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Let us glue up these two images, where we get closed loop thrust of water. Force $F_3$ has direction $-x$ and force $F_2$ has $x$ direction. What is the sum of all forces? Can it be more than zero? Speed of water is constant.Angles are the same.Half circle is not exactly circled at the ends due to the angles.

One more subquestion. What if speed of water is very high and we have quite big amount of centrifugal force?

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If I understand the question, you are basically talking about a stream of water flowing in a closed loop. Assuming the density is constant, and the speed of the water at each point is constant, then the total momentum of the water (which is the sum of the momentum of each little piece of water in the stream) is constant, and there must not be any net force on the water (by newton's second law).

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Yes. So no matter what the 3d geometry is behind the closed loop, the force is always zero? Sorry for my english. – Bojan Vasiljević May 31 '13 at 21:40

Newton's second law was originally formulated as $S_F$=$dp/dt$. P is momentum, which equals mass multiplied by velocity and that quantity divided by time. $S_F$ is the sum of the forces. Although $S_F$ is usually expressed as ma, to which it is mathematically equivalent, the original form is actually more descriptive, because it shows that a force is the change in momentum(how an object moves,if at all) divided by the change in time. Force, like momentum is a vector, and as a vector, it has direction. If two vectors of an opposing direction and equal magnitude are added, the result is zero always, no matter what the shape is. If the change in momentum is zero, then the force is likewise zero.

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