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Homework Statement:

Mark all the forces acting on the wedge (the coefficient of friction between the wall and the wedge is mu) the container is of depth h and breadth b.
The wedge is placed in a fluid of density d.
Relevant Equations:
 
(this problem is slightly modified from the original problem.)
There is a contact force(friction and normal force) between the wedge and the walls of the container and there is a fluid thrust acting on the side of the wedge in contact with the fluid( this force is normal to the slanted surface and the force per depth can be given by the are of the pressure diagram) there is also a buoyant force acting on this wedge in the upward direction.
$$F_{thrust} = \frac {dgh^2 b}{2cos\theta}$$
$$F_{buoyant} = dg((1/2) h^2 b tan\theta)$$
$$f = \frac{\mu dgh^2 b}{2}$$
$$F_{g} = mg$$
The normal reaction on the wedge is the horizontal component of ##F_{thrust}## and my confusion is, shouldn't ##F_{buoyant} ## actually not be a separate force rather it should be the vertical component of the force ##F_{thrust}##?
isn't the buoyant force a result of all the ##(da)P## (where ##da## is the differential area and P is the pressure at that point in the liquid)
If that's not the case what exactly is causing the buoyant force?
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