0
$\begingroup$

Let's say the hole is essentially at the bottom-most point and on the side, the shape of the container is cylindrical with flat surface on the floor. I know the thrust force from conservation of linear momentum parallel to plane is $$\frac{dm}{dt} \times (\text{relative velocity}) \, .$$ As a first approximation, the size of the hole could be neglected wrt the area of the base of cylinder. The problem I'm facing is the approximations on various fronts like velocity of fluid inside the container.

A rought attempt : https://imgur.com/a/3sWZy

$\endgroup$
  • 1
    $\begingroup$ What is the question ? $\endgroup$ – RedHelmet Apr 15 '17 at 16:14
  • $\begingroup$ time dependence of velocity of container. $\endgroup$ – Mrigank Apr 15 '17 at 16:34
  • 2
    $\begingroup$ For velocity of fluid at exit hole see Torricelli's Law. See also Is Torricelli's law "wrong" for big holes? - Tank draining problem. $\endgroup$ – sammy gerbil Apr 15 '17 at 16:39
  • $\begingroup$ Ok, so if I ignore bernoulli and do old fashioned energy conservation, the velocity of liquid at the top of the container would have two perpendicular components, one parallel to the plane from momentum conservation and other from volume conservation. If I measure the speed of the liquid coming out in ground frame, would the product of that velocity with area of hole be the rate of volume flow or the relative velocity of liquid coming out parallel to the plane should be used it it's place? $\endgroup$ – Mrigank Apr 15 '17 at 16:53
  • $\begingroup$ Sorry, the imgur link was having problems, I fixed the link. $\endgroup$ – Mrigank Apr 16 '17 at 5:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.