Atriangular plate with height 6 ft and a base of 7 ft is submerged vertically in water so that the top is 2 ft below the surface. express the hydrostatic force against one side of the plate as an integral and evaluate it. (recall that the weight density of water is 62.5 lb/ft3.)

Explanation:

For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).

With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:

γhxdy

we know that γ=62.5 lb/

from the drawing, we can determine the height (or depth under the water) of each differential area is given by:

h=8-y

x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:

when substituting the x and y-values given on the graph, we get that the slope is:

once we got this slope, we can substitute it in the point-slope form of the equation:

which yields:

which simplifies to:

we can now solve this equation for x, so we get that:

with this last equation, we can substitute everything into our integral, so it will now look like this:

Now that it's all written in terms of y we can now simplify it, so we get:

we can now proceed and evaluate it.

When using the power rule on each of the terms, we get the integral to be:

By using the fundamental theorem of calculus we get:

When solving we get:

what type of charge does it have? negative