A 2 -ft-diameter hemispherical plexiglass "bubble" is to be used as a special window on the side of an above-ground swimming pool. The window is to be bolted onto the vertical wall of the pool and faces outward, covering a 2 -ft- diameter opening in the wall. The center of the opening is 4 ft below the surface. Determine the horizontal and vertical components of the force of the water on the hemisphere.

The force exerted by a fluid on a surface is given by the pressure multiplied by the area over which the pressure is applied. According to the principle of fluid statics, the pressure at a certain depth below the surface of a liquid is \( P = \rho g h \), where \( \rho \) is the density of the liquid, \( g \) is the acceleration due to gravity, and \( h \) is the depth below the surface of the liquid. In this case, the density of water is approximately 1000 kg/m^3, the acceleration due to gravity is approximately 9.81 m/s^2, and the depth is 4ft. However, this exercise uses the Imperial measurement system, to avoid errors, let's convert 1000 kg/m^3 and 9.81 m/s^2 respectively to 62.4 lb/ft^3 and 32.2 ft/s^2.

The force exerted on the plexiglass bubble, as per above equation would be \( F = P.A = (\rho.g.h).A \), where A is the surface area of the hemisphere. The surface area of a hemisphere of radius r is given by \( A = 2 \pi r^2 \) . However, to get the force exerted by the liquid on the hemisphere, integration over the surface area should be performed since force varies over height from the surface of the liquid to the center of the hemisphere. It is a daunting task and an integration over the entire surface of hemisphere is necessary, but considering symmetry of the hemisphere and the principle of Archimedes, it is sufficient to compute the pressure at the centroid of the hemisphere. The centroid of a hemisphere is located at a distance of 3r/8 from the base, so h for the centroid h_c = 4ft + (3/8)*2ft = 4.75ft. Therefore, the force is \( F = (\rho.g.h_c).A = (62.4 lb/ft^3 . 32.2 ft/s^2 . 4.75 ft . 2 \pi (1ft)^2 = 2\pi . 62.4 lb/ft^3 . 32.2 ft/s^2 . 4.75 ft . 1ft^2 \)

Since the force is distributed symmetrically around the hemisphere, the horizontal components of the force on opposite sides of the hemisphere will cancel each other out, resulting in a net horizontal force of zero. The vertical component of the force equals the total force, which has already been calculated earlier.