Question

A triangular 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.)

Answer 1

Answer:

$62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56) } \, dy = 7875 lb$

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:

$\int\limits^a_b$γhxdy

we know that γ=62.5 lb/$ft^{3}$

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:

$m=\frac{y_{2}-y_{1}}{x_{2}-x{1}}$

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

$m=\frac{0-6}{7-0}=-\frac{6}{7}$

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

$y_{2}-y_{1}=m(x_{2}-x_{1})$

which yields:

$y-6=-\frac{6}{7}(x-0)$

which simplifies to:

$y-6=-\frac{6}{7}x$

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

$x=-\frac{7}{6}y+7$

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

$\int\limits^6_0{(62.5)(8-y)(-\frac{7}{6}y+7)}\,dy$

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

$62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)}dy$

we can now proceed and evaluate it.

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

$62.5[\frac{7}{18}y^{3}-\frac{49}{6}y^{2}+56y]^{6}_{0}$

By using the fundamental theorem of calculus we get:

$62.5[(\frac{7}{18}(6)^{3}-\frac{49}{6}(6)^{2}+56(6))-(\frac{7}{18}(0)^{3}-\frac{49}{6}(0)^{2}+56(0))]$

When solving we get:

$62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56) } \, dy = 7875 lb$