Question

A trough has cross section as shown in the figure. It is filled with water up to top. Find the thrust on a length $l$ on the left-side wall, bottom and on the right-side sidewall. Density of water is $\rho$ and acceleration due to gravity is $g$.

• A. lje: $\frac{\rho g l^2 b \sin^2 \theta}{2}$
• B. boom: $\rho g l \frac{b}{2} \sin \theta$

Answer 1

Answer: (B)

$\rho g l \frac{b}{2} \sin \theta$

Answer 2

The question asks for the thrust on three surfaces: the left-side wall, the bottom, and the right-side sidewall. The provided options are likely intended to be the answers for these surfaces.

Based on the common conventions and the structure of fluid mechanics problems:

1. Thrust on the right-side sidewall: The right-side sidewall is inclined at an angle $\theta$ with the horizontal. Let's assume the option "boom: $\rho g l \frac{b}{2} \sin \theta$" represents the thrust on this surface. For an inclined surface, the thrust $F$ is given by $F = P_{centroid} \times A$, where $P_{centroid}$ is the pressure at the centroid of the surface and $A$ is the area of the surface. Let $h$ be the vertical height of the water and $s$ be the slant length of the inclined wall. The area of the inclined wall for a length $l$ of the trough is $A = s \times l$. The centroid of this inclined surface is at a vertical depth of $h/2$. So, $P_{centroid} = \rho g (h/2)$. The thrust is $F = \rho g (h/2) \times (s \times l) = \frac{\rho g h s l}{2}$.

   If we assume that the option $\rho g l \frac{b}{2} \sin \theta$ is correct for the right sidewall, and that $b$ represents the slant length ($s = b$), then the vertical height would be $h = b \sin \theta$. Substituting these into the thrust formula: $F = \frac{\rho g (b \sin \theta) b l}{2} = \frac{\rho g l b^2 \sin \theta}{2}$. This does not directly match the option $\rho g l \frac{b}{2} \sin \theta$.

   There appears to be an inconsistency or a specific interpretation required for the term '$b$' in the option. However, if we assume the option $\rho g l \frac{b}{2} \sin \theta$ is the correct thrust for the right sidewall, it implies a particular relation between the geometry and the parameter $b$. Without further clarification or a corrected diagram/problem statement, we select this option as the most plausible answer for the right sidewall thrust based on the provided choices.

2. Thrust on the left-side wall: The left-side wall is vertical. Let its height be $h$. The area for a length $l$ of the trough is $A_{left} = h \times l$. The centroid is at a depth of $h/2$. The thrust on the left wall is $F_{left} = \rho g (h/2) \times (h l) = \frac{\rho g h^2 l}{2}$. The option "lje: $\frac{\rho g l^2 b \sin^2 \theta}{2}$" has $l^2$ in the numerator, which is unusual for thrust calculation where $l$ is the trough length. This suggests a potential error in the option or its intended application.

3. Thrust on the bottom: The bottom surface is horizontal. Let its width be $w_{bottom}$. The area is $A_{bottom} = w_{bottom} \times l$. The depth is $h$. The thrust on the bottom is $F_{bottom} = \rho g h \times (w_{bottom} \times l)$. The width of the bottom is not explicitly given, but from the diagram, it would be $w_{bottom} = b - h \cot \theta$, assuming $b$ is the top width. $F_{bottom} = \rho g h (b - h \cot \theta) l$. This expression is not among the options.

Given the ambiguity and likely errors in the question or options, and assuming the question intends to ask for one of the provided options as the answer for one of the surfaces, the option "boom: $\rho g l \frac{b}{2} \sin \theta$" is the most likely intended answer for the thrust on the right-side sidewall, despite the derivation challenges.