Question

A triangular lamina of area A and height h is immersed in a liquid of density Rho in a vertical plane with its base in the surface of the liquid. the thrust on lamina is​

Answer 1

The thrust on lamina is​  F = 2/3ρgAh.

Explanation:

x/b'=h/b

b'=b/hx

dA=b'dx = b/hxdx

p = ρgx

dF = pdA=ρgx b/h (xdx)

F=ρgb/h∫h0 x^2dx

=ρgb/h h^3/3 = 1/3ρgbh^2

=2/3ρgh(bh/2)

F =2/3ρgAh.

Hence the force of thrust on lamina is​ 2/3ρgAh.

Also learn about force of thrust

Define thrust?

https://brainly.in/question/2510974

Answer 2

Given that,

Area = A

Height = h

Liquid density = ρ

We know that,

Pressure :

Pressure is equal to the force divided by area.

In mathematically,

$P=\dfrac{F}{A}$

So, The force is

$F=P\times A$

The whole force of water will act at the centroid of the triangle.

This point is at distance of $\dfrac{2}{3}h$ from the top surface of water.

We need to calculate the thrust on lamina

Using formula of thrust

$F=P\times A$

Put the value into the formula

$F=\rho gHA$

Where, $\rho$ = liquid density

g = acceleration due to gravity

A = area

Put the value of H

$F=\rho g\times\dfrac{2}{3}h A$

$F=\dfrac{2\rho g hA}{3}$

Hence, The thrust on lamina is $\dfrac{2\rho g hA}{3}$