Doubt on fluids

A Hollow and baseless frustum of a cone is such that it has a semi Vertical angle Alpha now water is poured in it until the cone rising up find the minimum height such that the cone starts losing contact with the surface.

Q2) Find out net force on the walls of container.

In Q)2 is it a thin soap film or any other film?

It is a cylindrical container with one side concave inward

use torque method to find its a standard question
aporoch second by integration but will be typical to find

it may help you..
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For q2 , we need to calculate the net force or net horizontal force?
I’m presenting the solution fir both of these

Horizontal force = net force on a straight wall of height 2R as the liq is in equilibrium =( dgr)(2Rw)

Where d is density and w is width of the container (in you question L is given which is of no use , instead of that width of the container should be given)

Net vertical force = buyount force = {d(pie r² w)g } /2

There is no need of integration

I didn’t get what you meant !
Because Net torque on this curved wall will be zero

oops i was mis quoted sorry i just see in one vision remember i did such type question... i m wrong now pardon mee sir.. actully i want to talk about projected area concept i wrote torque instead of it..


For q)1 for the frustum to lose contact ,the force applied by water contact(normal)vertical component must be equal to the weight of the frustum.
Take the area of frustum projection vertically and find force by integrating with pressure(P.dA)
Then equate it with weight of frustum.

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Thanks a lot💙
It cleared the doubt

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