A plate in the shape of an isosceles triangle 3 feet high and 4 feet wide is submerged vertically in water, base doward, with the base 5 ft bellow the surface. Find the force exerted by the water on one side of the plate.

The force exerted by the water on one side of the plate is F = 24*pg  

Explanation:

From the given question, the first step to take is to find  he force exerted by the water on one side of the plate.

Solution

Given that:

Let the pressure the  at a depth of y ft be = pgy lb/Pa

the area of the atrip is given as = f(y)*delta(y) = 4/3*(y-2)delta(y)

Then

we combine with the range for y as = y E [2 , 5]

Thus,

F = 4/3*pg * integral from (2 , 5) [y(y-2)] dy

Recall that,

p = water density

g= gravity of acceleration

so,

F = 4/3*pg * integral from (2 , 5) [y^2 - 2y]dy]

F = 4/3*pg * [y^3/3 - y^2] [2 , 5]

F = 4/3*pg * [18]

Finally, F = 24*pg  

it was a fall down when it happens

explanation: