I'm struggling with the following White's Fluids Mechanics problem:

Gate AB is a quarter-circle 10 ft wide and hinged at B. Find the force F just sufficient to keep the gate from opening. The gate is uniform and weighs 3000 lbf.

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For the horizontal force I found the height for the center of gravity $h_{CG}=4ft$ and the area of the vertical plane where the surface is projected $A_{proj}=8*10ft^2$ and therefore have $F_H=\gamma_{water}*h_{CG}*A_{proj}$ = 19968 lbf. Then I found how many $ft$ below A was the horizontal force (center of pressure CP): $y_{CP}=-\frac{I_{xx}*\sin{90^o}}{h_{CG}*A_{proj}}=1.(3)ft$ and therefore the horizontal force is $h_{CG}+y_{CP}=5.(3)ft$ below $A$. Finally, the vertical force $F_V$ is equal to the weight of the missing piece of water above the gate and this is where my problem begins. The following image is from White's Fluid Mechanics Solutions and it illustrates what I'm doing to determine $F_V=8570lbf$ enter image description here

But then, in the solutions, the line of action $x$ of the vertical force is found by summing the moments around B: $\sum M_B=8570*x = 39936*4 − 31366*(4.605)$, $x=1.787ft$. My first problem is: to measure the torque around B shouldn't I use the sine of the angle between the length from B to the point of application of $F_V$, which I don't know yet?

My doubt is further instigated when measuring the torque of all the forces, around B. This is the force diagram drawn for this problem after determining the horizontal distances from B of all the forces ($F_V$,$F_H$ and $Weight$). The torque is therefore: $\sum M_B(clockwise)=0=F(8.0) + (3000)(2.907) − (8570)(1.787) − (19968)(2.667)$. All the moments are determined using the horizontal distances from the hinged point. Shouldn't it be, as I've said, the $||force||\times ||r|| \sin{\theta}$, where $r$ is the distance between B and the point of application of the force and $\theta$ the angle between them. enter image description here

Another problem-example where I have the same issue is this:

The 2 inch by 2 inch by 12 ft spar buoy from the figure has 5 lbm of steel attached and has gone aground on a rock. If the rock exerts no moments on the spar, compute the angle of inclination $\theta$

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After I found the weight of the wood spar buoy $W_{wood}$ and the buoyancy $B$ this is the following torque about A: $\sum M_A=0=W_{wood}*(6\sin{\theta})-B*(\frac{submergedlength}{2}\sin{\theta})$, these are the distances to the right of A of each force and my question arises once again. Thank you for your time.