Question

A light cylindrical vessel is kept on a horizontal surface. Its base area is a. A hole of cross sectional area a is made just at its bottom side. Find the minimum coefficient of friction necessary for sliding of the vessel due to the impact force of the emerging liquid. (a<< a)

Answer 1

Answer:

$\frac{2a}{A}$

Explanation:

If the height of the liquid in the vessel is h then the velocity of the emerging liquid from the bottom is given by

$v=\sqrt{2gh}$    (Easily proved by Bernaulli's theorem)

if the density of the liquid is $\rho$

The rate of mass of liquid emerging out of the vessel

$\frac{dm}{dt}=a\times \sqrt{2gh}\times \rho$

$\implies \frac{dm}{dt}=a\rho\sqrt{2gh}$

From the second law of Newton, Force is

$\boxed {F=\frac{d(mv)}{dt}}$

$\implies F=v\frac{dm}{dt}$

$\implies F=\sqrt{2gh}\times a\rho\times \sqrt{2gh}$

$\implies F=2gh\times a\rho$

This must be equal to the frictional force

i.e.

$\mu N=2gh\times a\rho$

$\implies \mu\times A\times h\times \rho\times g=2gh\times a\rho$

$\implies \mu=\frac{2a}{A}$

Hope this helps.

Answer 2

Answer:

Answer:

\frac{2a}{A}

A

2a

Explanation:

If the height of the liquid in the vessel is h then the velocity of the emerging liquid from the bottom is given by

v=\sqrt{2gh}v=

2gh

(Easily proved by Bernaulli's theorem)

if the density of the liquid is \rhoρ

The rate of mass of liquid emerging out of the vessel

\frac{dm}{dt}=a\times \sqrt{2gh}\times \rho

dt

dm

=a×

2gh

×ρ

\implies \frac{dm}{dt}=a\rho\sqrt{2gh}⟹

dt

dm

=aρ

2gh

From the second law of Newton, Force is

\boxed {F=\frac{d(mv)}{dt}}

F=

dt

d(mv)

\implies F=v\frac{dm}{dt}⟹F=v

dt

dm

\implies F=\sqrt{2gh}\times a\rho\times \sqrt{2gh}⟹F=

2gh

×aρ×

2gh

\implies F=2gh\times a\rho⟹F=2gh×aρ

This must be equal to the frictional force

i.e.

\mu N=2gh\times a\rhoμN=2gh×aρ

\implies \mu\times A\times h\times \rho\times g=2gh\times a\rho⟹μ×A×h×ρ×g=2gh×aρ

\implies \mu=\frac{2a}{A}⟹μ=

A

2a

Hope this helps.

Explanation:

Thanks me as much as possible my friend