net force = A1(h3*d*g - h4*d*g) -mgcosa = U(net) along +R

So, the net forces are 1. V(net) acting along the axial direction( of the cylinder

2. U(net) acting along the perpendicular direction of the cylinder.

Now, since the two net forces are perpendicular, so the net of these two forces act along the Rr plane. The net pressure force from the fluid is acting along the Rr plane and the net applied force is acting along the XY plane. Two forces acting at an different direction yet balancing each other. I feel if my calculations are not wrong, then the net pressure force makes an angle with the X axis and Y axis. Say, these angle is b. If we resolve these forces then we find, since the body is in equilibrium sum the net force along the X axis is 0 i.e F cosb = 0 and the net force along the Y axis is zero so F sinb = T (applied force).

Now, when i release the applied force, the net force from the fluid pushes the body upwards to the surface.

I would really appreciate if you can tell me where i made the mistake. Thank you again for your time and help.

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