Thrust on an triagular lamina emersed in a liquid.

[Navin]

Homework Statement

A triangular lamina of area A and height h is im mersed in
a liquid of density ρ in a vertical plane with its base on the
surface of the liquid. The thrust on the lamina is
(a) 1/2Aρgh (b) 1/3Aρgh

(e) 1/6Aρgh (d) 2/3Aρgh

Relevant Equations

F(bouyancy)= Vρg

I feel there is a really obvious flaw in my logic but i can't pin it down

So i have to find the thrust on the lamina which is basically force of bauyancy

F(bouyancy)= Vρg

Now volume of the triangular lamina is its rea into its hieght.

v = Ah
hence
F = Ahρgsome information i feel i didnt take into account:

-base on surface of liquidPlease guide me in solving this

 

[haruspex]

So i have to find the thrust on the lamina which is basically force of bauyancy

Deducing from the offered answers, and the fact that you are not given the thickness of the lamina, this is not a buoyancy question. It wants the normal on (presumably, one face of) the lamina.
It is poorly worded.

 

[Navin]

Deducing from the offered answers, and the fact that you are not given the thickness of the lamina, this is not a buoyancy question. It wants the normal on (presumably, one face of) the lamina.
It is poorly worded.

Okay so let's assume i have to find the normal force on one face, how do i find it. I have kinda hit a brick wall here :( can you give me a hint ?

 

[kuruman]

I think the triangle is immersed in the liquid so that its height is along the vertical and its base parallel to the surface. Find the force element $dF$ on a strip of area $dA$ parallel to the base and at distance $y$ from the base. Then add all such elements over the surface of the triangle.