A vertical plate is submerged in water and has the indicated shape.6 ft5 ft5 ft6 ftAn isosceles triangle is submerged in water pointed down. The base of 6 ft is horizontal and 6 ft below the surface of the water. The other two sides are each 5 ft.Express the hydrostatic force (in ft) against one side of the plate as an integral (let the positive direction be downwards) and evaluate it. (Assume that the weight density of this water is 62.5 lb/ft3.)
Question
A vertical plate is submerged in water and has the indicated shape.6 ft5 ft5 ft6 ftAn isosceles triangle is submerged in water pointed down. The base of 6 ft is horizontal and 6 ft below the surface of the water. The other two sides are each 5 ft.Express the hydrostatic force (in ft) against one side of the plate as an integral (let the positive direction be downwards) and evaluate it. (Assume that the weight density of this water is 62.5 lb/ft3.)
Solution
The hydrostatic force on a submerged plate is given by the integral of the pressure over the area of the plate. The pressure at a depth h below the surface of a fluid is given by the weight density of the fluid times the depth.
Let's denote the weight density of the water as w (which is given as 62.5 lb/ft^3), the depth as h, and the length of the plate at a given depth as L(h).
The plate is an isosceles triangle with a base of 6 ft and height of 5 ft. If we take a small horizontal strip of the plate at depth h with thickness dh, the length of the strip L(h) is proportional to the depth h.
Since the triangle is isosceles, we can see that L(h) = (6/5)h.
The area of the strip is then L(h)dh = (6/5)hdh.
The pressure at depth h is wh = 62.5h.
The force dF on the strip is the pressure times the area, so dF = whL(h)dh = 62.5h*(6/5)hdh = 75h^2dh.
The total force F on the plate is the integral of dF from h=0 to h=5.
So, F = ∫ from 0 to 5 of 75h^2 dh.
To evaluate this integral, we use the power rule, which states that the integral of x^n dx is (1/(n+1))x^(n+1).
So, F = [ (75/3)h^3 ] from 0 to 5 = (75/3)*5^3 - (75/3)*0^3 = 6250 lb.
So the hydrostatic force against one side of the plate is 6250 lb.
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