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Consider the configuration pictured below: a cart with a u-shaped pipe on it, with a fan on one end blowing air into the pipe.

The fan rotates and makes wind towards the left. The wind will be curved in the bent part of the pipe and will come out with changed direction. The pipe is attached to the cart.

There are factors to know the answer.

The factors are like air pressure, force that fan pushes the air, force that wind pushes the pipe, momentum change of air and cart.

Considering this, will this cart move or not, if move, which direction?

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Your configuration is equivalent to two famous examples: a sailboat with a fan pointing at the sail, and a leafblower pointing at an umbrella. Both of these have been proven experimentally, and both of them work, with the former verified by MythBusters.

They are both covered in detail in Blowing your own sail?.

In short, there are two sources of force acting on the cart: (1) the fan, which is pushing air towards the left, and (2) the u-turn of the pipe, which is receiving the air blown from the fan. If the u-turn bottom were completely flat, and it was redirecting the air that blows on it exclusively towards lateral directions (say, if you replace the u-shape for a T junction) then both of these forces would be equal. However, the u-turn is also redirecting air towards the right, which adds additional force towards the left.

As such, the cart will move towards the left.

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  • $\begingroup$ I don't think so $\endgroup$
    – Anonymous
    Commented Nov 17, 2020 at 11:35
  • $\begingroup$ I figured out based on the "Blowing your own sail?" If elastic, force on u-turn is twice bigger than force on fan. fan make air from stationary to move while u-turn change the direction. To make same speed but opposite direction use twice bigger impulse than momentum $\endgroup$
    – cavalist
    Commented Nov 17, 2020 at 16:25
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The cart will be stationary in this case as the fan pulls the air from the lower opening it will also push the air to outer opening simultaneously.

And as the speed of inlet and outlet air is same but in different direction so the momentum gained by one would be cancelled by other in this ideal case.

But if the momentum of inlet and outlet ai is not same then the cart would surely move

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  • $\begingroup$ What causes difference in momentum in real life experiment? $\endgroup$
    – cavalist
    Commented Nov 17, 2020 at 13:38
  • $\begingroup$ it will depend, if you take it ideal like no energy loss then the momentum will remain conserved $\endgroup$
    – Anonymous
    Commented Nov 17, 2020 at 13:39
  • $\begingroup$ otherwise like in real life it is not $\endgroup$
    – Anonymous
    Commented Nov 17, 2020 at 13:40
  • $\begingroup$ In experiment, cart move left. I know there is loss. Then why move left? $\endgroup$
    – cavalist
    Commented Nov 17, 2020 at 13:52
  • $\begingroup$ because on rebounding the air looses energy and its outlet speed reduces because of loss of energy. Now if you apply conservation of momentum then you will find your cart moving. $\endgroup$
    – Anonymous
    Commented Nov 17, 2020 at 13:54
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The air has to flow out from the top opening at the same speed as it entered at the bottom.

If it didn't, there would be an accumulation of air in the pipe, at higher and higher pressure. This would continue until the pressure became so great that the outflow speed again matched the inflow. Hence, the same amount of air has to flow out, at the same speed, as flows in.

Hence the forces to left and right are equal, and the cart will be stationary.

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  • $\begingroup$ Accumulation. I have same thought. But experiment is different. Why is that? $\endgroup$
    – cavalist
    Commented Nov 17, 2020 at 13:36

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