Question

The viewing portion of the rectangular glass window in a fish tank is 6969 inches wide and runs from 1.51.5 inchesinches below the​ water's surface to 46.546.5 inches below the surface. find the fluid force against this portion of the window. the​ weight-density of seawater is 64 lb divided by ft cubed .

Answer 1

2760 lbf
The issue here is that the pressure isn't constant for the entire depth of the glass window. But if you imagine a very thin horizontal slice of the window where it's so thin that the pressure is constant for that slice, The force on that slice is mass of the water times the gravitational acceleration times the depth times the width. So we solve the integral from 1.5 inches to 46.5 inches of g*m*l*y dy.
You'll notice that g, m, and l are all constants, so you can move them outside of the integral and then you just need the integral of y dy. which is y^2/2. Before we do that calculation though, we need to make sure of our units since we have inches and feet. For convenience, I'll convert the dimensions of the window into feet.
Due to the confusing mix of lbs being used as a measure of mass as well as force, in this case, we can ignore the value of g for solving this problem.
Top of window = 1.5/12 = 0.125 ft
Bottom of window = 46.5/12 = 3.875 ft
Width of window = 69/12 = 5.75 ft
F = L*m* (b^2/2 - t^2/2)
F = 5.75 ft * 64 lb/ft^3 * ((3.875 ft)^2/2 - (0.125 ft)^2/2)
F = 5.75 ft * 64 lb/ft^3 * ((15.015625 ft^2)/2 - (0.015625 ft^2)/2)
F = 5.75 ft * 64 lb/ft^3 * ((15.015625 ft^2) - (0.015625 ft^2))/2
F = 5.75 ft * 64 lb/ft^3 * 15/2 ft^2
F = 5.75 ft * 64 lb/ft^3 * 7.5 ft^2
F = 2760 lbf
Let's check this by redoing the problem using metric units.
1.5 in = 0.0381 m
46.5 in = 1.1811 m
69 in = 1.7526 m
64 lb/ft^3 = 1025.18 kg/m^3
So
F = g*L*m* (b^2/2 - t^2/2)
F = 9.8 m/s^2 * 1.7526 m * 1025.18 kg/m^3 * ((1.1811 m)^2/2 - (0.0381
m)^2/2)
F = 17607.95859 kg/(m*s^2) * ((1.39499721 m^2)/2 - (0.00145161 m^2)/2)
F = 17607.95859 kg/(m*s^2) * (0.697498605 m^2 - 0.000725805 m^2)
F = 17607.95859 kg/(m*s^2) * (0.6967728 m^2)
F = 12268.74661 kg*m/s^2
And since 1 newton is equal to 0.22481 lbf, we have 2758 lbf, which is the same result to within the number of significant digits.

Answer 2

The first thing you should know in this case is the definition of the pressure:
P = d * g * h
where
P = pressure.
d = density.
g = gravity
Then, the force is defined as the integral of the pressure by the area differential:
F = int (PdA)
The window area differential is:
dA = w * dh
where
w: width
dh: height differential
Substituting the expression of the pressure and the area we have:
F = int (d * g * h) (w * dh)
where w is 69 in.
F = d * g * w * int (hdh)
from 1.5 to 46.5 in
F = d * g * w * (h ^ 2) / 2
from 1.5 to 46.5 in
F = d * g * w (((46.5 ^ 2) - (1.5 ^ 2)) / 2])
F = (64) * (32,16) * (69/12) * ((((46.5 / 12) ^ 2) - ((1.5 / 12) ^ 2)) / 2)
F = 88761.6 lb * feet * (1 / s ^ 2)
answer:
the fluid force against this portion of the window is
F = 88761.6 lb * feet * (1 / s ^ 2)