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Chapter 7: Problem 1026

A triangular lamina of area \(\mathrm{A}\) and, height \(\mathrm{h}\) is immersed in a liquid of density \(\mathrm{S}\) in a vertical plane with its base on the surface of the liquid. The thrust on lamina is (A) \((1 / 2)\) Apgh (B) \((1 / 3)\) Apgh (C) \((1 / 6)\) Apgh (D) \((1 / 3)\) Apgh

Short Answer

Expert verified
The thrust formula on the triangular lamina is given by \[ F = (1/3)ASgh \], which corresponds to option (B).

Step by step solution

01

1. Calculate the pressure at the center of pressure

To find the pressure on the center of pressure, we need to find the distance of the center of pressure from the surface of the liquid. For a triangle, the center of pressure (COP) is located at the distance of (h/3) from the base and 'h' is the distance of the lowest point of triangle from the base. The pressure at the COP can be calculated using the formula: \[ P = Sg(h/3) \] where, P = pressure at the center of pressure S = density of the liquid g = acceleration due to gravity h = distance of the lowest point of the triangle from the surface of the liquid
02

2. Calculate the thrust on the lamina

Now we can find the thrust (F) on the lamina by multiplying the pressure at the center of pressure with the area of the lamina. \[ F = PA \] \[ F = (Sg(h/3))A \]
03

3. Simplify the expression for thrust

We can simplify the expression for thrust: \[ F = (h/3)ASg \] Now we see that the expression for thrust is matching with option (B) of the given problem. Therefore, the correct thrust formula on lamina is: Thrust F = \((1 / 3) \cdot A Sgh \) So, the correct answer is (B).

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