Physical problems?

4. The equations involved in calculating the thrust on an inclined plane are:
Thrust on inclined plane Fr = pg(hc).A
where p is the density of water (= 1000 kg/m^3), g is the acceleration due to gravity (= 9.81 m/s^2), hc is the vertical depth, from water surface, of the centre of mass of the plane, A is the area of the plane.

For a rectangular plane of height H and width W, the thrust acts at a distance (yr), measured in direction along the inclined plane from water surface, is
yr = yc + WH^3/(12.WH.yc)
where yc is the distance, measured in direction along the inclined plane from water surface, of the centre of mass of the plane, and
yc = hc/sin(a) where hc is the vertical depth from water surface of the centre of mass of the plane, and a is the inclined angle of the plane.

First consider the thrust acting on the left hand side of the gate,
Fr = 1000 x 9.81 x [4 - 1 x sin(60)] x (2 x 1) m = 61,489 N
The point of action of Fr is given by,
yr = yc + (1 x 2^3)/[12 x (1 x 2)yc]
where yc = [4 - 1 x sin(60)]/sin(60) m = 3.619 m
Hence, yr = 3.619 + (1 x 2^3)/[12 x (1 x 2) x 3.619] m = 3.711 m
Distance along the plane of Fr from the hinge A = yr - [(4 - 2.sin(60))/sin(60)] m
= 3.711 - [(4 - 2.sin(60))/sin(60)] m = 1.092 m
There the thrust on the left hand side of the gate is 61,489 N and acts at a distance of 1.092 m from the hinge at A.

Using the same method, the thrust on the right hand side of the gate Fr' can be found equals to 7,358 N and acts at a distance of 1.423 m from A.

Take moment about hinge A,
W.cos(60) x 1 + (Fr') x 1.423 = (Fr) x 1.092
i.e. W = (61489 x 1.092 - 7358 x 1.423)/cos(60) N = 113,351 N

5. You could use the same equations given above to solve this problem. Just try it yourself.