Question

A beaker of height h is kept open to atmosphere with liquid with varying density as kx fin the force at horizontal walls of beaker

Answer 1

The force at the vertical walls (assuming "horizontal" is a typo for "vertical") is $\frac{1}{6} kg W h^3$, where $W$ is the width of the wall.

Answer 2

The problem asks to find the force on the horizontal walls of a beaker filled with a liquid whose density varies with depth. The height of the beaker is $h$, and the density is given by $\rho(x) = kx$, where $x$ is the depth from the surface.

Interpretation of the Question: The term "horizontal walls" for a beaker is ambiguous. Beakers typically have vertical side walls and a horizontal base.

1. If "horizontal walls" refers to the base of the beaker: The base is a horizontal surface. The pressure at the bottom of the beaker (at depth $x=h$) is uniform.
2. If "horizontal walls" is a typo for "vertical walls": This refers to the side walls of the beaker, which are vertical. The pressure on these walls varies with depth. This is a more common type of problem when density is varying, as it requires integration over the height of the wall. Given the context of fluid mechanics problems, it is highly probable that "horizontal" is a typo for "vertical". We will proceed with this assumption.

Solution for Force on Vertical Walls (assuming "horizontal" is a typo for "vertical"):

1. Define Coordinate System and Density: Let the free surface of the liquid be at $x=0$, and the bottom of the beaker be at $x=h$. The depth is measured downwards from the surface. The density of the liquid at depth $x$ is given by $\rho(x) = kx$.

2. Calculate Pressure at Depth $x$: The pressure $P(x)$ at a depth $x$ in a fluid with varying density is given by the integral of $\rho g$ with respect to depth: $P(x) = \int_0^x \rho(x') g \, dx'$ Substitute $\rho(x') = kx'$: $P(x) = \int_0^x (kx') g \, dx' = kg \int_0^x x' \, dx'$ $P(x) = kg \left[ \frac{(x')^2}{2} \right]_0^x = kg \left( \frac{x^2}{2} - 0 \right)$ $P(x) = \frac{1}{2} kg x^2$

3. Calculate Force on an Elemental Strip of a Vertical Wall: Consider a vertical wall of the beaker. Let its width be $W$. Consider an elemental horizontal strip of this wall at a depth $x$ with a small thickness $dx$. The area of this elemental strip is $dA = W \, dx$. The force $dF$ exerted by the liquid on this elemental strip is $P(x) \, dA$: $dF = P(x) (W \, dx) = \left( \frac{1}{2} kg x^2 \right) W \, dx$

4. Calculate Total Force on the Vertical Wall: To find the total force $F$ on the entire vertical wall, integrate $dF$ from $x=0$ (surface) to $x=h$ (bottom of the beaker): $F = \int_0^h dF = \int_0^h \frac{1}{2} kg W x^2 \, dx$ $F = \frac{1}{2} kg W \int_0^h x^2 \, dx$ $F = \frac{1}{2} kg W \left[ \frac{x^3}{3} \right]_0^h$ $F = \frac{1}{2} kg W \left( \frac{h^3}{3} - 0 \right)$ $F = \frac{1}{6} kg W h^3$ This is the force on one vertical wall of width $W$. If the beaker's cross-section is not specified, the answer is given in terms of $W$.

Explanation of the solution: The pressure at a depth $x$ in a fluid with varying density $\rho(x) = kx$ is found by integrating $\rho(x)g$ from the surface to depth $x$, yielding $P(x) = \frac{1}{2} kg x^2$. To find the force on a vertical wall of width $W$, consider an elemental horizontal strip of area $W dx$ at depth $x$. The force on this strip is $dF = P(x)W dx$. Integrating this from $x=0$ to $x=h$ gives the total force $F = \int_0^h \frac{1}{2} kg x^2 W dx = \frac{1}{6} kg W h^3$.