Question

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

 

• A. $\frac{a}{A}$
• B. $\frac{2a}{A}$
• C. $\frac{A}{2a}$
• D. None of these

Hint:

The thrust force exerted by a fluid jet is given by $F = \rho a v^2$, where $\rho$ is the fluid density, 'a' is the jet's cross-sectional area, and 'v' is the jet's velocity. This is a direct application of Newton's second law in terms of momentum.

Answer 1

The Correct Option is B

Let h be the height of the liquid in the vessel.
The velocity of the liquid emerging from the hole (velocity of efflux) is given by Torricelli's law: $v = \sqrt{2gh}$.
The mass flow rate of the liquid through the hole is $\frac{dm}{dt} = \rho a v = \rho a \sqrt{2gh}$, where $\rho$ is the density of the liquid.
The emerging liquid exerts a horizontal thrust force on the vessel, which is equal to the rate of change of momentum of the liquid.
$F_{thrust} = v \left(\frac{dm}{dt}\right) = (\rho a v) v = \rho a v^2 = \rho a (2gh) = 2\rho gha$.
The weight of the liquid in the vessel provides the normal reaction force N. Since the vessel is light, we consider only the liquid's weight.
$W_{liquid} = (\text{Volume}) \times \rho g = (A h) \rho g$.
The normal force is $N = W_{liquid} = \rho g A h$.
To prevent sliding, the force of static friction ($f_s$) must be greater than or equal to the thrust force.
$f_s \ge F_{thrust}$
The maximum static friction is $\mu N$, where $\mu$ is the coefficient of static friction.
$\mu N \ge F_{thrust}$
$\mu (\rho g A h) \ge 2\rho gha$
Canceling $\rho g h$ from both sides (assuming $h>0$):
$\mu A \ge 2a$
$\mu \ge \frac{2a}{A}$
The minimum coefficient of friction required is $\mu_{min} = \frac{2a}{A}$.